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in this lesson we want to talk about angles in trigonometry all right so let's start off the lesson by talking about some basic definitions we'll think about the concept of a line a line segment and array so let's suppose that we have two distinct points and here on the screen you'll see that you have a point a and also a point B so those two distinct points can be used to determine a line okay so you see we've drawn a line through them we can also use those letters the A and the B to name the line if we want so we could call this line a like this or we could also put the A and we could put a line on top like this now I want you to notice that you have arrows at each end okay and that tells you that the line continues indefinitely in each Direction now similarly you have something known as a line segment with a line segment it's just a piece of a line or a part of a line so notice how you have an end point here with a and an end point here with B so so here it doesn't continue in definitely in each Direction it's a finite or countable length from the point a to the point B so this is your line segment your line segment AB okay you can also write it like this you can put AB with a little line on top and you're not going to have arrows there it's another line segment that's put on top of a b okay lastly we're going to talk about the concept of array so here you have one end point okay in this case it's going to be a and it's going to extend through this point B indefinitely okay so we see this Arrow here it tells us that it starts at a and continues through b indefinitely so this is going to be the ray the ray AB or you could write ab and you can do this All right so now I want to talk about an angle and basically an angle consists of two different Rays with the same end point so when we talk about the two rays these are the sides of the angle angle and the common end point is the vertex of the angle so if we look at our example here we have a ray AC so we have a ray AC and then we have another ray which is AB so another ray which is AB so these are going to be the sides for my angle okay additionally a is the common end point so that's my vertex now if I'm naming the angle then I can call it angle cab you can go in order like that or you go the other way you could say angle BAC okay or you could just use the letter of the vertex now we can write out angle each time or you can use this special symbol so you can use this symbol here and I could say okay this is angle C A or it's angle b a c okay notice how a the vertex is in the middle each time and then also I can just use the letter of the vertex so I can say this is angle a all right now associated with every angle is going to be its measure which is generated by a rotation about the vertex so the measure will be determined by rotating array starting at one side of the angle which is called the initial side so let me just highlight that for in this case is the initial side is here we'll talk more about the standard position later on in the course for right now our initial side's going to be right there okay so that's the starting position and then we rotate to the other position on the other side which is known as the terminal side which in this case is right here so you can imagine that you started with two rays that are on top of each other so this guy represents this guy in the end and I know I'm not drawing that perfectly well but basically you can think about starting here and then just rotating about the vertex to get here so this is your initial side where the two are on top of each other and then it rotates to the terminal side where you still have one Ray here or one side and then another ray up here now when we rotate in a counterclockwise Direction so meaning I start here and I rotate this way okay that's how I'm measuring well what happens is I generate a positive angle okay now if I go the other way if I go clockwise I'm going to generate a negative angle we'll talk more about this as we progress through the course but basically if I started this way and I went that way okay now I'm generating a negative Angle now you could imagine if I started here and I went this way around to generate the same terminal side okay well now I would have a positive angle because I'm going counterclockwise for the rotation okay so we'll talk more about this as we progress all right so when working with angles we're often going to measure our angle using the degree okay so this is a unit of measure for us we'll also see radians here shortly but for now when we work with the degree we assign 360° to one complete rotation of array so in other words what we're saying here is that the terminal side of the angle will correspond exact L to the initial side when it makes this complete rotation so you can see this image that we have here and you can imagine that we started out with two rays that were on top of each other okay and then we rotated it completely around so it's one complete rotation here and we assign the value to that angle 360° okay so that's one complete rotation now of course you're going to have a lot of scenarios where you make more than one complete rotation or less than one complete rotation so if for example you made 1 over 360 that amount of a complete rotation well we would just take this number in the numerator here which is one and we would say that we have one degree in this case okay we can also write this using our degree symbol so we can say it's one degree like that if you had for example 5 over 360 okay that amount of a complete rotation this would be now 5° okay that would be the measure of that angle and you can of course still write it like this 5° now when we work with angles we're going to have some terminology that we have to get used to so the first guy we want to talk about is an acute angle okay so this is where the measure of the angle is greater than 0 de but less than 90° so you see you have this Greek letter here Theta and in trigonometry we often use these Greek letters to represent an unknown angle so here this is just some unknown angle again Theta here we'll say that Theta is greater than 0° but less than 90° so this is an acute angle now we've all seen right angles in our algebra course especially when you're talking about the Pythagorean formula but a right angle in case you don't know you're going to denote this with this special symbol here okay and this is where your angle is exactly 90° okay so Theta here we can say is exactly 90° so next we want to talk about something known as an obtuse angle so this is an angle that is greater than 90° but less than 180° so here we can say that Theta is greater than 90° okay the right angle but strictly less than 180° which we'll find out in a minute is a straight angle so for the straight angle what we're going to have is Theta is going to be 180° and basically what you have here you can imagine you start out with again two rays on top of each other you make half half of a complete rotation okay so this guy ends up right here okay and you end up with what looks like a straight line right but basically you started with two rays on top of each other you made half of a complete rotation and now the two rays are facing away from each other okay and in terms of the complete rotation we know the degree measure is 360° this is half of a complete rotation so half of 360 is 180 so that's why Theta here is 180° for your straight angle all right so let's wrap up the lesson and talk about complementary and supplementary angles so so we'll start with complimentary angles and basically when the sum of the measures of two positive angles is 90° the angles are known as complement angles so here's a typical example you can see that the bigger angle here if I started from here and I rotated here this is a 90° angle again this is given to us by this symbol right here okay but we have these two smaller angles and we want to find out their measure so you have one that's given by this guy right here 15x + 12 degrees and another that's given by this guy right here the 12x - 3 degrees so to find the measure of each angle here we use the fact that again the complimentary angle sum to 90° so I'm going to take this measure here which is 15x + 12 I'm going to leave the degrees off for now then plus this measure here which is 12x - 3 I'm going to set this sum equal to 90 for 90° so we're going to solve this and basically 15x + 12x is 27x X then 12 - 3 would be 9 and this equals 90 so what I'm going to do now is just subtract 9 away from each side of the equation and of course this is going to cancel we get 27x is equal to 81 divide both sides by 27 and you're going to get that X is equal to 3 so let me write that up here we'll say that X is equal to 3 so we're not done we need to find the measure of each angle here and xal 3 just tells you what you need to plug in okay so don't stop there so would have 15 * this 3 + 12 okay so 15 * 3 is 45 45 + 12 is 57 so the measure here that you're looking for is 57° okay then the other guy is going to be 12x - 3 so 12 * 3 and then minus 3 12 * 3 is 36 36 - 3 is 33 so the measure of this angle here is 33° and you can see that the measure of 57° plus the measure of 33° gives you a measure of 90° which again is the measure of the bigger angle okay so now let's talk about supplementary angles so we already know that a straight angle here so if I started here and I rotated again this is a half of a rotation this forms a straight angle it's 180° okay half of 360° 180° that's how you remember your straight angle so when the sum of the measures of two positive angles is 180° the angles are known as supplementary angles so this example has basically the same in terms of how you set things up you have this angle here that's 9x - 5° so 9x - 5 you have this angle over here which is 5x + 3° so I just put those two as a sum so 9x - 5 + 5x + 3 so I've got my sum going there and I set it equal to 180 for 180° again that is the full angle right if I consider this all the way from here to here it's 180° and so what we do here is we just solve 9x + 5x is going to be 14x -5 + 3 is of course -2 so this equals 180 let's go ahead and add two to both sides of the equation we get 14x is equal to 182 we can divide both sides by 14 and find that X is equal to 13 so X is 13 again don't just put that in stop you need to plug in to find out your measures okay so 9 * 13 9 * 13 - 5 9 * 13 is 117 - 5 is 112 So this guy is 112° okay and then 5 * 13 is 65 let me write that out so 5 * 13 is 65 + 3 is going to be 68 okay so this would be 68° so again if you look at the sum of 68° and 112° you get your 180° which once again is the total measure of this angle right so if we think about the whole angle including this smaller angle and the other smaller angle is 180° so we would expect that 68° plus 112° to sum to 180 degrees and in fact it does in this lesson we want to talk about angle measures in degrees minutes and seconds all right in our last lesson we talked about some of the basics that are involved with angles and we also learned about the unit of measure known as the degree it's how we can measure our angles we also have radians but we'll talk about that later on so we learned that a complete rotation or one complete rotation is assign the value of 360° in terms of the measure of the angle that is produced so let's say I started with two rays that were on top of each other again I take this Ray and I rotate it completely around okay completely around so what happens is the measure of that angle is assigned a value of 360° now if I think about this in terms of 1 over 360 so let's let's say I made that amount of a complete rotation well this is now going to be 1 Dee if I made 5 over 360 that amount of a complete rotation that's 5° so on and so forth now what happens when we need to measure an angle where the amount of the rotation is less than a degree so it's a portion of a degree a part of a degree a fractional amount of a degree well what we're going to turn to here is going to be minutes and seconds so this is pretty easy to work with basically you start with the fact that if if you have one degree and you divide it by 60 this produces a unit known as the minute okay so this is going to give you one minute so this is your notation for the minute okay now you could also say that 60 Minutes is going to give you one degree right so these two statements are saying the same thing I can take one degree and divide it by 60 and get one minute or if I multiplied both sides by 60 okay and cleared the denominator well what I'd have have is 60 minutes equals 1° okay so that's the same thing there now also we can go smaller okay so we can take our minute so one minute and we can now divide this by 60 and we're going to get a unit of measure known as the second okay so we're going to say 1 second like this so that's going to be your notation for that okay and we can also again just multiply both sides by 60 and we can say that 1 minute is equal to 60 seconds okay so pretty easy to remember overall now Additionally you can start with 1 second okay and you can say this is equal to again 1 minute over 60 and this would be equal to what in terms of degrees it would be 1 degree divided by now You' have 60 * 60 which is 3600 okay so we'll use this later on this is something you probably want to write down in terms of these formulas so that you're going to be good to go when you're doing your homework all right let's go ahead and look at the the first problem we're just going to add two of these guys together it's very simple you just need to understand how the carrying process works so let me explain I'm just going to set this up vertically you don't have to but I think it's easier to do it that way so let's just set this up vertically so 19 degrees we got 50 and we got 17 here so what I have here is 37 Dees 15 minutes 45 seconds plus 19° 50 minutes 17 seconds okay so we're adding this together so we know when we do a vertical addition since we have a base 10 system if you get a digit that is 10 or larger you do a carrying process well here because of the way we convert things we're looking for a value of 60 or larger to carry so when I add 45 + 17 I get 62 right so basically this is saying that I have 62 seconds but again if we go back up here we know that 60 seconds equals 1 minute so really what I could say is that this is 1 minute and 2 seconds right so this is the carrying process you're going to use you're going to put two seconds down here and you're going to carry this one into the next column okay just like if you were doing a vertical addition it's no different it's just you're changing the way you think about things because you're using 60 now okay so in this one I've got 1+ 15 + 50 so 1 + 15 is 16 16 + 50 is 66 so again now we have 66 minutes and again if we go back we know that 60 minutes equals 1° so I can say that this is 1 degree and 6 minutes okay so I'm going to put my six minutes down here I'm going to carry my 1 degree up and over to the next column and now I'm good to go I do 1 + 37 which is 38 and 38 + 19 which is going to give me 57 so 57 degrees so the answer here would be 57 degrees 6 minutes and two seconds all right now let's move on and talk about a subtraction problem so you're going to find the process very similar to Vertical subtraction so let's just go ahead and set this up so we have 14 here degrees 18 minutes 30 seconds we have 9° 25 minutes 50 seconds so put my subtraction sign out here and what you're going to do is think about the fact that again I have 30 I have 50 so I want to subtract here so what I have to do is borrow right and think about the fact that 1 minute is equal to 60 seconds right so if I take one away from this and make this 17 minutes now I can take this and convert it into seconds so I can say 30 seconds plus 60 seconds would be 90 seconds so I'm just going to put 90 seconds there I'm just borrowing again just like I do with vertical subtraction same thing I just got to get used to that number of 60 okay so 90 minus 50 is going to be 40 so This Is 40 seconds then when I I get over to this column here I've got to do the same thing again so I've got to borrow I'm going to borrow here and put this as a 13 Dees and remember if I think about 1 degree it's equal to 60 okay 60 minutes so I'm going to add 60 to 17 that's going to give me 77 so 77 minutes and then 77 minutes minus 25 minutes is going to give me 52 minutes so now I move into my leftmost column I have 13° - 9° that's just going to be 4° So my answer here will just be 4° 52 minutes and 40 seconds okay let's find the complement and also let's find the supplement of the angle that's given so we know that complimentary angles sum to 90° we know that supplementary angles sum to 180° so if you're asked to find the complement of this angle what I would do is I would subtract 90 okay minus this 29° 15 minutes 6 seconds so of course you need a borrow here so you can just go ahead and write this as 89° okay and I'll put this as 60 Minutes you need to borrow again so let's just write this as 59 minutes and let's just go ahead and write this as 60 seconds so all I did was I took the 90 and I borrowed and I borrowed again so we can do our subtraction okay so we go through now we're good to go 60 - 6 is 54 so this is 54 seconds 59 - 15 is 44 so this is 44 minutes and then 89 - 29 is going to be 60 so this is 60° so the complement of this angle here is going to be 60° 44 minutes 54 seconds now similarly you can erase this part and up here what I'm going to do is I'm going to put 179 nothing else will change because if I put 180 there again I'd have to borrow and then borrow again okay so this is how you would set it up and once you've done this it's very easy you just go through and say okay well 60 - 6 we know that's 54 so 54 seconds 59 - 15 again we know that's 44 so 44 minutes and then 179 - 29 is 150 so 150° there so the supplement of this angle would be 150° 44 minutes 54 seconds all right so now what we want to do is go from this decimal form here this 114.0 n° to degrees minutes and seconds so this is a pretty common task and very easy to do so when the end you want something that looks like this you're going to have degrees you're going to have minutes and you're going to have seconds okay so you take the part in front of the decimal point the 114 that whole number part and just put it right here for your degree okay so it's 114 degrees you're good to go with that now you move on to the part after the decimal point so this 09 so 9° is what I have left over after I fill that out now you want to convert this into minutes first okay so remember 1 Dee is equal to 60 minutes so to get this in terms of minutes I need to multiply by 60 okay so I would take this and let me just erase this we don't need this so times 60 this would give me 5.4 okay 5.4 now this now is in terms of minutes so this is in terms of minutes I'm going to take the whole number part again and I'm going to just drag that up here so I'm going to put five here and now I'm going to say that I have4 minutes so remember when you think about it you have 1 minute is equal to 60 seconds okay so I'm just going to multiply this now this 04 by 60 to get it in terms of seconds and this is going to give me a value of 24 okay so when we convert this over what we end up with is 11 14.09 de is 114° 5 minutes 24 seconds and if you have a TI 83 or a ti84 it will actually do this conversion for you you want to type this number out first then you want to key up your angle button which is usually second and then angle and you want to go down to the DMS for degrees minutes and seconds when you hit that it's going to give you this out as an output okay so that's another way to do it if you want to use your calculator all right so to wrap up the lesson we want to look at this 119° 5 minutes and 15 seconds and we want to put this into a decimal form so how can we do this well we're going to start with this 119 here and then put a decimal point and we want to think about the 5 minutes how can we convert that and the 15 seconds how can we convert that well I want you to remember that again we think about it 1 degree is equal to 60 okay 60 minutes but also we said that 1° / 60 was equal to 1 minute okay so if I have five minutes okay if I have 5 minutes just think about the fact that that would be 5 degrees divided by 60 okay so I I would say that I could divide this by 60 okay and I can get this in terms of degrees well 5 divid 60 is going to be 1 over2 but I'm going to reduce this in a minute because I'll talk about this further you're going to end up with this nasty repeating decimal so just leave this alone for right now all right so to get this part in terms of degrees remember I told you in the beginning that basically if you had 1 second okay it was equal to 1 minute divided 60 okay and 1 minute divided by 60 we could say is 1 Dee divided by 3600 okay so to get this in terms of degrees we need to divide by 3600 so let's erase this so let's erase this and divide by 3600 okay so at this point you would really be adding these two amounts together and basically from those amounts you would get a decimal that you can put on here now in a lot of cases you're going to get a nasty repeating decimal for each of them it doesn't happen all the time sometimes it works out nicely but my suggestion to kind of limit that is to just get a common denominator going so I'm just going to multiply this by 60 over 60 this is going to be 300 + 15 over 3600 and this is going to be 3 35 over 3600 okay and this is going to work out to a nice decimal for us which is 0875 so 0875 and this is in terms of degrees so not bad overall just a lot of things to remember in terms of what the conversion is okay once you kind of have that memorized it's very very easy to do and a lot of times nowadays you can just do this on your calculator in this lesson we want to talk about angles in standard position and also we want to talk about co-terminal angles all right so an angle is in what we call standard position if its vertex is at the origin and its initial side lies on the positive X AIS so that's all it is that's the whole definition so in this example here you see we have a little 35 Dee angle and we can see that our vertex is at the origin okay and we can see our initial side is on the positive X AIS so this guy is in standard position now additionally when we have an angle that's in what we call standard position it's said to lie in the quadrant where the terminal side lies so in this case my terminal side lies in quadrant one okay so this guy would lie in quadrant one now Additionally you remember that these guys go counterclockwise so this is quadrant 2 okay so if you had an obtuse angle you'd be in quadrant 2 if you had an angle that was greater than 180° but less than 270 de you'd be in quadrant 3 and then if you had one that was greater than 270° and less than 360° should be in Quadrant 4 okay now something to keep in mind you might get an angle whose measure is some multiple of 90° so it could be 90g 180° 270° 360° you know so on and so forth well in this case we have something known as a quadrantal angle okay so you will not be in any quadrant okay so that might be a question for you on your test so another little quick example of this we have another angle that's going to be in standard position so again the vertex is at the origin the initial is on the positive xais this guy is a 160° angle so it's an obtuse angle and it's going to lie in quadrant 2 right because that's where its terminal side is okay so let's talk a little bit about getting an angle measure that's larger than 360° we already know that the complete rotation or one complete rotation of a ray represents an angle whose measure is 360° so we've seen this in pretty much every video at this point so we have two rays on top of each each other if I rotate this guy completely around so one complete rotation that angle measures 360° let me put 360° there and basically if we keep going we can generate angle measures that are greater than 360° so as an example here you can see that we have a 450° angle so a 450° angle and we generate this by making one complete rotation so one complete rotation so to write there okay where the terminal side would be on top of the initial side and then we keep going another 90° so you'll notice how this looks like a right angle or a 90° angle but basically we show here that we made one full rotation plus we went another 90° so to get this 450° we take the 360° angle again from one complete rotation and we add to that 90° we know from here to here is 90° because this is a right angle and so we add those two mounts together and we get at 450° okay so that's where this guy comes from all right so now let's talk about these co-terminal angles so the concept here is that coterminal angles are going to be angles that have the same initial side and also the same terminal side but differ by amounts of rotation so the measures are going to be different by some multiple of 360 okay and I'll show you how to calculate that in a moment but basically if we look at this first example you see that you have a 55° angle and you have a 415° angle so this same initial side the same terminal side to get this first 55° angle we start here and we just rotate here that's all we do but to get the 45° angle again you're going to make more than one complete rotation so you're going to go around once okay so that's 360° and then plus another 55 degrees so in other words you took 360 plus your 55 okay let me put my degree symbol in each case and this gives us the 415° that we see from that Angle now to show this without making a little diagram or a little graph what you could do is just subtract okay so you can either do 415° minus 55° and that would give you 360° or you could go the other way okay it's going to be negative but it doesn't matter you're looking for some multiple of 360° okay so if you did 55° minus 415° you would get- 360° now let's look at another example of this in this case you have one positive angle and one negative angle so this works as well you have the same initial side and the same terminal side so this guy is ative 45° angle and this guy is a 315° angle so we start here and we go again if you're going clockwise it's a negative angle so this is a 45° the other one we're going counterclockwise so it's positive and we go in this direction so we're going around and this becomes 3155 degrees okay if you were to subtract the 315 minus A45 so 315° minus a 45° remember minus Nega is plus a positive you would get 360° if you were to subtract - 45° minus 315° you would get 360° so again the way that you subtract is not going to matter okay you're just looking for some multiple of 360° to determine if these guys are co-terminal again same initial side same terminal side they're going to differ by some multiple of 360 Degrees that's all it is let's look at the first example so we have 110° and 470° a lot of you can eyeball that and see that the difference between those two whatever either be if you did 470° minus 110° obviously that would be positive 360° and if you did 110° minus 470° that would be- 360° but again you're just looking for some multiple of 360° it could be positive netive doesn't matter some multiple of 360° so in this case we know these guys are co-terminal okay for the next one we have 135° and we have 495 de so again you can just subtract you could do 135° minus a negative 495 de which would be 135° plus 495 de in this case that would give you 630° okay and if you don't know off the top of your head if that's going to be a multiple of 360° you just divide by 360 and see if you get a remainder so if I punch that up on my calculator I get 1.75 so to say you don't have a remainder remember you want something that's either a whole number or an integer right you don't want a decimal part or a fraction or something like that right you want a whole number or an integer so because we have one. 75 we know that 630 is not divisible by 360 so this guy is not some multiple of 360 so these guys are not co-terminal angles all right for the next one we have 335° and 2,135 de so again you can just subtract you can subtract in any order if I do 2,135 2,135 de - 335° that's going to give me 1800° okay and if I took 1800 and divided by 360 I would get five so we can say that their measures differ by some multiple of 360 okay so they are going to be co-terminal angles okay let's look at another example that you'll see in this section so they'll tell you to find the co-terminal angle that is between 0 degrees and 360° so for this guy there's kind of a slow way and a fast way the slow way is it just keep subtracting or adding depending on the situation you're given 360° until you get into the desired range so I can subtract off 360 degrees and then do it again and I'd be in the desired range I can kind of eyeball that and say okay well if I subtract it take some some nice round number let's say we take 400 so I take 400 twice okay that's 800 I know if I subtracted 1,20 minus 800 I'd have 220 so i' know I'd be about where I need to be so what I'm going to do is I'm going to take 360 and multiply it by 2 that gives me 720 so this is the amount I want to subtract away from here again it's the same thing as if I subtracted away 360 and then it subtracted away 360 again okay that's all I'm doing so I'm going to subtract away 1020 de minus 720° this is going to give me exactly 300° okay and we are in the range that we wanted between 0° and 360° okay the last example is 596 de again if you want to find a co-terminal angle where it's between 0° and 360° in this case you need to add because it's a negative 360 Degrees until you get where you need to be okay so I know if I add once it's not going to be enough again I can just use 400 and say okay 596 plus 400 just just approximating is going to give me netive 196 so I know I'm not in the range with that so I'd have to add again so if I add it again I'd be at 204 okay and that seems like it that's where I want to go Okay so what I want to do is say okay I need 2 * 360 which again is 720 okay so I'm going to take- 596 de and add 720° of course what you're going to do there is just subtract right so you would do 720° minus 596 de which would give me 124° right so in this case I'd have 124° as my answer this would be the coterminal angle with this guy okay that fits the criteria where it's greater than 0 degrees and less than 360 de in this lesson we want to talk about angle relationships and similar triangles all right for the first example we're going to look at a problem that involves vertical angles this is something very simple basically vertical angles are angles opposite of each other where the two lines cross so you can see here we have these two lines that cross and at this intersection point you have angles one and three that are across from each other okay so those are vertical angles and then you also have angle four and two that are across from each other so those are vertical angles right you can show it like this with matching arcs so I can just do two like this here and two like this here and what I'll do is I'll do one here and one here okay so angles one and three would have the same measure and angles four and two will have the same measure okay that's a property of vertical angles just saying that vertical angles are going to have equal measures if we want to find the measure of angles 1- 4 what we do is we say okay we're given the measure of angle 1 it's 10x + 35° we're given the measure of angle 3 it's 20x - 55° okay well what we can do is set angle one which is again 10 x + 35° equal to angle 3 which is 20x - 55° that's going to give me the value for x and then I can plug that back in and find out what is angle one what is angle three so I'm going to go ahead and just subtract and let me scroll down get some room I'll come back up to the diagram in a minute so I'm going to subtract 20x away from each side of the equation and let's go ahead and get rid of this and I'm going to subtract 35 away from each side of the equation so let's go ahead and get rid of this and then if I do 10x okay 10 x - 20x that's going to give me -10x and is equal to 55 - 35 is going to be 90 if we just divide both sides by -10 we're going to find that X X is equal to 9 okay so let's erase everything and if you're on a test or something like this you just have to plug into one of these because again these are vertical angles so the measure is the same okay so we know that X is 9 and if I plug in 10 * 9 is 90 90 + 35 is going to be 125 okay so the measure of angle one is 125° okay 125° and also the measure of angle 3 is 125° as well okay and you could plug PL in there and again if you plugged in a 9 20 * 9 is 180 180 minus 55 is 125 okay so that's where you get the 125 degrees now the other thing that you would note here is that these guys angles one and two are going to be supplementary angles right their measure is going to add up to 180 degrees if you kind of rotate this you can see that this guy right here again this is a straight line okay this is a straight line so if I took the measure of angles one and two let me just do this as a full Arc here okay this would be 180 degrees right it would be a straight angle so we know that the measure of angle two can be found by taking 180 okay degrees and subtracting away 125° and that would give us 55° okay and again because angles 2 and four are vertical angles I can just put this as 55° as well okay so our answer for this problem is that we have 125° for the measure of angles 1 and 3 and then 55° for the measure of angles 2 and 4 all right so let's look at an example that involves a transversal that intersects parallel lines so we see that eight angles are formed here you have 1 2 3 4 5 6 seven8 okay those angles there have certain properties so the ones I want to draw your attention to we have interior angles so you can think about this as being between the parallel lines or some people say inside of the parallel lines so those would be the Ang Les 3 4 5 and six okay when we have alternate interior angles they're going to have equal measure okay so angles three and six those would have equal measure and then also let me kind of change this up also angles four and five would have equal measures okay so if we look at the alternate exterior angles so the exterior angles would be you can think about this is outside of the parallel line so angles one and two and then seven and eight the alternate exterior angles will have equal measures okay so we can say that would be angles 1 and 8 and then also we can say angles 2 and 7 okay so we can use these properties to find out the value for all eight angles here so if we start out with the fact that angle two has this measure 11x - 53° and angle 8 has this measure 4x + 8° well let's think about the fact that angles 2 and seven are alternate exterior angles right alternate exterior angles so they would have the same measure okay so I can really write that the measure of angle 7 is 11x - 53 okay degrees okay so we know that angles s and 8 together form a straight angle right or an angle that's 180° so in other words angles seven and 8 are supplementary angles so what I can do here I can add 11x - 53 + 4x + 8 and set this equal to 180 again for 180° I can solve for x I can go back and plug in and find out what is the measurement of each angle 1 through 8 okay so what I'm going to do here 11x + 4x is going to be 15x and then 53 + 8 is going to give me - 45 so minus 45 this equals 180 okay so if I add 45 to each side of the equation let me scroll down and get some room going then what am I going to have this is going to cancel and if I do 2 180 + 45 I'm going to get 225 and this over here is still 15x if I divide both sides by 15 I'm going to get that X is equal to 15 okay so let's erase this and we're just going to plug in you can get rid of this okay we're going to plug in a 15 starting here so what is 11 * 15 that's going to be 165 and if I subtract away 53 I get 112 So I know that the measure of this guy is 112 de again alternate exterior angles okay alternate exterior angles have the same measure okay so this guy two and seven have the same measure so this is 112 degrees here remember that vertical angles have the same measure so that means that seven and six have the same measure okay so this is 112 degrees two and three have the same measure so this is 112 degrees okay so at this point you know four of your angles and again you can figure out the rest you can plug in here if you want you can plug the 15 in and do it that way or you can just take 180 and subtract away 112 and get 68 and fill out the rest right because again if I plugged in a 15 there 4 * 15 is 60 60 plus 8 is 68 so this is 68 degrees or again you can use the fact that seven and8 are supplementary angles so you can subtract 180 minus 112 to get the 68 as well now again8 and one are alternate exterior angles so they have the same measure and then again you can also say that one and four are vertical angles so they have the same measure and eight and five are vertical angles so they have the same measure okay and there's lots of different ways to go about getting these values but you can see that for angles 1 4 5 and 8 the measure of the angle is 68° and then for angles 2 3 6 and 7even the measure is going to be 112° all right so let's look at some examples now that involve triangles so what you need to know here is that the sum of the measures of the angles for any triangle is going to be 180° so with this example here we're given angles 1 2 and 3 and the measure of angle 1 is 6X + 1° the measure of angle 2 is 10x - 56° and the measure of angle 3 is given it's 43° so all we really need to do is set these guys as a sum and it will be equal to 180° okay so if we set up a sum here so we'd say 6 x + 1 + 10 x - 56 then + 43 and this will be equal to let me drag this down because I'm going to run out a room here and let me just get some room going so this will be equal to 180 okay and we just solve for x very simple so 6X + 10 x is 16x 1 - 56 is going to be- 55 then- 55 + 43 is going to be -12 so I'll put -2 here this equal 180 if we add 12 to both sides of the equation we're going to get that 16x is equal to 192 you get some room going here to finish things off I'm going to divide both sides by 16 and get that X is equal to 12 so X is 12 here so let me erase all this so we said that X was 12 so for angle one if I plug in a 12 there 6 * 12 is 72 72 + 1 is 73 so the measure of angle 1 is 73° okay and you can put it here as well this is 73° for the measure of angle 2 it's going to be 10 * 12 which is 120us 56 which is going to be 64 okay so this will be 64° and I'll write that in here this is 64° we're already given the fact that this is 43° and again if you sum these amounts you're going to get 180° all right for the next two problems we're going to see typical examples where we're dealing with similar triangles so hopefully you remember this from geometry but essentially if you have similar triangles then corresponding angles must have the same measure okay then secondly corresponding sides must be proportional so that means the ratios of their corresponding sides must be equal okay so for the first problem we're going to deal with the statement that says that corresponding angles must have the same measure okay so we have these two similar triangles and I'm just going to call this first one triangle ABC so I'll say triangle ABC is similar to Triangle DF now what you want to do is match up the corresponding angles for this problem you'll notice that the corresponding angles have matching arcs okay so we have one Arc here and one Arc here so angle a corresponds to angle D and then you have two arcs here and two arcs here so angle B corresponds to angle e and then you have three arcs here and three arcs here so angle C corresponds to angle F now as we just said corresponding angles if you have similar triangles they must have the same measure so what that tells me is that if I know the measure of angle f is 144 degrees then I also know the measure of angle C is going to be 144° as well if I know that the measure of angle a is 18° that tells me that the measure of angle D is 18° as well now what I don't know here is the measure of angle B and the measure of angle e so let me fill out this guy right here I'll just put e and I'll put a question mark and for this one I'll put B and put a question mark now how can we figure this out remember when we think about the sum of the measures of the angles of any triangle it's going to be 180° so what I could do is take 180° and subtract away the sum so in each case it's 18° plus 144° so this sum represents the angle measures for two angles and 180 Dees would be the sum of all three so if I take this one subtract away the sum of the two I'm going to get the missing one so let's put this as equal to so 18° plus 144° that would be 162° so this let me just slide this down would be 180° minus 162° which gives me 18° so if we're trying to find the measure of angle B it would be 18° and again because angle B corresponds to angle e they're going to have the same measure so this is 18° as well so if you wanted to know the angle measure for all of these guys well well in this bottom triangle the measure of angle a is 18° the measure of angle B is 18° as well and the measure of angle C is 144° and then in the top triangle the measure of angle D is 18° the measure of angle e is 18° and the measure of angle f is 144° all right for the last problem here let's look at an example with similar triangles where we're asked to find some missing side lengths so this is a very easy problem to solve just deals with proportions and essentially what you're going to do here is look at these two guys and generally you're going to be given a statement that says something like triangle ABC is similar to Triangle DF now we already know about finding the corresponding angles you could do it based on the way that I read the two triangles so again I said triangle ABC is similar to Triangle DF so that tells me that a is going to correspond to D B is going to correspond to e and C is going to correspond to F or again you can use those those matching arcs so you have a single Arc here and a single Arc here so B and E are going to correspond you have a triple Arc here and a triple Arc here so A and D are going to correspond and then a double Arc here and a double Arc here so C and F are going to correspond now when it comes to side lengths a lot of times you just have to write things down if you're given a not so easy scenario here it's very easy you can match things up but sometimes the triangles are flipped all the way around and stuff like that so it's not so easy so you might want to write the statement out and say triangle a b c is similar to Triangle in this case DF and then you could just match things up so I can say this side AB is going to correspond to this side de so here's my side AB here's my side de those are corresponding sides then I have the side BC that's going to correspond to the side EF so here's my side BC and my side EF you can see those are corresponding sides and then lastly you can say this side AC is going to correspond to the side DF so here's my side AC my side DF those are corresponding sides now we know if we're working with similar triangles that corresponding angles must be equal and corresponding sides must be proportional so again this just means that the ratios of their corresponding sides must be equal so let's use that to set up a little proportion so I know that this side AB corresponds to this side de okay so what I'm going to do is write the ratio of the length of this side AB to the length this side de so I'm going to set that up as a fraction I'll just say this is AB so that's going to denote the length of this side two this will be de so that'll be the length of this side I'm going to say that's equal to the ratio of let's go ahead and say the length of the side BC so the length of the side BC to the length of the side EF so the length of the side EF so let me write that in there and again we know that BC and EF are corresponding ing sides and I'll also say this is equal to the ratio of the length of the side AC so that's going to be right here so a to the length of the side DF so DF is going to be right there so once you have this set up I would go through and plug in what you can and then you're just going to solve for what you don't have okay now before I even do that a lot of people are going to ask why did I choose to put the ab on top and the de on the bottom here and it actually doesn't matter you could flip this but you have to stay consistent in this case I wrote the information from the smaller triangle over the information from the bigger triangle but I could have also set this up as D over AB is equal to EF over BC is equal to DF over AC okay so as long as you stay consistent you're going to be good to go so I'm just going to keep it like this and I'm just going to replace what I can so for the length of this side AB it's 15 so I'm just going to replace this with 15 for the length of the side de it's 30 so I'm going to replace that with 30 now 15 over 30 is2 so let's just simplify that and say it's a half now for the length of the side BC it's 30.5 so let me put that in and then for the length of the side EF that's an unknown right so we don't know the length of that side then for the length of the side AC that's 24 so this is 24 and for the length of the side DF that is unknown let's grab this real quick and I'm just going to solve one of these at a time so I'm going to pick the one on the left and say 1 12 is equal to 30.5 this is over again the EF is the length of the side EF so cross multiply and you would get EF is equal to 2 * 30.5 is 61 so the length of the side EF is 61 so you can come back up here and just put that this is 61 right here or if you wanted to notate this you could say EF this just stands for the length of the side EF that's going to going to be 61 okay let me put DF here and we'll come back down so I'm going to set 24 over DF equal to a half you wouldn't want to set this equal here because you're not going to be able to solve so I'm going to cross multiply again so this is DF is equal to 2 * 24 is 48 so the length of that side DF is going to be 48 so coming back up you can put 48 here and 48 here in this lesson we want to talk about the trigger ometric functions all right so to Define our six trigonometric functions we're going to start with an angle in standard position and we're going to name this guy using the Greek letter Theta so you have your angle Theta there remember in standard position your vertex is at the origin your initial side lies on the positive xaxis so this is your initial side and then this guy over here is called your terminal side so what we're going to do from here is we're just going to pick some point okay that's not the origin that lies on the terminal side so this point that I've chosen here could be any but I've chosen this one this is going to be the point which is -6 comma 6 so -6 comma 6 now for the formulas you want something generic so what I've done is I've just labeled this as Point p with the coordinates X comma y of course you can fill that in with whatever you have here we have -6 for x and positive 6 for y now additionally what we need to do to get started is we need to know about something called R okay so R is just going to take the place of d in the distance formula so you remember that the distance formula the distance between two points in the coordinate plane is equal to the square root of you have your x sub 2 - x sub 1^ 2ar plus you have your y sub 2 - y sub 1ty squar okay so look at what we've done here we have this point again generically X comma y are the coordinates and we have this guy right here which is the origin most books will call this the point O okay for origin I'm just going to call it the point l because I don't want you to get confused the coordinates there zero and zero okay it looks too similar so Point L here Z comma 0 is my origin I also have another Point here to form a right triangle okay which is point Q which has the coordinates X right because it's got the same X location and then zero right because it's on the X AIS so X comma 0 so we can see that we can form a right triangle and obviously you could use the Pythagorean formula to find the distance from point P to point l or the origin or you could just plug into the distance formula which is much easier now what we're going to do is we're going to replace the D with an R okay and I'll explain why that occurs in a future lesson let's get to the unit circle first I want to just keep this simple additionally since we're working with a point that is 0 comma 0 I can replace this with a zero and this with a zero okay and now I no longer have an x sub one and a x sub 2 or a y sub one and a y sub 2 so I can just drop this notation okay we know that x - 0 is X and Y - 0 is y so I can just really say this is x^2 okay plus y^2 under this radical Okay so we've made that quite simple now in terms of the definition for r all right so let's take a look at the six trigonometric functions that we're going to work with remember this R here that you see in these formulas refers to what I just wrote It's the square otk of x^2 + y^2 now as we move forward in the course we'll see different ways to remember this I know a lot of you think about the SOA TOA okay if you've taken trigonometry before but for right now we're just going to keep this really really simple we'll get to that later on so the S of theta is going to be equal to y r then the cosine of theta is going to be equal to X over R then the tangent of theta is going to be equal to Y overx where X is not equal to zero the cosecant of theta is going to be equal to R over Y where again Y is not equal to 0 the secant of theta is going to be equal to RX where X is not equal to zero and the co tangent of theta is going to be equal to X over Y where Y is not equal to zero all right so let's go ahead and look at an example let's say that we're given the point 9 comma 9 and basically it's going to be the terminal side of the angle Theta in standard position is going to pass through this point again 9 comma 99 would be the coordinates and we want to find the values of the six trig functions here so let's do a quick sketch a lot of people like to do that to see what's going on it's not completely necessary because you're just going to be some things in but just to get an idea of what's going on so if we highlight this point here again this is 9a9 that's going to be our Point Let's just call it P okay and basically what we're looking at is our angle here Theta okay so we want to find the six trigonometric functions based on this guy so let's go back up and let's remember that this is our x value so our x value is going to be nine our y value is going to be9 just pull it straight from this ordered pair here okay so this is your X and this is your y okay make sure you bring the signs along for the ride then your r value is what remember this is found by taking the square root of x^2 + y^2 so the square root of x^2 x is 9 9 squ is 81 plus y^2 Y is 992 is also 81 so you could really write this as 2 * 81 to make it a little bit cleaner okay because you know that 81's a perfect square the square root of 81 is 9 so you could really say this is 9 * the < TK of 2 to find all the trigonometric functions you're just going to plug in what you're given okay that's all it is so we go back up and there'll be easier ways to remember this but for right now we're just going to reference this so s of theta is y over R so s of theta s of theta is y okay which is -9 over R which is 9 * < TK of 2 okay well we know we want to WP rationalize this denominator first off we can simplify this because this would be1 but we want to rationalize this so let's multiply the numerator and denominator by the < TK of two and this would give us what it would give it negative < TK of 2 over two okay so that would be your s of theta so let's erase this and just kind of make this a little bit cleaner okay so we'll bring this down and bring this a little bit over and so now let's look at the cosine of theta so this is X over R so the cosine of theta okay so this is X over r x is 9 and again R is 9 * < TK two and basically you're going to get the same thing as this when you rationalize it would just be positive okay so you would cancel this with this and get a one here and you'd multiply the numerator and denominator by the square < TK of two and essentially you would have what you'd have the square < TK of two over two so this is the square < TK of two over two okay so the next one is the tangent of theta okay so y overx X so the tangent of theta okay so what is y That's -9 what is X that's positive 9 so that would be1 all right so next we have these guys that we can find by just flipping things okay so we know that the cosecant of theta can be found by flipping the S of theta okay because this is r y this is y over R so all I want to do for the cosecant of Theta is just flip this guy okay so I would have 2 over the < TK of Two And if you're smart before you rationalize that guy you'll go ahead and flip it right because now we have to rationalize again so I'm going to multiply this by the < TK of two over the < TK of two and that will give me what you have a negative out in front 2 * the < TK of two over this would be two and you can cancel this would cancel with this you're just left with the negative square < TK of two okay so the negative the negative square T of 2 so then we go back and we see that the secant of theta is rx so I'm just flipping the cosine of theta so I'm just flipping the cosine of theta and again the only thing that was different between this one and this one is the sign so I know that the secant of theta okay the secant of theta is going to be equal to the square < TK of two again you can go through this on your own you would flip this and let's just do it real quick so that nobody's confused it's 2 over the < TK of two and you multiply this by the < TK of two over the < TK of Two And so what do you get you get a two down here okay that's going to now cancel with this two up here you're left with just square of two okay so that's how we got that now the last one is the cotangent of theta and that is found by flipping the tangent of theta okay in this case the tangent of theta was1 so if you think about it you really say that's NE 1 over one flip it you still have 1 over negative 1 which is still Nega 1 so my co tangent of theta will be -1 as well all right so let's look at another example we're going to change things up just a little bit so you might also be asked to find the trigonometric function values of an angle when we know the equation of the line that coincides with the terminal Ray so I want you to think for a second we all know that the equation of a line in standard form that passes through the origin looks like this so it's something ax + b y = 0 so for example this X x - yal 0 if I didn't consider this restriction that would be a line that passes through the origin okay but what we're going to do here is we're going to restrict the X values to either be X is less than or equal to zero or X is greater than or equal to zero okay what we're going to get as AR aray with an end point at the origin okay so let's look at an example of this down here okay and what you see is that we have x - Y is equal to Zer okay that line would be drawn if I extended it past this point let me kind of highlight this in a different color if I extended it past this point 0 comma 0 but we're going to restrict the X values to be greater than or equal to zero so we're only going this way so this is the end point and we go this way so this is your terminal side okay so you might see this as well all you need to do is pick a point that's not the origin so in this case I've picked the point 3 comma 3 okay and one way you can do that if you haven't drawn this guy you can just pick an x value that lies in the range again don't pick zero so let's say you chose three plug in a three for X so you know that X is three solve for the unknown y what's it going to be in this case you can eyeball and see it's going to be three as well but again if you had to solve you would just say Okay 3 - y equal 0 let me go ahead and add y to both sides of the equation and this would cancel and I get 3 is equal to Y okay so X is 3 Y is three this is going to make it really easy for us to crank out those six trigonometric functions so what I'm going to do here is go through them more quickly so let's just write them out so we have S of theta and this is equal to our y value over our r value and we'll get R in a minute then we have our cosine of theta okay so this guy is going to be the X over The R okay then you have your let me slide this down a little bit so I went a little too far over so we have the tangent of theta this is going to be equal to your y over your X now we can crank this one out right now because X is 3 and Y is three let me label this this is X this is y so 3 over 3 we know would be one so let's just put that down right now okay then we know that the cosecant of theta okay the cosecant of theta is found by just flipping this guy so it's going to be R over y then the secant of theta okay the secant of theta is found by flipping this guy so it's R overx and then the coent of theta is found by flipping this guy in this case you flip one right if you had one over one and you flipped it you still have one over one so it's just one so what is r in this case well you would have let me scroll down and get some room going so R is equal to the square Ro T of x^2 + y^2 okay so X is 3 Y is 3 3 2 is 9 so you'd have 9 + 9 or 18 right so 18 is 9 * 2 9 is a perfect square it's 3 * 3 so pull that out and you'd say this is 3 * aunk of two okay so what I'll do is everywhere there's an R I'll just replace it with 3 * < two so 3 * < TK two and 3 * < of two and you can do it here as well but what's going to happen is we're going to crank these out really quickly because these are going to be the same and these are going to be the same okay because we have the same X and Y value so X and Y as I just said are the same so I'm going to put a three here and a three here and you can put a three here and a three here okay and very quickly we can see that this cancels and this cancels so this is going to be theun of two in each case okay and we can write that more cleanly s > of two and let's write that more cleanly Square < TK of two and in this case we need to rationalize right so we'd have one over the < TK of Two And if I multiply by theare < TK of two over the sare otk of two well down here I would have two okay and up here I would have the square root of two so let's write this as the square root of two over two and this is the same right so this is the square < TK of two over two all right so these were more simple examples as we progress we'll see things that are much more challenging in this lesson we want to talk about reciprocal identities signs of trigonometric function values and also how to find missing function values when we're given one value in the quadrant of the angle all right so let's begin our lesson by talking about the reciprocal identities so in the last section we talked about the six trigonometric functions we found that some of these functions are reciprocals of each other so keep in mind when we use the value of R okay R by definition is equal to the Square < t of x^2 + y^2 okay and we talked about that in the last video so we see that s of theta is equal to Y / R okay and then cosecant of theta is equal to R / Y and then in this case Y is not equal to zero we don't have to have that restriction for R there because R is always going to be a positive number okay but we do need it here because y could be zero so these guys the S of theta and the cosecant of theta are reciprocals so if you know one you automatically know the other by just flipping the fraction okay then the same thing goes when we look at cosine of theta which is X over R and secant of theta which is rx again X is not equal to zero here then again when we look at the tangent of theta which is YX X is not zero and the cotangent of theta which is X over y Y is not equal to zero those guys are reciprocals of each other as well so in your book when you're talking about the reciprocal identities let me erase this real quick they'll write them in a specific way so we can say that the S of theta okay is equal to 1 over the cosecant of theta okay then you could also say the cosecant of theta is equal to 1 over the S of theta remember if you have one divided by something okay all you're doing is you're ending up taking the reciprocal so if I had 1 divided by 1 / y over R you could really say this is 1 over one times the reciprocal of this guy so R over y so this ends up being R over y okay so that's all that is it's just fancy notation so let's erase this real quick and let's keep going so we have the S of theta is one over the coant of theta the cosecant of theta is 1 over the S of theta then you say the cosine of Theta is equal to 1 over the secant of thet okay and then the secant of theta again is just 1 over the cosine of theta let's write these down here so the tangent of theta is equal to 1 over the coent of theta and of course lastly the coent of theta is going to be equal to 1 over the tangent of theta okay so that's all this is it's not very complicated when you see this in your book the topic of reciprocal identities you take the two that are reciprocals okay one will be over here one will be in the denominator you can switch the positions again the S of theta and the cosecant of theta those guys are reciprocals so the S of theta is equal to 1 over the cosecant of theta similarly the cose of theta is equal to 1 over the S of theta and you see the same thing with the cosine of theta and the secant of theta and with the tangent of theta and the coent of theta so let's look at a quick example of this so we have that the secant of theta is equal to 9es or 9 /2 what is the cosine of theta well well again the secant of theta and the cosine of theta those guys are reciprocals of each other so if this guy is N9 Hales all I have to do is flip it and it would give me 2 9ths all right let's take a look at another example so here we have the S of theta is equal to the negative of theun of 6 over3 what's the cosecant of theta again these guys are reciprocals so I can just flip them and say this is 3 over the negative square T of six okay what I'm going to do is I'm just going to put my negative out in front and then I'm going to multiply the numerator and denominator by the square < TK of 6 so I can rationalize the denominator so this will be equal to the negative of we'll have 6 * 6 is going to be six and you'll have 3 * of 6 up here in the numerator of course you can cancel this would cancel with this each is divisible by three so this would just be one so you can just get rid of it and 6 divided 3 is two so your final answer here would be the negative of the < TK 6 over 2 all right let's talk about the signs of the trigonometric function fun values so we know that when we defined the trigonometric functions we gave r as the distance from the origin or the point 0 comma 0 to the point x comma y so since this distance is going to be undirected okay the value of R is always going to be positive so if you look at something like sine of theta okay we know that this is defined as y/ R okay so since R is always positive we think about the sign coming from just whether or not Y is positive or negative so if you look at your coordinate plane remember there's four quadrants in the first quadrant you have the X and Y values that are positive in the second quadrant the X values are negative the Y values are positive in the third quadrant both the X and Y values are negative and in the fourth quadrant the x value is positive the y- value is negative okay so where is y negative it's negative in quadrants three and four so that means if I'm in quadrant three or four the S of theta is is going to be negative because R is always positive right you would end up with a negative over a positive which gives me a negative okay and I should have put a negative there but if you look at this little table I have up here this summarizes everything for you in the first quadrant where X and Y values are positive all your six trigonometric functions are positive in the second quadrant where X is less than z and y is greater than zero your s of theta is going to be positive and then the reciprocal the cosecant of theta is going to be positive as well as we move into the third quadrant where again X and Y values are both negative you get just that tangent of theta and cotangent of theta again those guys are reciprocal those guys are positive everything else is negative and then in the fourth quadrant where X values are positive and Y values are negative you'll see the cosign of theta and also the secant of theta again those guys are reciprocals those guys are positive everything else is negative so an easy way to remember this because trying to remember this chart is kind of hard is a little pneumonic device all students take calcul okay so just take the first letter from each so all a students s take T and C for calculus so all students take calculus and this goes counterclockwise so this is for Quadrant One this is for Quadrant two this is for Quadrant three and this is for Quadrant four okay so the a just stands for all so all six trigonometric functions in quadrant one are going to be positive the S is for sign okay so the S of theta okay is going to be positive and you know the reciprocal of that will be positive as well the reciprocal of the S of theta is going to be the cosecant of theta so s of theta and cosecant of theta are positive in quadrant 2 then in quadrant 3 you have T4 tangent of theta and then also the reciprocal the cotangent of theta so those guys are positive everything else is negative and then in quadrant four you have a c for cosine of theta and the reciprocal of that is the secant of theta so again those two are positive everything else is negative all right so let's use what we just learned to look at a few examples we're going to identify the quadrant of the angle based on these conditions so we have that the cosine OFA is greater than Z and the cose of theta is less than zero let's turn to our all students take calculus once again quadrant one quadrant 2 quadrant 3 and Quadrant 4 the way I look at it is I just say okay well cosine of theta is greater than zero so where does that occur it occurs in quadrant one and also Quadrant 4 remember the C is four the cosine of theta and it tells you that the cosine of theta and the secant of theta are going to be positive everything else is negative so let me just write below this that it's quadrant 1 and Quadrant 4 that satisfy that condition then what satisfies this condition the cosecant of theta is less than zero remember the cosecant of theta is going to be the reciprocal of the S of theta okay so the S stands for the S of theta and that's in quadrant 2 so we know that in Quadrant One everything's positive in quadrant 2 the S of theta and the cosecant of theta is going to be positive and everywhere else so in quadrant 3 and Quadrant 4 the S of theta and the cosecant of theta would be negative so this condition is met in quadrants three and also quadrant four let's just erase this and now all you want to do is just say okay well which quadrant satisfies both conditions okay in some cases you might have more than one in this case we only have one it's quadrant four okay so if you wanted to answer this you could just erase these okay you don't need them anymore and you could just write out Quadrant 4 or a lot of teachers will just let you write the Roman numeral of the quadrant just say Quadrant 4 like that for your answer okay another type of example you might get is to determine the signs of the trigonometric functions of an angle when again it's in standard position and has a given measure here it's 110° so this is really easy to do this is something where you might want to make a little sketch so if this guy's in standard position again this is going to be your initial side here's your terminal side we know that if it was touching your terminal side was touching the y- axis like this it would be a right angle be 90° if it goes a little further than that but not far enough to where it's touching here right if it was touching the x- axis You' have 180° angle or a straight angle so it's in between those two so we know it would be in quadrant two right this would be your 110 Dee angle there so in quadrant 2 again all students take calculus very easy the S tells me that it's s of theta and also the cosecant of theta so you can say the s of theta and the cosecant of theta those guys are going to be positive okay and then the other guys are going to be negative so you might want to write all these out or you could say all others are negative depending on your teacher and what they ask for let's just go ah and write them out we'll say the cosine of theta will say the secant of theta okay those guys are negative and then we have our tangent of theta and our coent of theta those guys are negative as well so only s of theta and Co cant of theta are positive for this example all right another one of these we have a negative 150° angle remember when you have a negative angle and this guy's in standard position you're rotating clockwise now instead of counterclockwise so this is negative 150° again you're starting here and you're rotating clockwise so again if you rotated this way okay and you got to right here if it's a negative angle that would be 90° if it's positive it would be postive 270° but here it's negative so this would be 90° this would be 180° so we're between those two again we're going to land in quadrant 3 here so that's your negative 150° angle so in quadrant 3 again just do all students take calculus so in quadrant 3 you have a t for the tangent of theta and then also the coent of theta those are going to be positive everything else will be negative so you have your tangent your tangent of theta and also your cotangent your cotangent of theta those guys are positive and then the rest are negative so s of theta and the cosecant of theta those guys are negative and then also your cosine of theta and also your secant of theta okay those guys are going to be negative as well so for this example the only two that are positive are the tangent of theta and the cotangent of theta all right to wrap up the lesson let's look at a very common type of problem this is where you're given one trigonometric function value and you're either given the quadrant or you have to figure out what it is based on the information you have and basically you want to come up with all six trigonometric function values so here we're told that the secant of theta is 19 over 11 and the S of theta is less than zero okay so you can start this off in a few different ways and there's also multiple ways to do it but based on what we know so far we know that the secant of theta we know that the secant of theta is going to be equal to my r value over my x value okay so if this is 19 over 11 then I know that my r value is 19 and I know that my x value is 11 okay so that's important so I know I'm going to need Y and how do I find y remember that R by definition is the square < TK of x^2 + y^2 so let's work on this really quickly okay we'll come back up to this in a moment so let's just paste this in here so we have a lot of room to work we know that X is 11 11 s would be 121 and I know that R is 19 so I can replace this and say that I have 19 is equal to the square < TK of 121 + y^2 if I want to get y by itself I first have to clear this radical to do that I'm going to square both sides squaring 19 gives me 361 squaring this side tells me that this would cancel with this and I'm left with the radicand which I'm just going to change the order and say this is y^2 + 121 and now to isolate y because I'm trying to solve for y okay so to isolate this I'm just going to subtract 121 away from each side of the equation this would be 240 over here over here this is gone I'm just going to have y^2 so let me erase all of this and I'm going to also flip the order around and put y^2 on the left and 240 on the right again that doesn't matter that's just a personal preference what I'm going to do let me just kind of scooch this down a little bit just write this over here because we need some room I'm going to take the square root of this side and I'm going to go plus or minus the square root of this side so 240 if you take the square root of that you can simplify right because 240 is 16 * 15 so I could say this is 16 * 15 squ of 16 is four okay so I can just pull that out and say this is 4 * the of 15 okay but you do have this plus or minus out in front that is a bit problematic here so we need one y value it can't be pos4 * 15 and then also -4 * 15 you need one of the other okay so let's go back up and let's go ahead and write all the information that we know so X is equal to 11 we know that R is equal to 19 and right now we know that Y is equal to plus orus 4 * theun of 15 but we need to get that down to one single value okay and we're going to do that with the use of this guy right here okay so we're told that the S of theta is less than zero so where is the S of theta less than zero remember if you use your all students all students take calculus once again this is quadrant 1 2 3 and four okay the S is for the S of theta and the A is for all okay so in quadrants one and two the S of theta is positive in quadrants three and four it's negative okay so let's write three and four here three and four here okay and let's think about the C of theta remember the secant of theta and the cosine of theta those guys are reciprocals okay the cosine of theta is going to be positive because this guy is positive in quadrants one and four so quadrants one and four so again when you get this you're looking for what is the quadrant that matches both okay the quadrant that matches both is going to be quadrant four all right so if we know that this gu's in quadrant four then what are the Y values in quadrant four we know if we go all the way back up here that in Quadrant 4 your yv values are negative okay they are negative you are below the x axis so that tells me because the Y values are negative that I don't want the positive I want the negative 4 * theare of 15 okay so once you figured that out the rest of it is just plugging things in okay knowing your ratios having those ready and then plugging in as we progress we're going to go through one more lesson and we'll go through a section where we start talking about the SOA TOA okay and we'll remember this a little bit easier for right now we're putting it in terms of x r and Y okay so we have our s of theta let's just start with that that is the Y value over the R value then my cosecant of theta okay is the reciprocal of this so this is the r over the Y okay and then let me write out the rest so we have our tangent of theta our tangent of theta this is going to be our yalue over our x value and then our cotangent of Theta is going to be the reciprocal of this so it's X over Y and then our secant of theta okay is given to us but it's RX okay RX but in this case we already know what it is it's 19 over 11 okay then our cosine of theta okay is X over R but again it's the reciprocal of this guy so we can just flip this and say it's 11 over 19 okay so let's plug some things in and let's kind of get some room going here and let's see so y and it's a little bit off the screen here but we'll just go back and forth so Y is -4 * < TK 15 and R I know you can't see it let's scroll back up it's 19 okay so R is 19 so let me just fill that in everywhere that way we don't have to keep going back and forth okay so those are the only two places we need the 19 okay then y let me fill that in everywhere so this is -4 * > 15 and let me fill that in here -4 * > 15 and also here so 4 * 15 and X let's just scroll up real quick that's going to be 11 okay so we're done with that so let's go back up here put 11 and then go up here and put 11 for the cosecant of theta and the coent of theta you need to rationalize the denominator so for the cosecant of theta let's just do this down here I'll make the whole thing negative I'll say that I have 19 over 4 * theun 15 I'll multiply by the < of 15 over thek of 15 of 15 * of 15 is of course 15 okay 15 * 4 is 60 okay so you really can't do anything further to simplify so let's just erase this scooches down a little bit so I have more room I'll put the negative of you have 19 * the < TK of 15 over 60 all right so that one's done and now I just need to rationalize this denominator so I've got 11 over let's put the negative out in front we have 4 * > 15 so we'll multiply by the < TK of 15 over the < of 15 again so we have 15 * 15 which is 15 15 15 * 4 is 60 so let's erase all this put equals I'll make the whole thing negative my denominator will be 60 my numerator will just be 11 * the of 15 so 11 * the < TK of 15 and I'm all done in this lesson we want to talk about the Pythagorean and quotient identities we'll look at the ranges of trigonometric function values and also we'll see some examples where we're finding missing function values so before we jump into the lesson for reference s let's go through the six trigonometric functions again I want you guys to remember the way we've defined this we've said that R is equal to the square Ro T of x^2 + y^2 okay so remember that so I can also say if I Square both sides that R 2 is equal to x^2 + y^2 so we'll be using this in a moment so I want you to recall that we defined s of theta as y/ R we Define cosecant of theta as R over Y where Y is not allowed to be zero now you'll notice that again s of theta and cose of theta those guys are reciprocals remember your reciprocal identities that we talked about in the last lesson then we have our cosine of theta which is X over R our secant of theta which is rx X is not zero again these two guys are reciprocals then you have your tangent of theta which is YX X is not equal to zero and your cotangent of theta which is X over y y is not equal to Z again these guys are reciprocals now our first topic of today's lesson is going to be the Pythagorean identity so you'll usually see these in your book in this exact manner so sin^2 thet plus cosine 2 th = 1 your tangent s Theta + 1 is equal to your secant s Theta and then 1 + cang S Theta is equal to cose 2 Theta so how do we get the first one so we have our s^ SAR theta plus our cosine 2 Theta is equal to 1 we'll see other ways to do this when we get to the Circle for right now we're just going to divide everything by R 2 okay by R 2 and we know that R 2 / R 2 would be 1 so when we look at this guy we can say this is now going to be x^2 over R 2+ y^2 over R 2 is equal to 1 okay so now using the rules of exponents remember I have a two here and a two here as an exponent and a two here and a two here as an exponent so I can pull that outside of a set of parentheses right using my power rules so I can say this is X over R okay this amount here would be squared then plus you have y over R this guy here would be squared this equals one well by definition we just saw that x over R is cosine of theta okay and then y over R is s of theta if we go back up again s of theta is y r cosine of theta is x r so I could replace this guy right here by saying it's cosine of theta being squared and I could replace this guy let me just kind of move this down I could say this is s OFA being squared and this equals 1 okay so I could apply my exponents and use the commutative property to get into this form here I can write this as cosine squ Theta okay and I can write this as Sin Square Theta okay and then I could just switch the order because again this is just addition so when you add two things together the order doesn't matter so that's how we get to this form right here so let's look at the other ones so we have tangent s Theta + 1 is equal to secant squ Theta so to get this let me just erase everything you're just going to divide everything by x s now okay quite easy so divide this by x s this by X2 this by X2 and we know that this guy is going to be one all right so we know we would have 1 plus you'd have your Y and I'm just going to write this as y overx this guy squared because we know we do that step anyway is equal to R overx this guy squared what is y/x we know that this is the tangent of theta so we go back up again y X is the tangent of theta so you can replace that you can say this is 1 plus the tangent of theta this guy squared is equal to what is rx well RX is the secant of theta okay so we go back and we say this is the secant of theta let me make that a little bit better the secant of theta this guy is being squared and again you can apply your exponents and just say this is the tangent squar Theta okay and this is the secant squar Theta okay and then the way we had it was tangent s thet + 1 is equal to see squ thet so all we'd have to do is just rearrange this and say okay I have the tangent s Theta + 1 = secant 2 thet let's look at the last one which is this guy right here so 1 + cang S Theta equal cosecant s Theta and to get this one I'm just going to divide everything by y squared and again it's going to be the same trick overall so now this is going to be one so I would write this as X over y this guy would be squared plus this would be one this equals you'd have R over y this guy squar okay so X over y we know that's coent right so cotangent of theta X over Y is the cotangent of theta so let's go ahead and say we have cent of theta and I'm just going to put Cent square of theta we know it would end up being that anyway so then + one equals what's R over y so R over Y is cosecant of okay so I'm going to say this is cose squar of okay and again if I go back all I have to do is switch the order of these two so I'll just write this as 1 + Cent 2 Theta is equal to cosecant SAR Theta okay so that's where those three guys come from not very difficult to derive overall but something you might see in your book and get a bit confused about all right so now let's move on and look at the quotient identities these are a lot easier to understand you have S of theta over Co of theta gives you the tangent of theta remember s of theta is y/ r and cosine of theta if I divide it by that it's X over R well what do we do here this is y r times the reciprocal of this which is rx okay so the RS cancel and you get y/x which by definition is the tangent of theta so the other one is the cosine of theta over the S of theta gives you the cent of theta well before you even do this remember the tangent of theta okay which is YX and the cotangent of theta which is X over y those guys are reciprocals well look at what I did here I just took the reciprocal of just what I was working with I had s of theta over cosine of theta now I'll flipped that I have cosine of theta over s of theta so of course I'm going to get the reciprocal of what I got before which is the cotangent of theta okay so now all I would have is X over R / y / R this is X over R * the reciprocal of this which which is R over y okay so these cancel and you get X over Y which again is the cotangent of theta all right so one last thing before we start doing these problems in a lot of cases you need to know the ranges of the six trigonometric functions because you'll go to solve a problem and you might get something that's not possible but you think that it is possible and we'll see that on the first example I'll show you exactly what I mean so when you look at your trig functions you have this range when you think about s of theta and cosine of theta the value needs to be from -1 to 1 including negative 1 and one but in that range okay when you work with tangent of theta and cotangent of theta it could be any real number okay it can be any real number when you work with the secant of theta or the cosecant of theta well then you could really say this is just that the absolute value has to be one or larger right because you can have from negative Infinity up to an including negative 1 and then also from one out to positive Infinity okay so basically the absolute value needs to be one or larger okay it can't be less than that all right so let's look at our first example here we're going to use the identities to find cosine of thet and S of theta given the fact that the tangent of theta is 7/3 and the S of theta is less than zero meaning it's negative so first and formost think about the quadrant you're going to be in remember that all students take calculus so we have our four quadrants one two three I'll put four marks there should be three so three and then four okay so all students take calculus all right we know that t stands for the tangent of theta and also the cotangent of theta the reciprocal so it's positive in quadrants three and also in quadrant one because everything's positive in Quadrant One S of theta is negative okay in quadrants three and four because the S stands for the sign of theta so we know that we're in quadrant 3 because it meets both criteria right it's going to be positive for the tangent of theta and negative for the S of theta so we will be in qu three okay so keep that in mind when you get stuff to know whether it's positive or negative now the other thing is and this is a very common mistake remember we just said that the tangent of theta is equal to the S of theta the S of theta over the cosine of theta so what students do is they say okay well s is negative and I know that tangent is positive so what they'll do is say okay well this is going to be -7 over -3 okay over -3 and they think they're done with the problem this is a very common mistake why doesn't it work out this way remember the range for S of theta and cosine of theta is from -1 to 1 including those two numbers so this is not possible you can't have A7 and you can't have a -3 so throw that out okay so what we're going to do is we're going to take our Pythagorean identity that includes the tangent of theta and we're just going to plug into that so we have the tangent squar of theta okay + 1 equals the secant squar of theta okay so you might be saying why would we plug into that we're trying to find cosine of theta and S of theta remember the secant of theta is the reciprocal of the cosine of theta okay so if we find the secant of theta we can flip it we'll have cosine of theta okay and once we've done that we know that our other identity involves s of theta and cosine of theta we can plug into that and figure out our s of theta so let's just go ahead and knock this out we're given that the tangent of theta is 7/3 so you would want to plug in here okay remember this guy is being squared so when you plug in here make sure that you square this guy it's very important okay so we come down here once you've plugged in the rest of this is really really easy you're just doing some basic algebra 7 s is 49 3^2 is 9 then you're going to add one to that so get a common denominator let's just say this is 9 over not okay this is equal to^ s ofet okay so let's go down a little bit more I'm going to flip this around and put this guy on the left so secant squar of theta is equal to 49 + 9 is going to give me 58 and this is over 9 now to get this in the form of secant of theta okay what I want to do since this is being squared I want to take the square root of each side remember on this side I want to go plus or minus okay the square root of 58 over 9 I'm only going to need one of these remember I'm in quadrant three that's what we did that to begin with I'm in quadrant three in quadrant 3 the secant of theta is going to be negative so through out the positive one you just need the negative okay that's all you want so let's scroll down a little bit and we'll say the secant of theta okay is going to be equal to the negative of this would simplify because the square root of 9 is three so I'll say the square OT of 58 over three okay so we're not looking for the secant of theta of course we want the cosine of theta and I know this is a bit tedious but the cosine of theta is found by flipping this guy so this is equal to the negative of three over the square < TK of 58 now when you have an answer with the square root in the denominator you typically want to rationalize the denominator so let's multiply this by the < TK of 58 over the < of 58 and that would give me an answer of the negative of 3 * the sare < TK of 58 over 58 okay so that's my cosine of theta I also wanted to find s of theta okay so s of theta what is that equal to remember sin squ theta plus cine s Theta is equal to 1 well I can plug in for the cosine squ Theta I'm just going to plug in the non-rational version here so I'm going to put -3 over < TK 58 okay you don't need to rationalize one because this is not the final answer let's copy this because we're going to run out of room go to a fresh sheet here and let me erase this we don't need that and scroll down a little bit more so if I Square this let's just go Ahad and put sin^2 theta plus if I Square this I'm going to get positive of course 9 over 58 and this equals 1 okay so all I got to do is subtract this away from each side of the equation so I'll have sin^2 Theta is equal to 1 - 9 over 58 and of course we just get a common denominator going so I'm going to say this is 58 over 58 and 58 - 9 would be 49 so this is 49 over 58 so this is 49 over 58 okay so to get sine of theta remember you're taking the square root of each side so do you want the positive or the negative square root remember sine of theta is going to be negative in quadrants three and four we're in quadrant three so it's going to be negative so I want the negative square < TK of 49 over 58 okay let me make that reach all the way and of of course this is equal to the negative of you have theun of 49 is 7 so we'd have seven up here over you'd have the Square < t of 58 and to make this a little bit cleaner because again we want to rationalize the denominator s of theta is going to be equal to you have the negative of 7 * theare < TK of 58 over 58 right because I'm going to multiply this by theare < of 58 over thek of 58 and I'll be done so s of theta is equal to of 7 * 58 over 58 all right so let's take a look at another example we want to find s of theta and tangent of theta given the fact that the secant of theta is -2 and the cotangent of theta is greater than zero or you could say positive so the first thing is what quadrant are we going to be in well where is the secant of theta negative it's going to be negative in quadrants 2 and also three okay and then where is coent of theta where is that positive it's positive in quadrants 1 and three okay so three matches both conditions there so we'll say this is in quadrant three okay so this is going to be in quadrant three so when we look at this we know that our s of theta will be negative and our tangent of theta will be positive okay so keep that in mind when you're solving this problem so the first thing is we're given the secant of theta it's -2 so let's plug in to this guy we have the tangent squar Theta + 1 is equal to the secant SAR thet so I'm just going to plug in for this remember it's squared so it would be -2 s -2 both the negative and the two being squared if I did that that would be positive4 so let's just erase this and put four in here okay I'm going to subtract one away from each side of the equation so this is gone and this would be a three so what you have here is tangent squar Theta is equal to 3 if I want to find out what tangent tangent of theta is I'm going to take the square root of each side remember we already said this guy is going to be positive so I don't need the plus or minus I just want the principal square root here so it's just the principal square root of three okay so that's my tangent of theta so now that we know what the tangent of theta is how can we find the S of theta well one thing we can do we can flip this guy since the tangent of theta and the coent of theta those guys are reciprocals I can say my cent of theta is just flipping this guy remember if it's not displayed as a fraction just write it over one and then flip it so you can say this is 1 over the < TK 3 we're not going to rationalize this here because this is not going to be our final answer so what I'm going to do now is I'm going to plug this in I'm going to plug this in for our Pythagorean identity where it says one plus our coent SAR Theta is equal to cose SAR Theta remember the cosecant of theta and the S of theta those guys are reciprocals so I can plug in here find out what this is flip it okay and I'll have S of theta so let's go through and plug in we'll have one plus I'm going to plug this guy in here 1 over < TK of 3 this guy is going to be squared and this equals my cose squar Theta okay let's get some room going and so now if I Square this I'm going to get what I'm going to get one squar is one square of 3 squ is three so this is just 1/3 this is just 1/3 to add this to one I'm going to get a common denominator and say this is 3 over3 so this is basically going to give me 4/3 so 4/3 is equal to and let me write this in another way so cosecant 2 thet is equal to 4/3 okay so to solve for this guy remember the cosecant in quadrant 3 is going to be negative so I'm going to take the square root of this side I'll say the cosecant of theta is equal to over here I want the negative square root of 4/3 so the cosecant of theta is going to be the negative of the square root of four is two and then over the square < TK of three so now let's cut this we'll go to a fresh sheet and we'll just paste this in right here so remember if I want the sign of theta I just flip this guy okay so if I flip this guy I'm going to have the negative of the square < TK of 3 over 2 and that's my answer all right let's look at one that involves the quotient identities so we have find S of theta and tangent of theta given the fact that our cosine of is equal to -2 * 2 3 and S of is negative so cosine of theta is negative here where is cosine of theta negative well it's negative in quadrants 2 and three where is s of theta negative it's negative in quadrants three and four so again we have a situation where we're in quadrant three okay and it doesn't always work out this way we had three problems today where they're all in quadrant three but it's going to vary based on the problem you're given so we know we're in quadrant 3 from here we're going to say that we know our sign would be negative it's given to us and we know our tangent is going to be positive because we're in quadrant three okay so how do we find our s of theta where we can use our Pythagorean identity the one that says cine 2 theta plus sin^2 Theta equal 1 and I know this is normally reversed but I'm writing it this way to give myself some room because I want to plug in here so I'm going to plug in I'm going to plug in for cosine Theta okay so I'm going to say this is -2 * the < TK of 2 over 3 remember this guy is squared it's so important that you don't forget to square things okay so if I Square this guy the negative becomes positive so you can forget about that 2^ 2 is four the square root of 2^ SAR is two okay four * 2 would give me 8 and then down there you have three which is squared which is n so I have 8 nths plus sin^2 th = 1 okay let me erase this we're just going to kind of stick around here because I'm going to need some of that information so I'm going to be going back up so let me just kind of drag this up a little bit so that we stay right here okay so what I'm going to do now is I'm going to subtract 8 nths away from each side so I'm just going to erase it from over here I'm going to write this as 9 9ths so I have a common denominator and then we're going to subtract away that 8 9ths okay 9 9 - 8 9th is 1 nth okay so 1 nth remember s of theta is negative so if I take the square root over here I want s of thet is equal to the negative square < TK of 1 nth now the square otk of 1 nth is the square Ro of one which is one over the square < TK of 9 which is three so it's 1/3 all right so this is negative 1/3 13 all right so let me erase this real quick let me just going to drag this over here and I want to find tangent of theta now given the fact that I know what cosine of theta is and I know what s of theta is remember the tangent of theta the tangent of theta is defined as the S of theta / the cosine of theta okay so what I can do it's going to be hard to fit this in this limited real I have but I would have that the tangent of theta let me just write that over here so the tangent of theta is equal to the S of theta is Nega the ne of 1/3 / by the cosine of theta which is the negative of 2 * < TK 2 three so let me just copy this and we're going to go to another sheet we'll paste this in here real quick so this is equal to what it's equal to the negative of 1/3 times flip this guy the negative of 3 over 2 * < TK 2 of course the negatives are going to cancel forget about those the threes are going to cancel so what I'm left with here is 1 over 2 * < TK of two let's rationalize we're going to multiply by < TK of two over < TK of two and this gives us the < TK of two over > 2 * 2 is 2 2 * 2 is 4 so < TK of two over 4 would be your tangent of theta in this lesson we want to talk about trigonometric functions of acute angles and also we'll look at co-function identities so previously we used angles in standard position to define the six trigonometric functions so what we're going to do in this lesson is we're going to Define our six trigonometric functions as ratios of the lengths of the side of right triangles so we're going to kick things off by looking at an acute angle and standard position this is something we've already seen so we've seen this diagram before so this acute angle here this angle Theta is in standard position okay and we've just picked a point on the terminal side we're going to call it X comma y That's our point but again this has actual coordinates in this case it would be 8 comma 8 okay but we're just going to refer to it as X comma y now generically we know that this distance from here to here can be represented by just saying the letter Y okay because this right here is y and down here this coordinate is going to be zero right because we're on the x- axis so I can take Y and subtract away zero and I would get y so that's my distance for this vertical leg here of this right triangle that we formed then this horizontal distance from here to here remember this is a coordinate of X okay this is a coordinate of X and this is is a coordinate of zero so really I could do x - 0 and just say that the distance from there to there we could say this is X okay and then from here to here let me highlight this from here to here we know that is the hypotenuse right so we call this R we call this R and we'll get more into that as we progress through the course in terms of why we call it R right now we just call it R Okay so we've already defined the six trigonometric functions using these letters right so we've said that the S of theta is y / R we've said that the cosine of theta is X over R we said that the tangent of theta is y/x and then using the reciprocal identities we can get the three remaining ones so the cosecant the cosecant of theta is going to be the reciprocal of this which is R over Y and keep in mind if your denominator ends up being zero okay it's undefined so I'm not going to write that but y can't be zero there then here you would have the secant of theta okay and this would be R overx and then lastly you'd have your cotangent of theta which is going to be X over y okay so let's explore a different way to look at this now so instead of calling this guy y okay we're going to refer to it as opposite because it's opposite of this angle here okay now what you want to pay attention to is where this guy is because in this case this is the opposite side but if I Chang the angle up if I was measuring this angle here well then the opposite side would now be here so you got to pay close attention to this I'll show you some examples where we switch things up and the answers are going to change so you got to keep in mind where the opposite side is so this guy is opposite of your angle Theta and then this guy right here is adjacent to it okay and then of course the hypotenuse stays the same the hypotenuse is always opposite of your right angle your 90° angle okay so what we can do now is say that okay since this was Y and this was X before and this was R well if we have S of theta okay we said before this was y over r y over R now we're going to say it's what well Y is the opposite so I'm going to put op p p for opposite and this again is the measure of that so the length that you're given and then you have your R your hypotenuse so I'm going to put H YP for hypotenuse so opposite over hypotenuse okay and I'll give you a pneumonic device for this in just a moment the SOA so then you have your cosine of theta okay this is X over r or in terms of this it's the adjacent okay the adjacent over the hypotenuse okay and then you have your tangent of theta which is going to be what it's Y overx which is your opposite okay your opposite over your adjacent over your adjacent and then of course you can flip these guys and you can get your cosecant of theta so your cosecant of theta and I'm going to run out of rum here so let me actually copy this real quick and let's go to a fresh sheet and just paste this in so we have more R so then I can say that cosecant of theta which is the reciprocal of this is going to be the hypotenuse over the opposite and then your secant of theta which is the reciprocal of this would be your hypotenuse over your adjacent and then lastly your coent of theta is going to be the adjacent over your opposite okay so how do we remember this well recall when you work with something like the order of operations you have your PEMDAS right your please excuse my dear Aunt Sally well you have a similar pneumonic device here it's the sooa okay so the so TOA okay so the S stands for sign the C stands for cosine and the T stands for tangent okay so this is pretty much what everybody uses to remember these guys so S for S it's the opposite over the hypotenuse so opposite over hypotenuse C for cosine the adjacent over the hypotenuse so adjacent over hypotenuse the T for tangent the opposite over adjacent so opposite over adjacent so this is something you want to write down and think about constantly until you have it memorized it's going to save you a lot of time and then of course if you know your reciprocal identities it's very easy to get the cosecant the secant and the coent by just flipping guys all right so let's go ahead and take a look at an example so with this guy right here we're just going to find the six trigonometric functions so let me just write them out we want s of theta cosine of theta and let's just do tangent of theta for right now we know we can get the other three by just flipping these guys okay so what is s of theta again let me go back to the so okay so s of theta it's y over R but let's think about it in terms of this we're going to say opposite so opposite over hypotenuse okay cosine of theta we're going to do adjacent adjacent over hypotenuse and tangent of theta we're going to do the opposite the opposite over the adjacent okay so what's the opposite here what's the hypotenuse what's the adjacent so we know that across from the right angle okay which is the longest side of a right triangle is your hypotenuse okay so this is the hypotenuse here okay and if you're confused about these things when you first start go ahead and take the time to label things figure out where everything is so this guy right here is adjacent this is your angle here Theta okay and I made that terribly so this is your angle here Theta this guy is adjacent to this okay or next to it so this is going to be your adjacent your adjacent and then opposite of this you're going to have this guy right here so this is your opposite once you figured this out you're just plugging things in and you're done okay that's all you need to do so what is the opposite it's two what is the hypotenuse it's 2 * < TK 5 well of course you want to rationalize the denominator there let me just kind of scooch this down a little bit so we have some room so I'm going to multiply this by square < TK of 5 over Square < TK of 5 the twos are going to cancel so I'm going to have the square < TK of 5 over square of 5 * of 5 is 5 okay so that's my S of theta my cosine of theta is the adjacent okay which is four over the hyp hypotenuse which is 2 * < TK 5 the four and the two would cancel to two up there so again you're just going to multiply by the < TK of 5 over the < TK of 5 and what does that give me well 2 * 5 2 * of 5 over squ of 5 * square of 5 is five so the last one down here we have the opposite over the adjacent so the opposite is two the adjacent is four so this is one2 now we can flip these guys again and figure out our other ones so let me erase this all right so for the cosecant of theta remember that's the reciprocal of this guy the S of theta okay so I'm going to use the one before we rationalize the denominator to make it easy so 2 * < TK 5 over two I just took this guy right here okay and I flipped it it's all I'm doing you can flip this guy too but then you have to rationalize the denominator again it's just double work okay so we've already done that so cancel this with this you just get the square root of five then the next one we would have the secant of theta okay and that would give me what the secant of theta would be the reciprocal of your cosine of theta so again take the one before we rationalize the denominator and I'm just going to say this is the < TK of 5/ 2 and nothing really else you can do with that so lastly I'm going to flip the tangent of theta okay to get the coent of theta and so if I flip 1/2 I just get two over one or two all right so let's look at another example we're going to have this the same right triangle but now we're going to change the angle so before our Theta was here okay so we were talking about in reference to that Angle now the angle has changed okay so this is our Angle now and so this guy right here now is the adjacent okay so this is the adjacent and this is still the hypotenuse that is not going to change because it's always across from the 90° angle or the right angle and now if you look opposite of this this is going to be this side over here okay so we've got our adjacent our hypotenuse and our opposite so our s of theta again so s so s is opposite over hypotenuse so what's the opposite it's four what's the hypotenuse it's 2 * the < TK of 5 you do need to rationalize the denominator here so let me cancel this with this and get a two here okay so two over the < TK of 5 so this multiplying by the < TK of 5 over the < TK of 5 and you're going to get 2 * the < TK of 5 over 5 now I want you to notice something s of theta here is the same as cosine of theta here okay we're going to talk more about this in a moment but I just want you to notice that all right so I can flip this guy to get my cosecant of theta okay let's just do that now and I'm going to start with it in this form and I'm going to say this is the square < T of 5 over two okay let's erase this so the next one I want and I'm just going to tighten this down so we kind of make that a little bit neater so let's move this down a little bit okay so the next what I'm going to do the next one I'm going to do is going to be my cosine of theta okay my cosine of theta again this is the C part of this so cosine is adjacent over hypotenuse so where is the adjacent that's two what's the hypotenuse it's 2 * thek 5 before I rationalize the denominator let's cancel this with this and put a one up here so this is is times thek of 5 over theun 5 and this gives me the < TK 5 over 5 okay now I want you to notice that the cosine of theta here okay with this angle up there is the same as the S of theta here okay with the angle here again we'll talk more about this in a moment so let me flip this guy right here before we rationalize the denominator the secant of theta is going to be the square < TK of 5 okay over one but just square root of five so let's just kind of tighten this down a little bit so we don't have stuff all over the place so let's move this down and then lastly we're going to have our tangent of theta and then our coent of theta okay so what's the tangent of theta that's the TOA part right so to a opposite over adjacent so the opposite here and I erase this I apologize this is the opposite here opposite is four adjacent is two so four over two or two okay and if I flip this guy remember if you're flipping a nonf fraction number or a whole number on Integer if you want to think about it that way you just write the number over one and then flip it right so it would be 1/2 so this is 1/2 and again if I look at the tangent of theta here it's going to match the coent of theta here so the tangent of theta here is a half and the cotangent of theta here is two those guys are flipped so now the tangent of theta is two and the cotangent of theta is 1 12 so this is going to lead us to an important result and we're going to talk now about something called co- functions so to talk about these co-function ident at least the first time you see them it can be a little bit challenging to understand so just stay with me for a moment I'm going to show you when you get to the examples they're very very easy to figure out okay so even if you don't fully understand the concept you can work the problems very very easily so the first thing is that we noticed that the S of theta from the first example was equal to the cosine of theta from the second example so I have the same triangle here I've just marked it with different letters there's no actual numbers here okay so we have this lowercase a this lowercase b and the lowercase C that's representing the lengths of the sides here okay so if we wanted let's say we had this measure of angle a so this is angle a this is what we used in the first scenario if we wanted sign of a here okay if we wanted s of a well what is this going to be well it's opposite over hypotenuse your opposite here is the lowercase a your hypotenuse here is lower case C okay and then if you wanted the cosine okay of B now okay of B so I'm changing the angle up I'm here now you're going to notice it's going to be the same thing it's going to be a over C and again lowercase a lowercase C because now this is your adjacent okay and this is still your hypotenuse remember cosine is adjacent over hypotenuse and you'll notice that again your s of theta is theun of 5 over five this is the first example where this is our angle in the second example the cosine of theta here's our angle here is > of 5 over5 so this relationship is always going to be true for the two acute angles of a right triangle okay so since the sum of the three angles in any triangle is 180° and our angle C here remember this is our right angle it's exactly 90° so that means because the sum of all three angles is 180° the sum of the measures of angles b and a is going to be 90° right these are complimentary angles we've talked about this before if they sum to ° two angles they're complimentary angles so because these two angles A and B are compliment and our s of a is equal to our cosine of B we say the function s and cosine are co- functions we also say that tangent and coent are co- functions and then we can say secant and cosecant are co- functions as well and if you want to go back if you look at the tangent of theta here it's a half and if you look at the coent of theta here it's a half okay if you look at the cosecant of theta here is < TK of 5 and if you look at the secant of theta here it's Square < TK of 5 so how can we use this for practice problems these are your co-function identities you just need to memorize them it's something you're just going to plug into so your s of your angle a is equal to your cosine of 90° minus your angle a so for example if it was s of 50° well you would just say this is equal to cosine of let me make this a little bit better so cosine of what well you would say 90° minus 50° that's 40° okay and you can punch this up on a calculator and you see that those are the same same deal with cosine of A and S of 90° minus a secant of a and cosecant of 90° minus a so on and so forth you can copy these down but basically if we look at a problem like this the cosine of 48° and we want to write it in terms of its co- function well again you just say well this is going to be equal to the sign of you take 90° 90° minus the measure here which is 48° okay and what's 90° - 48° well it's going to be 42° okay so we can say cine of 48° is equal to S of 42 Dees okay so this number here and this number here those guys because these are again complimentary angles they have to add up to 90° so that's what you want to make sure of and then make sure your co- functions there so cosine and S tangent coent secant and cosecant so on and so forth let's do another one so we have the tangent of 11° so again I would do coent okay of 90° minus 11° 90 - 11 is going to be 79 so this would be cotangent okay of 79 Dees there so the tangent of 11° is equal to the cotangent of 79° again you want to check to make sure that this Plus plus this is going to give you 90° okay so let's look at an example that involves solving equations using co-function identities for this guy right here we're going to find one solution for the equation okay and we're going to assume all the angles involved are acute angles so you'll see a problem like this or a few problems like this in this section and solving it is very easy remember I told you that this plus this okay is going to equal 90 degrees so all I need to do is take the inside Parts the 2 * Theta + 20° and add this to the 3 * Theta + 5° and set this equal to 90° okay I'm going to solve for Theta okay my unknown angle measure so what I'm going to do here is I'm going to combine like terms so 20° + 5° is 25° and 2 * Theta + 3 * thet would be 5 * Theta so 5 * Theta + 25° is equal to 90° okay so just like I'm solving an equation just like if I had X there I want to isolate this guy so I want this by itself so how do I do that I'm going to first subtract 25° away from each side of the equation so this is going to be gone I'll have 5 * Theta is equal to what's 90° minus 25° it's going to be 65° okay so to wrap this up what I'm going to do is divide both sides by five so that I can isolate Theta and I'll say that Theta is equal to 65 ID 5 is 13 and of course you have degrees there for the units so I would say Theta is equal to 13° now if you want to check this all you would do is plug in so 2 * 13° is 26° 26° + 20° is 46° so s of 46° is equal to cosine of what 3 * 13° is 39° 39 degrees plus 5 degrees is 44 degrees so is this true yes it is right s of 46 degrees is equal to cosine of 44 degrees again this plus this has got to be 90° and of course you have your s and cosine in this lesson we want to talk about trigonometric functions of non-acute angles and also we'll learn about reference angles all right so let's start off with a concept of a reference angle this is something that's very important in trigonometry so first from the definition we're going to say that every non- quadrantal angle in standard position will have a positive acute angle known as its reference angle so when we talk about the reference angle for an angle Theta it's written as Theta Prime so you'll see here we have this Theta Prime okay so this is your reference angle notation now when we talk about the reference angle for an angle Theta then basically this is the positive acute angle that is made by the terminal side of our angle Theta in the x axis so I'm going to go through this in each quadrant in every case I'm going to be thinking about an angle angle Theta that is between 0° and 360° if you get something that's outside of that you're just going to find the co-terminal angle that's between 0° and 360° and you'll apply the little formulas that we're going to look at here so what I'm going to start with is an angle in quadrant one so if I think about the positive acute angle that is made from the terminal side of this angle Theta in the xaxis well easily I can see that Theta here is going to exactly be equal to Theta Prime so if you're in quadrant 1 then you can say that Theta is equal to Theta Prime so your angle Theta is going to be its own reference angle if you're in quadrant 2 you have to do a little bit more thinking so now I'm thinking about this angle right here okay so this is my Theta Prime let me use a different color that's going to show up here so this is from the terminal side of this angle Theta to the xaxis so I'm always thinking about going to the x-axis so that angle has this measure of theta Prime so how do we get this what I would think about is the fact that if I took this angle measure here which is a straight angle the measure is 180° so I'm going to say in quadrant 2 I'm going to start with that straight angle which is 180° and I'm going to say that I need to subtract away the measure of the angle Theta right so I'm going to take this part away right that's what I'm going to subtract away and that's going to leave me with my Theta Prime that I'm looking for let me get rid of all this so that's more clear so I'm going to subtract away my angle Theta whatever it is and let me slide this down here and I'll say that Theta prime or your reference angle is going to be equal to 180° minus Theta again this is in quadrant 2 all right if we're in quadrant 3 well now we have to think about this guy right here so this is your angle here and I only want this part right here so coming from the x axis going down to that terminal side that's going to be my Theta Prime here so in this particular case in quadrant 3 we are going to start with our angle Theta so let me say Theta Prime is equal to our angle Theta whatever it is and we are going to subtract away this straight angle here so this part right here that's what we want to take out and that's going to leave us with this amount of rotation right there so I'm going to take away that straight angle which is 180° so in quadrant 3 your Theta prime or your reference angle is your Theta minus your 180° all right now let's take a look at an angle that's going to be in quadrant four so if we think about one full rotation so this guy right here that's going to be 360° so if I'm in Quadrant 4 I would start with 360° and then to get this angle Theta Prime here which is going to be this part right there well all I need to do is subtract out this angle here Theta right I'll be left with my Theta Prime so let me actually slide this down just a little bit and I'll say that Theta Prime is equal to 3 360° minus my angle Theta okay so we have summarized everything and basically when you get to the examples once you've gone through how it works you can just use this little table after you work enough examples of this or get deeper into trigonometry this is something you're just going to have memorized and you won't have to consult a table or draw a sketch or anything like that so I'm going to start out with 330° and I want to find the reference angle so I know that this is in Quad 4 so I look through my little cheat sheet here and in Quadrant 4 again my Theta Prime is equal to 360° minus my angle Theta in this case it's 330° and so this is just going to be 30° for the reference angle and again you can always make a sketch if this is not clear for you so this would be a 330° angle and a lot of you can just think about that and see that I would need 30 more degrees to get back to the x-axis so so that's your reference angle there but again you can also do it where you think about one complete rotation is going to be 360° so 360° and then you subtract away the part that you're not going to want to measure so that's your angle Theta there and in this case that's your 330° and that leaves you with your reference angle which is this part right here which is 30° right so let me put Theta Prime is equal to this okay for the next one we have - 250° so as I mentioned earlier if you're not between 0° and 360° find the co-terminal angle now you could do it this way and a lot of people do but I think finding the co-terminal angle first is just worthwhile because it's just going to make the process easier you can just use this so for - 250° I would start by just adding 360° to find my co-terminal angle and that's going to give me 110° so now I'm thinking about a Theta of 110° that's going to be in quadrant 2 so in quadrant 2 your Theta Prime is equal to 180° minus this 110° and that's going to give me 70° so again if you want to look at a little sketch we think about an angle with a measure of - 250° we're going to show that with a clockwise rotation so I'm going to start here and go this way so that guy would be - 250° but again you just work with a co-terminal angle right so that's 110° angle same initial side same terminal side and so let's get rid of this and we'll just work with this guy and again you can think about this logically if I start here and I rotate 110° counterclockwise well then I need to go another 70° to get to the xaxis right 110° plus 70° is going to be 180° or form that straight angle again the other way you could think about it is your Theta Prime here is going to be your straight angle which is 180° minus the part of this rotation that you don't want so that's this part right here which is your 110 degree angle and so that's 70° okay so that's your Theta prime or your reference angle all right let's look at one more so now we have 560° and again I would find the co-terminal angle that's between 0° and 360° so 560° let's say - 360° and that's going to give me exactly 200 degrees so this angle right here is going to be in quadrant three so in quadrant 3 we would say Theta Prime is going to be Theta which is 200° minus 180° so my reference angle Theta Prime is 20° so coming to our little sketch again if you wanted to think about it as a 560° angle that's one complete rotation which is 360° and then another 200° to get to that point right there but I just like to work with a co-terminal angle I think it's much easier in this particular case you think about the entire rotation here so let me put Theta Prime is equal to that's going to be 200° and then you want to subtract out the part that you don't want again I'm thinking about from here the x-axis to here so that's going to be me taking out this straight angle here which is going to be 180° so that gives me my reference angle or this rotation from here to here of 20° all right so now what we're going to do is talk about trigonometric function values of special angles so we have these 30° 45° and 60° in terms of the angles that we're going to want to memorize the sign cosine and tangent values for and then also your reciprocal functions so your cosecant your secant and your cotangent so I'm going to start off with a triangle that you should know from geometry if you haven't taken geometry before it's not that big of a deal so this triangle is known as a 30 de 60° 90° triangle so this is a special right triangle and so the side that's opposite of the 30° angle has length one the side that is adjacent to this 30° angle has length square of 3 and then the hypotenuse here has length two so if I wanted to figure out the sign of 30° and then the cosine of 30° and then the tangent of 30° let's go ahead and set those up so for the S of 30° let me make this s a little bit better again it's opposite over hypot so the opposite side here has length one and the hypotenuse has length two so this is 1 12 then for the coine of 30° it's going to be adjacent which is square < TK of 3 over hypotenuse which is 2 now for the tangent of 30° it's going to be opposite which is one over your adjacent which is square < TK of 3 and of course we would rationalize the denominator there so multiply by theun of 3 over the > of 3 so this is the of 3 over 3 so Square < TK of 3 over 3 now you could find your reciprocal functions if you wanted to so you could take the reciprocal of this to get your cosecant of 30° you could take the reciprocal of this to get your secant of 30° and you can take the reciprocal of this to get your cotangent of 30° I'm not going to waste time doing that I'm going to give you a table where everything's presented I'm just showing you where this stuff is coming from I'm going to do the 60° angle as well so let me go s of 60° and we know from our previous lesson that that should be equal to the cosine of 30° so this should be the < TK of 3 over 2 and if we check that well if I look at this 60° angle here well the opposite is now going to be theun of 3 and the hypotenuse is still two so 3 over two so that checks out then the cosine of 60° again we know that should be the S of 30° so this should be 1/2 so if I look at my cosine of 60° it's the adjacent which is 1 over over the hypotenuse which is two so I do get 1/2 then for my tangent of 60° this is going to be equal to what well again I would just think about my opposite which is square < TK 3 over my adjacent which is 1 soqu of 3 over 1 is just going to be the square < TK of three so the other angle that we didn't do was the 45° angle so this is an isoceles triangle it's the 45° 45° 90° that's a special right triangle from geometry again if you've never seen this before it's not that big of a deal I'm just showing you where the things come from so if I want to write out the S of 45° the cosine of 45° and then the tangent of 45° so for these two it's going to be the same because let's say I start with this guy right here well it's going to be the same as if I worked with this guy right here it doesn't matter they're both 45 degree angles so if I take this one right here and I say opposite over hypotenuse it's going to be one over theare < TK of two and if I rationalize the denominator multiply by the > two over theun of two so this is the of two over two so this is theun of 2 over two now let's say I start here and I say what's the cosine of 45° well now it's going to be adjacent which is 1 over your hypotenuse which isare of two well I just did that so if I rationalize the denominator it's > two over two and it wouldn't have mattered if I would to use this angle here I would get the same result okay now the tangent of 45 degrees you're going to think about your opposite so let's use this one so the opposite is one and then over your adjacent which is one and again if you pick this one well now the opposite is one and your adjacent is one so it does not matter it's one over one which is one all right so what I've done here is actually summarized all the information into a nice little table so we already know where this information comes from we looked at the S cosine and tangent for 30° 45° and 60° for this table over here where we have the cotangent the secant and the cosecant for 30° for 45° and 60° it's just found using the reciprocal identities this is something you definitely want to commit to memory these angles are going to come up over and over and over again so it's just going to make your life a lot easier now there's something you can search for called the hand trick if you want an easy way of memorizing this all right now let's talk about how to find the trigonometric function values when you have a non-acute angle but your reference angle is still going to be 30° 45° or 60° so we're going to be a to use that table that we just looked at so I'm going to start with this special right triangle that we looked at earlier I've just flipped it around such that you can draw this if you want to it'll make it easier to understand when I put this on the coordinate plane so this is your 30° angle I just want you to remember that the side opposite of this has length one and then your side adjacent to this is going to have length square of three and then your hypotenuse that length is two so what I did here was slap this on the coordinate plane so you see that this point right here is the origin and you see see that this point right here we're going to find the x value and the Y value and we'll talk about that in a second but again if you wanted to find the sign of this angle which is 150° well basically you would just find that point on the terminal side and remember you have S of 150° is equal to Y over R right s of theta is y over R let me write out cosine also so cosine of 150° is X over R and then we have tangent of 150° this is y X so we'll look at this in a moment and we'll compare it now I want you to notice that you have this reference angle here which is 30° right again if you're in quadrant 2 you have this entire guy here which is a straight angle that's 180 degrees and you subtract away the part that you don't want to measure so that's 150° so 180° minus 150° is your 30° so that's the amount of the rotation left to get from the terminal side to your X AIS okay so I want you to think about what would the X x value be what would the Y value be what would the r value be so the R value is simple it's just going to be two so R is equal to two for your y value again opposite of the 30° angle this side has length one so if I'm starting here and going here why I've traveled one unit up and again you see the one right there so the Y value here is going to be one when we think about the x value notice that we're going left right if we go to the left of the origin on the coordinate plane the X values are negative so this length right here we know that side adjacent to 30° is length square of 3 but again because we're on the coordinate plane this is the negative of the S of three for this x coordinate right there so let's put this right here so the negative of theare < TK of three like that okay so let me actually write this all out so the x value here is the negative of theun of 3 the Y value here is one and the R value here is two we're going to think about these values here and then we're going to compare them to these trig function values for this reference angle which again is 30° so we have y over R so Y is 1 and R is 2 then here we have X over R so X is the negative of the square < TK of 3 over R which is 2 and then here we have y overx so Y is 1 and X is the negative of the of 3 so let me actually rationalize the denominator here so times the 3 over the 3 so this would be the negative of the < 3 over 3 now I want you to think about these function values that we talked about with a 30° angle the S of theta is a half the cosine of theta is 32 and the tangent of theta is 3 3 so coming back here you'll see that everything matches in terms of absolute value okay so this leads to an important result what's going to happen is if you want to find something in quadrants 2 three or four you just first find the reference angle okay so if it's a reference angle of 30° 45° or 60° you can use your little table we'll talk about other values later on but for right now that's what we're going to work with so that's your first step so you would figure out in this case what is the sign of 30° then you would apply the appropriate sign based on your sign rules right so we know in quadrants 1 and two sign is positive so here the sign of 150° is a half because again the sign of 30 Dees is a half and it's going to be positive in quadrant 2 so you don't have to make a sign change there but in quadrant 2 cosine is negative right so that's why the cosine 30° which was the of 3 over2 is not going to exactly match the cosine of 150° you have to apply that negative there to account for the fact that cosine is negative in quadrant 2 because in this case your X values are to the left of the origin and so they're going to be negative then when we think about the tangent of 150° again in quadrant 2 you're going to be negative for your tangent value so you have to take your tangent of 30° which is of 3 over3 and then slap the negative on there to get your answer here so let's go ahead and take a look at some examples all right so we're going to start out with cosine of- 600° so again I'm going to first find the co-terminal angle between 0° and 360° just to make my life easier so let me put - 600° I'm going to need to add 720° to that so that's going to give me 120° so this is what I'm going to work with so the first thing is if I have 120 degree angle my Theta Prime which is my reference angle is going to be again I'm in quadrant 2 so 180° minus 120° which is 60° so this is my reference angle let me write Theta Prime is equal to 60° and we can erase this scratch work cine of -600 De is equal to the cosine of this 120° now here's where you have to apply a little bit of thinking and you want to say well this is in quadrant 2 is cosine positive or negative it's negative in quadrants 2 and three so I'm going to put equals here and then I'm going to put the negative of the cosine of this reference angle which is 60 degrees remember when you talk about things in quadrant one so the 30° the 45 degrees or the 60° everything is positive right all students take calculus in quadrant one they're all positive so that's why you have to apply this negative here to this result so the cosine of 60 de if you go back to your table is going to be 1/2 so what we'll do is say this is equal to the negative of again the cosine of 60° is a half so your answer here for the cosine of 600° is going to be2 okay let's look at one more so here we have the secant of 495 de so again I would first find my co-terminal angle so 495 de let's subtract away 360° and that's going to give me 135° so when I think about 135° angle again I'm going to be in quadrant 2 in my reference angle my Theta Prime is 180° minus 135° which is 45° okay so that's what I'm working with there so let me get rid of this and I'll just state that the secant of this 495 Dee angle is equal to the secant of this co-terminal angle which is 135° so this is equal to if I'm in quadrant 2 secant is going to be negative right remember see is the reciprocal of cosine they're both going to be negative in quadrants 2 and three so I'm going to put this negative out in front to account for that and then I'm just going to think about the secant of 45° so coming up to my table if I want the secant of 45° it's going to be the square root of two so I'll say this is equal to don't forget your negative so the negative of the square root of two in this lesson we want to talk about finding trigonometric function values using a calculator we'll also discuss using inverse trigonometric functions to find angles Okay so so the first topic here is going to be very very simple I just want to make sure everybody's on the same page with using their calculator if you're using an online calculator you can pretty much just skip this part of the lesson if you see something like the cosine of 60° we already know from this special right triangle where we have 30° 60° and 90° if we want the cosine of 60° well it's adjacent which is 1 over the hypotenuse which is two right as a decimal this is 0.5 so what I want to everybody to do is pull out their calculator that they're going to use for this course and it should have a s key a cosine key and a tangent key you want to figure out what mode you're in first so most calculators have a radian mode we're going to talk about radians later on and a degree mode you want to select degree mode so you should be able to just hit the cosine key on your calculator so it would look like that and then basically you would have some parentheses that would open up and you just want to type in 60 don't worry about trying to find the degree symbol if you're in degree mode then it knows you're working with 60° so you hit the enter key after that so you hit your enter your enter key after that and you should get a result of 0.5 if you're not getting 0.5 then either you typed in the 60 wrong or you're in radiant mode so it's very important if you're not getting the right answer to investigate and figure out what's going on a common thing that books will tell you to do is always before you start working hit s of 90° in your calculator so that's going to look like this you're going to hit the sign key and then you're going to put 90 inside of the parentheses and again you're just going to hit enter and then that should produce a value of one if you get a value that's something like 894 you know you're in radiant mode and you need to switch it all right so what we're going to do is just run through a few basic problems again just making sure you're on the same page with me so we have the sign of 170 degrees so I would just hit my sign key I would just hit my sign key and then in the parentheses type in 170 and then you're going to hit your Enter key and this should give you let's say approximately 0. 1736 48 1777 that's where my calculator cuts off usually the book will tell you where they want you to round to so let's just say we want to round to the nearest 10,000th so that's four decimal places so 1 2 3 4 remember this is the 10th The Hundreds the thousands and this is the 10,000s so if I think about it the digit that follows is a four so the six would be unchanged and you can just get rid of all these digits that follow so the sign of 170° is going to be approximately 01736 all right let's take a look at another example so here we have the secant of - 62° now there's a few things here first off you might not have a secant key on your calculator the calcul calator I use does not have a secant key so you would say the secant of some angle Theta again using your reciprocal identity you could say this is one over the cosine of theta since most calculators do have a cosine key so you can convert this over and say this is one over the cosine of this - 620 de now another thing you might be concerned with you've seen throughout the course if I got a negative angle or an angle whose measure was greater than 360° I would always find the co-terminal angle between 0° and 360° now you could do that but if you're using a calculator you don't have to this will automatically calculate the value for you so you're just going to type one then the divided by key hit your cosine key when the parentheses come up just type in - 620 again don't worry about the degree symbol hit enter and we're going to get approximately let me just make a little border here if I pick up my calculator I'll say it's NE 5.75 and then I'm going to have 877 0483 now with this guy again if you wanted to round to the nearest 10,000th that's pretty common for most books Let's see we have the tenths the hundredths the thousands and then the 10,000 so this is the digit that follows so the Seven Falls in the category of five or greater so I would add one to this seven and then I would cut all these off so this will turn into an eight so your answer here for the SE of 620 de is going to be approximately - 5.75 88 okay let's look at cent of 670 de so again you might not have a cotangent key but more than likely you have a tangent key so again from the reciprocal identities the cotangent of theta is equal to 1 over the tangent of theta so really you could just say this is equal to one over the tangent of your 670 degrees like this and again you can just follow those steps with your calculator hit one hit divided by hit your tangent key inside the parentheses type in 670 and then close the parentheses and hit enter and so this is going to be approximately let's say negative I'm going to go 0 83 9099 and then we'll have 6 312 and again if you wanted to round to the nearest 10,000th again forget about the fact that it's negative I know that throws off a lot of people you've got 1 2 3 4 decimal places so this guy right here the digit that follows is a nine so add one to this zero that's going to be one and then you can cut all these digits off so rounding this to the nearest 10,000th you're going to have 0.8 391 all right let's spend a little bit of time now introducing the inverse trigonometric functions so these are going to be a little bit confusing for right now when we get further into the course we're going to spend a lot of time developing the whole thought process behind how and why these guys are going to work for right now I just wanted to think about something simple let's say you had the sign of 30° and again we have our special right triangle here so we can immediately see that the opposite is one and the hypotenuse is two so we know this is 1/2 and again you can type that into your calculator and verify you would get 0.5 or half as a decimal now what if you typed into your calculator s inverse so it's going to look like that or you might see Ark sign that's another way to denote it but basically if you took this part right here or your answer from this so let's say I put that in the parenthesis this is going to give you this angle measure back so in other words it's undoing what you did here here you took s of your angle which is 30° and you want to find the opposite over the hypotenuse so that gave me a half here you have the inverse sign and you're taking your opposite over your hypotenuse and it's going to kick your angle measure which is 30° back out to you so in other words you could say that you have S of theta is the opposite over the hypotenuse and in this case you have the S inverse which is going to be you plugging in the opposite over the hypotenuse and you're going to get your angle Theta back but there's a problem here okay and it's kind of a big problem but again we'll talk about this more later on in the course if you typed in let's say s of 390° so this is an angle that's co-terminal with 30° it's going to be in the first quadrant and it's going to have a reference angle of 30° well you would get 0.5 or 1/2 now I just saw that if I did Sin inverse of 1/2 I get 30° so I'm not getting my 390° back that I'm looking for so what's going on well the problem is that sign is not a one: one function we talked about this earlier in our pre-calculus course where for a one:1 function to exist you have to have for each x value a given yvalue and for each yvalue a given x value so for each X there's one y and for each y there's one X well that's not going to be true with your sign function so what happens is you have to restrict The Domain in order to make it a onetoone function and that leads to the fact that if I plugged in 1/2 here I'm only going to be able to get something in quadrant 1 so that's 30° if I needed to find something else well then I'm going to use the concept of a reference angle and I'm going to think about where sign is positive so in other words I know that sign is going to be positive in quadrants 1 and two so really anywhere in quadrants one or two where I would have a 30° reference angle well basically my sign value is going to be a half if I did s of let's say 150° that has a 30° reference angle so that's going to be a half if I added 360° to that I would have 510° and if I did the sign of 510° I would also get a half so on and so forth so if you ever get a question and we'll see them later on in the course where it says something like the S of theta is equal to a half well there's an infinite number of angle measures there where if you take the sign value of it you're going to get a half so generally speaking when you see these questions especially in the first part of trigonometry they're going to tell you that Theta is going to be between 0 de and 90 de so you're going to be in that first quadrant where 0 degrees and 90 degrees will be included so let's look at a little example here so we have S of theta is equal to 4 fths and here Theta is restricted like I was just talking about to be greater than or equal to 0 de and less than or equal to 90° okay so for this one again we would just use our inverse sign function so s inverse of this guy right here so this four FS so it's like I'm plugging in the opposite over the hypotenuse and I'm looking for my Theta and this is going to come from Quadrant One so I'm going to be good to go in terms of this restriction here so let's punch this into the the calculator so hit your inverse sign key hit four and then divide it by five close your parenthesis and let me move this over and I'll say Theta is equal to this and I'll say this is approximately let's write out what we have so 53131 and then 0 2 35 so let's again round this to the nearest 10,000th so you have 10ths hundreds thousands this is the 10,000 so the digit that follows is a zero so we can just cut all these off and this is in terms of degrees so our Theta here is going to be about or you can say approximately 5313 01 de and again if you didn't have this restriction here you could really say that Theta could be any angle in either quadrant 1 or quadrant 2 where sign is positive where the reference angle is equal to this guy right here right so there's an infinite number of solutions if you take this guy away this restriction all right so similarly we have the cosine of theta this is equal to 27 so here again we're going to say that Theta is greater than or equal to 0° and less than or equal to 90° so since this is positive if we use our inverse cosine function we're going to get something in quadrant one so we'll be okay so I'm going to say that I want my Theta to be equal to the inverse cosine and some people write that as Arc cosine like this and basically you're just going to plug this in here so this 27ths and this is going to spit out my angle my Theta that will be in quadrant one so let's hit our cosine inverse button and then inside the parentheses type 2 / 7 close your parenthesis and hit enter and this is approximately let's go ahead and say I'll just write out everything it gives me so 73. 39 84504 that's where my calculator stops if we want to go to the nearest 10,000th we have the tenths the 100 the thousands the 10,000s well this guy's a five here so let me add one to this and we'll make this into a five and then you can cut these off and this is in terms of degrees so we would say that our Theta here is approximately 73.3 n85 de now again if you did not have this restriction here then Theta could be any angle that's in either quadrant 1 or Quadrant 4 that's where cosine is positive that has this reference angle in this lesson we want to talk about solving right triangles and also look at some word problems that involve angles of elevation and also angles of depression all right so let's just jump in and look at our first example when we talk about solving a right triangle we're just trying to find all the missing angles and all the missing sides so in this particular case we have our triangle here we're given one side that has a value of 14 and then we're given one angle with a value of 37° so I know the measure of angle a is 37° I know the measure of angle C because it's a right right angle is 90° you can write that in there and so if you wanted to figure out what the measure of this angle is well you know that the sum of the three angles is 180° so you could say that 180° minus your 37° plus 90° this guy right here would be the measure of your angle B so let's go ahead and crank that out real quick so 37° + 90° is 127° and if I subtract 180° - 127° I get 53° so the measure of this angle here is going to be 53° okay so you can write it like this or some people will say that the measure of your angle B is equal to 53° like this so whatever notation your teacher prefers that's how you want to give it so I'm just going to erase this so we have some room all right so now we want to find these missing sides so we have this guy right here and this guy right here okay so I always use the information given so that I can get a more accurate answer in almost every case you're going to have to round okay and just get with your teacher or your assignment to know what you need to round to so for these guys I'm going to round to the nearest 10th just to keep it really simple I'm going to start with the fact that I was given this angle here which is 37° okay so I know that s of 37° is equal to what it's opposite it's opposite over hypotenuse okay okay so the opposite side here is this lowercase a this lowercase a and the hypotenuse here is given as 14 remember the hypotenuse is always across from your right angle okay so this guy is going to be 14 okay so how can we solve for this a here okay the lowercase a we just multiply both sides by 14 to clear the denominator okay so we'd multiply by 14 over here and byy 14 over here and this guy cancels with this okay same thing we been doing forever so what is 14 * s of 37° well if you punch that up on your calculator you're going to get about 8.4 again if we go to the nearest 10th so this guy right here I'm just going to say that a is approximately 8.4 okay so it's approximately 8.4 and I'm just going to write in that it's 8.4 here again we know this is an approximation okay so what about this lowercase b here what's that going to be again when you work with these problems try to use the information that you're given because in some cases you're going to have to approximate your angle so you won't have this guy available to you as an exact measure so I'm going to go with the 37 degrees that they gave me and I want to think about how can I get this guy right here so we would use cosine now because the side here which is lowercase b this guy is the adjacent side and this guy is the hypotenuse so if I did cosine okay cosine of 37° well now this is equal to again the adjacent the adjacent over the hypotenuse and the adjacent is what it's going to be this lowercase b and the hypotenuse is 14 okay so same deal very easy to solve this multiply both sides by 14 and what are we going to get well we know that this would cancel with this and 14 * the cine of 37° again if you're rounding to the nearest 10th you're going to get about 11.2 okay so let's just say that b here I'm going to put 11.2 again we know this is an approximation it's not an exact value so we're going to put that b is approximately 11.2 so we've solved this triangle okay we have a value of 8.4 and a value of 11.2 for those two unknown sides again those are approximations you've got your 14 which was given you've got this angle is 37° which was given we know this is 90° that was given and then this guy we figured out was 53° all right so let's look at a different type of problem now we're solving a right triangle but we're given two sides so if we want to start out by finding let's say the measure of this angle a so the measure of this angle a what can we do well we're given the adjacent side okay we're given the adjacent side let me just write that down this is equal to four we're also given the hypotenuse remember this is always across from the right angle so we're also given the hypotenuse this is 15.4 here okay so what trigonometric ratio is the adjacent the hypotenuse is cosine right the cosine function so we can go ahead and say that cosine of this angle a is going to be equal to we would have the adjacent which is four over the hypotenuse which is 15.4 remember we're not done here okay we need a solve for a which is our angle measure how can we do this we talked about in the last lesson the fact that you could use the inverse of cosine okay to figure out what the measure of this angle was in other words I'm going to take my inverse cosine function and I'm going to plug this guy in okay the ratio in so 4 over 15.4 and what I'm going to get out of this is going to be a value that's a measure of my angle a okay and this is going to be approximated I'm going to do to the nearest 10th so if you punch this up on a calculator and you round to the nearest 10th you're going to get about 74.9 so let's just say this is about 74.9 okay and this going to be in terms of degrees so let me put that there and let me just erase this we don't need this anymore I'm going to say that this is approximately 74.9 okay in here I'm just going to label this and say this is 74.9 degrees but again this is an approximation so let me put my degree symbol there and then let me label this as 90° we already know that because of the right angle symbol now what's going to be the measure of this again this is just an approximation but essentially we know that all three angles here with sum to 18 or the measure of angle a plus the measure of angle B would some to 90° so I can say the measure of my angle B here is approximately 90° minus 74.9 de which is 15.1 de so let me go ahead and just write that in here as 15.1 de okay so we have our three angles now what I want to do is I want to think about how I can find this guy right here okay so the value of this lowercase a the unknown side so what I can do there is just use my Pythagorean theorem we know that the length of one leg which is four squared plus the length of another leg which is going to be a in this case so that's going to be a that guy squared is equal to the length of the hypotenuse which is 15.4 squar okay now again you're going to be estimating this so 4 S is 16 then plus a^2 = 15.4 if I square that I'm going to get 237 23716 so let me erase this we're going to be solving for a so I'm going to subtract 16 away from each side of the equation let me scroll down I'll come back up so minus 16 over here minus 16 over here this is going to cancel so you're going to have your a^2 is equal to If I subtract away 16 over here I get 22116 let me erase this and let me just drag this up so we have enough room to work let's move this up here and so what I want to do now is solve for a now normally you would say a equal plus or minus the square root of this guy right but you don't need the minus here you just need the principal square root so a is going to equal the principal square root of 22116 so that's approximately going to be 14.9 okay approximately going to be 14.9 so let me scroll back up and we'll write this as our answer I'm just going to put 14.9 here so 14.9 and just to make sure everyone knows it's an approximation I'll say a is approximately going to be 14.9 okay in some of these problems you're going to have units you might get inches or yards or feet or whatever you're working with just make sure you include your units in this case we're not given any all right let's look at a little word problem now so when Max is 123 ft away from the base of the school's flag pole the angle of elevation to the top of the flag pole is 26 de 40 minutes we want to find the height of the flag pole if his eyes are 5.30 ft above the ground so when you first see problems that involve the angle of elevation or the angle of depression it's a little bit confusing what you want to do to clarify things is always draw a little picture Okay so that's what I've done here and so basically you can see that this part right here remember his eyes are 5.30 ft above the ground so five 5.3t above the ground so that's what this part is right here if you think about this entire thing okay let me change the color here this is what we're looking for this height going all the way down so it includes this 5.30 feet plus the measure of this side a okay so we need both of those we need to add them together okay so what we're going to do is we're going to work on this part first so how can I find this unknown side a well I'm given this angle measure here which is 26° 40 minutes okay so if I know that I have this guy right here which is the adjacent side okay and I want to find this part right here which is the opposite side well what trigonometric ratio gives me opposite over adjacent well I can use my tangent function right so I can say that the tangent the tangent of this 26° 40 minutes is equal to the opposite side over my adjacent side which is 123 feet okay I'm just going to put 123 just to work with the numbers for right now okay so what we want to do is just multiply both sides by 123 let me scroll down and get some room going so I'm going to multiply this side by 123 and this side by 123 we know that this cancels with this and I'm left with just a okay and it kind of looks like a nine so let me make that more clear so I would say that I'm just going to flip this over and say a is equal to or actually let's put approximately because we're going to have to estimate this if I did 123 times the tangent of 26° 40 minutes I would get about 61.8 ft so about 61.8 ft okay now and I can just put ft to abbreviate that now if you're confused about how to enter this into your calculator you have on a TI 83 or 84 whatever you're using you can hit second and then angle okay and you can choose CH degrees and you can choose minutes okay that way you don't have to convert it into a decimal and then type it in it's a little bit easier and it'll be a little bit more accurate all right so a is approximately 61.8 ft now we are not done we haven't solved the problem yet let me erase all this and let's go back up and let's say that this guy is 61.8 feet but again I'm trying to find the height of the flag pole which goes all the way down okay remember his eyes are 5.3 ft above the ground okay so I've got to add these two guys together to get my final answer so we're going to do 61.8 ft plus 5.3t okay and this would give me this would give me 67.1 ft in this lesson we want to talk about converting between radians and degrees so far in our course we have strictly measured our angles using degre de now as we progress through this course and math in general it's going to be more useful to measure angles using radians instead of degrees this is going to allow us to treat trigonometric functions as functions with domains of real numbers so what exactly is a radian well a radian is going to be based on the radius of a circle so I want you guys to recall that the distance from the center of a circle to any point on the circle is known as the radius so typically we will denote this with a lowercase R so let's look at this little graph we have here we have a circle that's already drawn I have made a point at the center of the circle and I have made a point just somewhere on the circle okay it's right here but it could have been right here it could have been right here anywhere on the circle if I draw a line segment from the center to some point on that Circle the distance is always going to be be the same and it's represented with that lowercase R so if I drew this guy right here this would have a distance of r or if I drew this guy right here this would have a distance of r as well so let's say that we make a copy of our radius and we just extend this vertically so we'd say that this has a distance of R and now let's Bend this onto our Circle again to where from right here okay starting right here going all the way down to this point and from right here going to this point here those both have a distance of R so we understand what we're looking at here now let's suppose that we have an angle where the vertex is at the center of the circle and it intercepts an arc on the circle that's equal in length to the radius of the circle so you can see that's the exact situation we have here we have the center of the circle that's the vertex of our angle and it intercepts an arc okay it intercepts an arc that is exactly equal in length to the radius of the circle so in this case occurs we can say that our angle Theta is exactly equal to one radian so one radian now let's extend this concept to two radians what would an angle of measure of two radians look like well we have this here so now you can see that the angle would intercept an arc that's equal to twice the radius of the circle this is 2 * R right so from right here from right here to right here is R okay the radius this is two * R so from right here to right here okay and I can highlight this in so this is two times the radius so in this particular case we can say that Theta is equal to two radians okay two radians so on and so forth you could extend this concept and say if the angle would intercept an arc that's equal to half of the radius of the circle well then Theta would be half of a radian again so on on and so forth so let's talk a little bit about how we can get a formula going to convert between degrees and radians so in general if our angle Theta is a central angle of a circle of radius R okay and Theta intercepts an arc of length s then we can obtain our radian measure of theta by saying that Theta is equal to we would have S again this is the length of our Arc divided R this is going to be terms of radians okay so if we go back to this example here well the length of the arc here is exactly R so if we said Theta is equal to S over R radians well now if I substitute in an R there okay you would have R over R which is one so you would get exactly one radian if I come to this example here again if you say Theta is equal to s/ R radians and I plugged in 2 r there so if I plugged in 2 R here so 2 R well what's going to happen is the RS are going to cancel and you'd have two radians so two radians so that's where that comes from so now let's think about this in terms of a 360° angle and we'll use this to build a little conversion formula so we know that a 360° angle corresponds to one complete rotation or one complete circle so I can just kind of go over this this is one complete rotation okay so we know that Theta here is 360° okay but in terms of radians what is it well let's think about the fact that the formula for the circumference of a circle or the distance around the circle C for circumference is equal to 2 * piun * R okay so this is a Formula you should know from basic geometry if you don't there it is so basically if we think about the fact at a 360° angle which again is one complete circle intercepts an arc equal in length to 2 pi 2 pi times the radius of the circle well if I plug this guy in here what do I get well in this particular case Theta is equal to we would have 2 * piun * R over R this number of radians so what's going to happen here is that the radius and the radius those are going to cancel from numerator to minator so that's gone and I'm left with 2 pi radians okay so for a 360° angle it's 2 pi radians so let's go ahead and erase this and put 360° equals 2 pi radians okay so we're all in agreement on that let me just copy this real quick and let's go to a fresh sheet so let's paste this guy in here I'll just drag this down here let me just write this in a neater format so 360° is again equal to 2 2 pi radians okay all right so what we're going to do now is actually divide this okay by two on each side so if I divide this by two and this by two what I'm going to get half of a 360° angle or half of a full rotation is half a rotation right so it's 180° angle or straight angle so we would have that 180° is equal to the twos would cancel there right so I would just have Pi radians and this is the relationship that you're going to see in most textbooks so let's erase this real quick and let's drag this up here and I'm going to show you a few different ways you can think about this now when you work with some type of conversion problem generally you want to set up a fraction that's equal to one where what you want to end up with is in the numerator so in other words for these problems the way I would do it is I would say okay you can say 180° over Pi radians okay like this if you're trying to convert to degrees okay or if you were trying to convert to radians you could say that you had Pi radians Pi radians over 180° okay so either of these would work it's just a matter of what you're trying to cancel am I trying to cancel out degrees am I trying to cancel out radians what you want to end up with needs to be in the numerator so let's keep that in mind so let's go to the first problem here and we have 150° so again the relationship is that you have 180° is equal to Pi radians so I'm just going to write that down this is something you should memorize so that you always have that in your head okay so if I want to convert to radians I want degrees in the denominator okay so what I'm going to do is I'm going to multiply this by pi radians okay Pi radians over 180° okay let's erase this we don't need this any anymore for right now and let's just think about what's going to cancel well degrees would cancel so you can get rid of that okay and so we know our answer would be in terms of radians and now I'm just going to go through and say okay well 150 is divisible by 30 right if I took 150 and it divided by 30 I would be left with exactly five if I divided 180 by 30 I would be left with exactly six so what I'm going to have is 5 * piun over 6 radians so 5 piun over 6 radians okay so very easy to convert these guys and you can erase this right just so you can show the relationship let me just kind of scooch this over so we can make it nice and neat for our presentation so 150° is again equal to 5 pi over 6 radians let's look at another one so we have 260° again I'm just going to multiply this by 180° is equal to Pi radians so I want degrees in the denominator so I want to get rid of that so I'm going to say that pi radians is going to go in the numerator and 180 de is going to go in the denominator so I'm just going to cancel this with this those are gone and I would say that okay 260 and 180 they're each divisible by 20 okay so I can say that 260 divided 20 is 13 and 180 ID 20 is 9 so this would be equal to 13 Pi okay over 9 this number of radians okay all right so now we have 157 piun over 36 this is of course in terms of radians so let me just go ahead and write that in real quick so this is in terms of radians and we want to convert this into degrees so again I'm going to multiply now I would want radians to be in the denominator here and degrees to be in the numerator so 180° would go in the numerator and then we would put Pi radians in the denominator again if you just remember this relationship 180° equals Pi radians again this is because these two guys are equal to each other this fraction has a value of one we're just using it to kind of cancel what we don't want in this case we don't want radians so that's going to cancel we're going to get rid of Pi in each case and so this is going to be equal to what well you'd have 180 / 36 which is going to give you 5 so you have -57 * 5° so this would be 785 de again if you wanted to make this more presentable if you're turning this in for a test or something like this then you could erase all that and just put let me put radians in here to be clear so radians and we'll put is equal to- 785 de and I'll just erase this here so -57 piun over 36 radians is equal to - 785 de okay let's do one more again a very easy concept I'm going to say this is in terms of radians so 43 piun over2 radians I'm going to multiply this by what I want radians in the denominator so I want 180 degrees in the numerator again I want this in terms of degrees so I'm going to put Pi radians in the denominator and of course this cancels this cancels and so what I end up with is what 180 divid 12 is going to be 15 so this will be 15 and that'll be gone and then 43 * 15 is 645 so this would be 645 de again if you want to clean this up a little bit you just erase all this and let me put this back and we'll just slide this over we'll say that 43 pi over 12 radians is exactly equal to 645 de in this lesson we want to talk about how to find the Arc Length on a circle and also we'll look at how to find the area of a sector of a circle all right in the last lesson we learned how to measure our angles using radians instead of degrees so now what we're going to do do is we're going to take our little formula and we're getting get a formula for finding the Arc Length on a circle so let's recap what we learned in the last lesson so if our angle Theta is a central angle of a circle of radius R and Theta intercepts an arc of length s then we can obtain our radian measure of theta as we say that Theta is equal to S over R okay now in this first example what I've done is I've made this Arc Length equal to R exactly so this guy right here is the same as the measure of our radius okay so when this happens you can plug in an r in the place of s r over R is of course one right if we said this is R over R this is equal to exactly one so the measure of your angle Theta in this particular case is one radian now in many cases you're not going to have a measure of one radian okay so now now our Arc Length here is going to just be represented with s okay so it's no longer equal to R so in this case we just go back to that little ratio and we say that Theta is equal to S over R okay so s over R but what this is telling me is that I could rearrange this equation okay and I could solve for S I could solve for S given the fact that I know the measure of my angle Theta in radians and I know R which is the length of my radius so what I do is just multiply both sides by R okay by R and I'm going to get my formula which is that s again that's the length of the arc here is going to be equal to this cancels so I would have R * Theta okay and Theta has to be given in radians for this to work if you get something in degrees you just need to convert it or you could use another formula from geometry which I'll cover in a little while all right let's go ahead and take a look at an example so we have an angle here which is marked as Theta and Theta is going to be equal to in terms of radians pi over 4 okay and then we have our radius R okay so R is going to be equal to we're going to have 12 miles here so I'm just going to put Mi to abbreviate miles so what we want to do for this problem is find the length of this Arc here okay find that length so again how do we do this we're just use our little formula we're going to say that s okay I'll mark this with s is equal to what it's equal to R this radius measure okay which is 12 miles in this case times Theta okay and Theta in this case is pi pi over 4 okay so very very simple math here I'm going to cancel this 12 with this four and get a three and basically what i' have here is 3 Pi miles okay so s here is equ Al to 3 Pi miles and you could write out miles or you could put Mi doesn't really matter I think I'll just stay consistent and put Mi there and leave it as an abbreviated version now depending on the context of the situation you might want to give it an exact answer like this or you might want to punch Up 3 * pi into a calculator and get an approximation and write it that way just depends on what your teacher's asking for okay let's take a look at another example so let's suppose that now our angle Theta our angle Theta here okay is given as 135 degrees okay so now we're working with degrees and then R the radius let's say that this is 12 in so 12 in again I'm just going to abbreviate that so what do we do here well one strategy if you want to use the formula that we just talked about is to First convert this into radians so remember how to do that I radians is exactly equal to 180° so I would say 135° times remember how to do this you want the units that you want to cancel in the denominator so I'm going to put 180 Dees down here and I'm going to put Pi radians up here you don't need to write radians so I'm going to cancel this with this so degrees are gone so the answer I get here is going to be in terms of radians now okay so let's think about 135 and 180 what's going to be common between the two well each would be divisible by 45 if I divided this guy by 45 I would get exactly three if I divided this guy by 45 I would get exactly four so this turns out to be 3 pi over 4 in terms of radians okay so I'll just erase this real quick and I'm just going to erase 135 degrees and put 3 pi over 4 for radians and now we're basically good to go again we're trying to figure out what is the length of this guy okay and I'm just GNA Mark that and say that this is s okay so s is going to be equal to again just multiply these guys together so 3 pi over 4 * 12 in okay and again what's going to cancel here 12 cancels with four and gives me three so you're going to have 3 * 3 which is 9 * pi * in so you can say this is n okay this is 9 pi in again you can write that out or leave it abbreviated whatever you want to do all right so before we move on I just want to talk a little bit about how you can do this with it being in degree mode okay it's a very simple formula all you want to do you probably remember this from geometry but you just want to take 135° and divided by 360° so in other words you're taking 135° which is the measure of this angle okay and dividing it by 360° which is the measure of one complete rotation right the angle that's formed by a complete circle then I would just multiply it by remember the circumference of the circle is the distance all the way around okay so the formula for that is going to be 2 * pi * R okay so if we do 135° divided 360° your units would cancel 135 and 360 are each divisible by 45 if I divide this by 45 I get three if I divide this by 45 I get eight okay so let's just write this as 38 let's just write this as 38 times this guy so now I know that I can cancel this eight with this two that would give me a four down here the r here if I plug in for R it's 12 in right so this is time 12 in so what else can I do I can cancel this 12 with this four and I'm going to get a three and If I multiply across I'm going to get 3 * 3 which is 9 okay time Pi * inches so I would still get an s that's equal to 9 pi in so it's the same thing and we can write it out and say that it's the same thing as this so s is equal to 9 pi in right so those guys are the same result so either way you want to do this is going to give you again the same result it's just a matter of which way you prefer all right what we want to do now is talk about how to find the area of a sector of a circle so we've already seen that a complete circle or one full rotation forms an angle with a measure that's 36 de or in terms of radians it's going to be 2 pi radians so you can see this guy one complete rotation or one complete circle it's 2 pi radians okay now if we think about the formula for the area of a circle most of us know this from geometry the area of a complete circle here is pi * the radius squared so now what if we wanted the area for a piece of the circle okay so not the whole thing so a sector of a circle is the portion of the interior of a circle intercepted by a central angle so visually you can see on the screen this looks like a piece of Pi okay so how do we find how do we find the area for this guy right here well let's think about the formula and I'm just going to put a for area let's think about the formula for a whole circle again it's Pi * R 2 okay so now that's for a 360° angle we're again in radians 2 pi okay so what I do here is I think about the fractional amount of this guy so I can take Theta okay which is the measure of this angle in radians and divide it by 2 pi which is the measure of again a complete circle or one full rotation in terms of radians so if I multiply this fractional amount here by this area of a circle formula I'm going to be good to go okay now notice that you can cancel this with this and I can say that the area is going to be equal to what let's go ahead and say this is 12 time we'll do theta * R 2 and there's different ways to write this but most textbooks will write it in this way so that's how I'm going to present it now what's important here is that Theta be a central angle okay and that the Theta also be given in terms of radians if it's not again you've got to convert from degrees to radians to use this formula all right let's take a look at an example so we have our angle we have our radius R Theta is going to be given to us as 3 Pi / 2 the radius R is going to be given to us as 9 M you can write out meters or you can just put M for meters that's an abbreviation and again what are we trying to find we want to find the area of inside of here okay and I'm not going to highlight this perfectly but you get the idea okay so we're trying to find that area and let me just erase that so how do we do this again we're just going to use the formula so again the formula here you can write a for area we'll say this is 12 * R 2 * your Theta okay so I'm just going to plug in what is r in this case it's 9 M okay so your area would be equal to 12 times I'm going to plug in 9 M M is for meters and both of those are going to get squared remember you got to square your units as well and then times your Theta which is 3 pi/ 2 let me scroll down a little bit and get some room going so my area here is/ 12 times the 9 gets squared it's 81 The Meters get squared so you could write this as square meters or Meer squared or you can just put M squ for short so M squar and then times your 3 pi/ 2 okay so can we cancel anything no it doesn't look like we can so we're going to say that the area the area is going to be equal to we're going to say that 81 * 3 is 243 okay then times your pi and then this will be over 2 * 2 is 4 and this will be in terms of square meters or you can just write M squar again m is the abbreviation for meters and we know it's being squared okay so you could write meters out and square it like that you could put square meters or meters squared if you want to write it out any of those would work but make sure that you square the units and you show that in some way all right let's take a look at one more so suppose now that Theta is given to us as 60° and the radius R is going to be given to us as yard so I'm going to put yd to abbreviate yards so let's go ahead and say that the area again is equal to 12 times your Theta times your radius squared now again Theta has to be given in terms of radians okay so we need to convert this guy into a radian measure so I'm going to take 60° and multiply it by remember Pi radians equals 180° so I'm going I'm going to put Pi up top I'm going to put 180° at the bottom and I'm just going to cancel okay so this would cancel with this and leave me exactly three okay so this would be pi over 3 this would be pi over 3 okay so let's erase this and let's think about this now so we're going to have the area is equal to 12 time you've got your pi over 3 time the radius which is 13 yards being squared so don't forget about those units it's very important so we're gonna say that the area is equal to can anything cancel no so I'm just going to go ahead and say that 13 squared is 169 so this would be 169 time Pi okay I'll deal with the yards in a moment over your six and then yard squared again you could write square yards you could write yard squared however you want to do this I'm just going to put yd for the abbreviation of yards and then I'm going to put a squared like this so the area here would be 169 pi/ 6 square yard yards or again yards squared in this lesson we want to talk about the unit circle and circular functions all right back in our algebra course we learned how to graph a circle we also learned how to work with the equation of a circle so we have here a unit circle and the center here is at the origin and the radius is going to be one so the equation for this circle is going to be x^2 + y^2 is equal to 1 so again this has a center at the origin so there's my center it's at the origin it's going to be the point 0 comma 0 and my radius is going to be one so that tells me that the distance from the center to any point on that circle is going to be one unit okay so from the center here to this point here if I move one unit to the right this is the point 1 comma 0 okay so I've moved just one unit to the right if I move from the center one unit up this is the point 0a 1 if I move from the center one unit down this is the point 0 comma 1 and lastly if I move from the center one unit to the left this is going to be the point -1 comma 0 okay but the main thing to understand is that the distance from the center to any point on that circle is one unit Now using the unit circle to work with our trigonometric functions will lead to some interesting results that allow us to quickly solve problems as we move forward in trigonometry and also into calculus so let's just get started here on our unit circle we have an angle that's in standard position so basically this is the initial side and this is your terminal side so your terminal side of this angle Theta okay is going to lie in quadrant one and notice we have a little Point here and this point is going to be labeled as X comma y just generically you could use a comma b or whatever you want and that point there is going to lie both on the unit circle and on the terminal side of our angle so if we want to define something like s of theta so s of theta what would this be well we know when we started this course we had a def definition that was y/ R but you could also use opposite over hypotenuse it doesn't matter you're going to get the same thing here but basically we know that this guy right if we look at this guy from here to here this has a length of Y right because this coordinate is y this coordinate is zero right so this vertical distance here this vertical distance here is y right Yus Z that's all you're doing so if I think about this R the radius which is my hypotenuse here okay is going to be one right because we're on the unit circle the radius is one so again the distance from the center to any point on that circle is one so s of theta is going to be y/ R which is y over 1 which is just y okay so sign of this angle Theta is just going to be the y-coordinate of this point right here okay and then the same thing is going to happen when I work with cosine of theta remember this is X over r or again you could say adjacent over hypotenuse okay and in this case the length here from right here to right here is going to be X okay so it's just x / R which is X over 1 which is X so the cosine of theta is just going to be the x coordinate for this point right here so what we can do is really label this point right here as the cosine of theta and then the S of theta so again this is your x coordinate and this is your y chord all right so let's quickly talk about the concept of circular functions basically all we're going to be doing here is swapping out Theta for S okay so how does this work well when we're working with the unit circle remember that the radius is one the formula for our angle Theta in radians remember it comes as Theta is equal to S which is the length of the Ark okay divided by R which is your radius well in this case the r is one okay the r is one so Theta will be equal to S so when we're on the unit circle our trigonometric functions of our angle Theta measured in radians found Again by choosing some point x comma y on the unit circle can now be re written as functions of the arc length s some real number okay so when we show it in this way the functions are known as circular functions so you'll probably see this in your book They're going to use this letter s instead of theta right so they'll start saying s of s is equal to Y and then cosine cosine of s is equal to X and then you can get the rest of them using these so you can say tangent of s is going to be y X and then you can get the rest using the reciprocal identities right so the cosecant of s is 1 over y the secant of s is 1/x and the cotangent of s is X over y all right so let's talk about how we can set up our unit circle now so basically in quadrant one we're going to rely on these trigonometric function values of special angles right we've talked about this before so we know and I'm just going to go through a few of these so the S of 30 degrees is a half and we know the cosine of 30° isun of 3 over2 so let's just take that real quick let's go down to a sample unit circle it's not all filled out just part of it so I can show you where things are coming from and if we look at a 30° angle which is in terms of radians is Pi / 6 we see that the x coordinate is a of 3 over2 and the y-coordinate is 12 remember because we're on the unit circle we have this property this x coordinate here is the cosine of this angle you could say it's 30° or you could say it's pi over 6 radians whatever you want to use it doesn't matter and in this case the sign here is going to be your y-coordinate so the S of 30° is 1/2 that's your y-coordinate again the cosine of 30 degrees is your x coordinate isqu of 3 over2 okay so using the same logic you can see that if I'm working with a 45 degree angle or Pi over4 radians this would be my cosine and this would be my sign okay if I'm working with a 60° angle or pi over 3 radians this guy right here this would be my cosine the 1/2 and this would be my S the < TK 3 over2 okay so the point that's associated with that angle okay the x coordinate is the cosine the y-coordinate is the sign okay so very important to remember that so once we have this guy set up there's a lot of different tricks or techniques to remember these three different values here some people use the hand trick there's all kinds of stuff out there I'm going to show you what I learned along time ago there's a lot of videos that do it this way as well what you'd want to do is set up in each case a denominator of two so a denominator of two a denominator of two okay so that's going to be everywhere you'll notice you have two two two two two2 that's very easy to remember okay then the next thing you want to do you'll notice that if you ignore the square root symbol you start here with one and you're counting down and you start here with one and you're counting up so I'm going to make all the numerator like this so 1 2 3 and then I'm going to go 1 2 3 okay if you get lost about where you want to put the one in start and go down all you need to do is think about okay well if I'm going from right here to right here okay if I'm making a right triangle this x distance is greater than this y distance okay so that tells me that I would want the three here and I know it's square root of three but just just forget about the the square root symbol for now I know this would be greater than this so I would start with a three here or you could say I'd start with a one here however you want to remember that you're going to have a one a two and a three and then a one a two and a three okay so that's a little technique you can use there are many others that you can go with now once you're finished with that you want to square root all of the numerators okay and if you have the square root of one remember that's one so you can skip those so I'm going to have square root of this square root of this square root of this and square root of this okay remember the one here and the one here Square Ro of one is just one so you can just just ignore it okay so now we have all of these guys you can wrap it in parentheses to make it official if you want but that's how I get these points there so now that I have all of these points what I want to do is reflect across the y AIS to get these points over here so let me show you this and basically if I'm reflecting across the y axis think about this for a moment if I go across the y- axis like this the y-coordinate is the same okay so you'll notice here you have 3 over 2 and here you have squ of 3 2 so that's the same the x coordinate is now its opposite right because I'm on the opposite side of the Y AIS over here x values are negative over here x values are positive okay so this is negative 1/2 now instead of 1/2 this is negative < TK of two over2 instead of < TK of two over two this is negative 3 over2 instead of of 3 over2 okay so it's very easy to just reflect These Guys across okay and I know these lines I'm drawing are not perfectly straight but just pretend they are so once you've done that now you can reflect These Guys across the x axis and you'll have all your points okay so now if you think about a point that's up here if I reflect it across the x axis the x value will be the same but the Y value will be its opposite so if I reflect this guy right here down there and I know that Line's not perfectly straight but you have2 and you have 1/2 so the x value is the same then you have of 3 over two and then the negative < TK of 3 two so that's all we're doing here to get these values again if you can just remember these values in the first quadrant okay you're going to be good to go you can use the reflection okay across the y- AIS and then across the x- axis to get all the remaining points okay so let's go ahead and take a look at just a few examples so I want to look at the S of 3 pi over 4 and then I want to look at the cosine of 3 pi over 4 and then lastly I want to look at the tangent of 3 pi over 4 okay so we know that we could solve this using methods without the unit circle right we already know how to do this but using the unit circle it's going to be a little bit quicker so I'm going to find 3 Pi over4 on my unit circle so where is that going to be 3 Pi over4 is going to correspond to a 135° angle so it's this guy right here and you can see that you have the negative of the square Ro 2 over 2 so that's your cosine let me write that out your cosine of 3 piun over 4 will be equal to the negative of the < of 2 over 2 and then you have your s of 3 pi over 4 is equal to theare < TK of 2 over two okay so let's copy this real quick come back down here I'm just going to paste this in so let me just rewrite this so this would be the < TK of 2 over 2 and this guy would be let me just erase this I can just drag this over it's the > 2 over2 okay so how do I get my tangent here well remember the tangent is found by taking the sign and dividing it by the cosine so I would take the square < TK of 2 over two and divide it by this guy but if I'm dividing fractions I want to multiply by the reciprocal so I'm going to flip this guy I'm going to put the negative out in front I'm going to flip this guy and put two over the < TK of two and you're going to notice that this will cancel with this this will cancel with this and you're left with -1 okay so my tangent of 3i over4 is1 all right let's take a look at another example so suppose we had something like the sign of let's say 13 pi/ 6 and let's also do the secant the secant of 13 pi/ 6 and let's also do the cosecant the cosecant of 13 pi/ 6 okay so if you get an angle that measures that's larger than 360° or 2 pi in terms of radians again what you want to do is find a co-terminal angle that's between 0° and 360° or zero and 2 pi radians so if I'm working with radians what I'd want to do let me just kind of divide this workspace up here and I'll just move this over for right now what I want to do is say I have 13 pi over 6 radians If I subtract 2 pi away I would get a common denominator so I'd multiply this by 6 over 6 so this would end up being what it would end up being 12 12 over 6 right 12 over 6 is the same thing as two so what I'd have here is pi over 6 radians okay so I would look at pi over 6 on my unit circle and that's going to be right here okay so this is pi over 6 again if I went one more revolution around I would end up with an angle that has a measure of 13 pi/ 6 radians or 390° okay so my coine of pi/ 6 is going to be the square < TK of 3 over2 and then my S of pi/ 6 is going to be 12 okay so let's go ahead and copy this and so I'm going to write that the S of 13 piun over 6 is going to be 1 12 okay because it's exactly equal to this here here then for the secant it's the reciprocal of my cosine okay so this would be equal to I would flip this guy I would have two over the < TK of 3 okay you just erase this you don't need this anymore and so what I do here is just rationalize the denominator so times the > of 3 over the < TK of 3 so this would give me 2 * the < TK of 3 over 3 okay then for my cosecant I'm going to flip I'm going to flip my sign right so so if I flip that the reciprocal of 1/2 is just going to be two all right let's just do one more example so I'm going to look at the sign of let's do -23 pi over 4 let's also do the cosine of the same guy -23 pi over 4 and let me just kind of move these up and out of the way and then let's also do the coent okay the coent of Let's do -23 3 pi over 4 okay so now that we have a negative angle remember if you have a negative angle you want to do the same thing you want to find a co-terminal angle in terms of radians it would be between Z and 2 pi radians so what I'm going to do is I'm going to add 2 pi radians until I get into that range so let's go to a fresh sheet and say I have -23 pi over 4 if I add 2 pi radians well 2 pi to this again I want to commment denominator so I would multiply this by 4 over 4 so you could say this is8 okay eight and then four down here okay so 8 Pi over4 now so would this get me where I need to be no so I could add another 2 pi radians so I could add another 8 pi over 4 again all I'm looking for is for this guy to not be negative right it's got to be between 0 and 2 pi okay so if I add 8 plus 8 that's 16 that's not going to get me there so I need to add another one so let me kind of scooch this down so let me add another 8 pi over 4 and that would get me there right so 8 + 8 + 8 would be 24 24 - 23 would be 1 okay and you can do this with one common denominator and say that you have -23 pi plus again 8 + 8 + 8 is 24 so this would be 24 Pi all over the common denominator of four so this would end up giving me 1 Pi or just pi over four in terms of radians so this is my co-terminal angle so let's go find that guy so in the unit circle if I look at pi over 4 in terms of radians or 45 degrees in terms of degrees it's of 2 over2 and then of 2 over2 so that's really easy to remember so I know the sign of this guy would be the sare < TK of two over two okay and the cosine of this guy would be the square < TK of two over two now what about the cotangent remember the tangent is s divided by cosine well the cotangent you flip that it's the cosine divided by the sign but in this case I have the same thing divided by itself either way you go either for the tangent or for the cotangent so in each case this is just going to be equal to one in this lesson we want to talk about linear and angular speed so previously we talked about how to solve motion word problems using our distance formula so that's the D for distance is equal to R which is the rate of speed time T which is the amount of of time traveled so we're going to relate this formula to linear speed which we'll talk about in a moment basically if we know our rate of speed and we know our time traveled we know our distance so for example if I plugged in 80 miles per hour here and I plugged in let's say three hours here what would be my distance travel so we sit in a car we're moving at 80 M hour we do this for 3 hours we know that we would do 80 * 3 or 240 this would be 240 miles okay so this is the distance traveled now let's suppose we didn't know the rate of speed but we did know the distance traveled and we knew the amount of time it took to travel that distance so now what I could do is I could solve this formula if I had D is equal to R * t i could solve it for R right and I could plug in and find out what R is so this would give me this cancels the rate of speed is equal to the distance traveled divided by the amount of time it takes to travel that distance so let's erase this and let's think about these numbers again just as a simple example so suppose my distance again is 240 and this is going to be in terms of Miles let's suppose my time again is 3 hours so now if I have this information given what's my rate of speed again 240 divid 3 would give me 80 okay and in terms of units miles and then hours you'd have miles per hour so miles per hour okay so very easy to understand this concept just think about all the times that you've traveled on a road trip and you've calculated how far you're going to be at what time okay so let's erase this and I'm just going to drag this up here and I'll talk about how this is related to linear speed now so for linear speed let's look at our little diagram here I want you to suppose that some point P okay so this is my point P it's going to move at a constant speed along a circle with this radius of R okay and with the center of O Okay so the measure of how fast the position of p is Chang changing is known as the linear speed okay so instead of R we're going to use V okay for velocity and instead of D now because we're traveling along a circle remember when you think about the distance or the length that you're traveling on a circle this is the Arc Length right so we use S for this so now I'm going to have s over t Okay so that's how these two guys are related instead of R we use V for velocity instead of D for distance we're going to use S for the Arc Length that's tracing by this point P at the time T to put this in more simple terms the linear speed is just the speed at which a point on the outside of some object travels in a circular path around the center of that object now when we talk about angular speed we're talking about how fast the measure of the angle is changing so again if I think about this point here P right so this is the terminal side of the angle p o if I'm rotating P around this circle the measure of the angle is changing okay so how fast the measure of that angle changes that's our angular speed okay so this is given by the formula you're going to use this letter Omega okay and I'm terrible at writing it but this is the Greek letter Omega and it's equal to Theta okay which is the angle measure in radians divided by T which is your time now there's one more thing I want to talk about before we get to the problems you might want to make a little substitution okay based on what you're given in the problem you might want to make a little substit tion here and instead of saying s The Arc Length remember the Arc Length s is equal to your Theta okay your angle Theta that's in radians times your radius that length R okay so if we plug in there we could say V is equal to you would have your Theta * R okay over T now we know from fractions that I could really rewrite this as R times your Theta over T so now we can make a little substitution because Theta over T okay is this guy right here okay it's the Greek letter Omega which again is our angular speed so I can put this as my angular speed here so the linear velocity or the linear speed however your book says it is going to be equal to the length of your radius times your angular speed so let's go ahead and look at an example so the tires of a bicycle have radius 15 in and are turning at the rate of 230 revolutions per minute how fast is the bicycle traveling in miles per hour okay so the first thing you want to do for a problem like this is calculate the angular speed so again I'm going to try to make the Omega okay this is equal to again Theta over t Okay so the first thing you might notice is that you're given this rate here of 230 revolutions per minute okay when you think about one revolution it's one complete circle or 360° or 2 pi radians since we're working with radians here we want to put 2 pi up here okay this is one complete Revolution or one complete circle we have 230 of these so I'm just going to multiply this by 230 okay so again this is one complete circle or one complete Revolution or one complete rotation however you want to think about that okay so it's 2 pi okay or 360° if you want to relate it back to degrees then we have 230 of these guys so that's why I'm multiplying by 230 now I'm going to divide this by the time is given as per minute so per minute means 1 minute there okay so this is in terms of radians so you can write this as radians okay right here so this is 2 pi * 230 or you could say 460 460 Pi radians per minute now the next thing we want to do is think about the time units that we want it's going to ask for this in terms of miles per hour we have radians per minute okay so we want to convert this to radians per hour so let me erase this so we have a little bit more room to work because this can get quite messy so let's move this down a little bit and I want to multiply this now by what I want hours in the denominator I want one single hour because I want per hour okay so if you always think about what you want go ahead and put that there first and then how do I get there I want minutes to cancel what's the relationship between hours and minutes well 60 Minutes 60 Minutes is equal to 1 hour okay so now this is basically equal to one okay so what I can do is I can cancel this with this and I'm left with the units that I want in my denominator so let's go ahead and multiply 460 by 60 which is 27,600 don't forget the pi there it's easy to forget it pi and then you have radians okay and then per one 1 hour okay so that's your angular speed 27,600 Pi radians per hour okay so let's erase this I don't need these intermediary steps anymore and let's just bring this up here and I'm just going to say it's 27,600 pi and then I'll put radians per hour okay I'm just going to abbreviate there now let's think about now the linear speed okay in terms of miles per hour power so how are we going to get this so V is your linear speed remember the formula we looked at when we plugged in here we said that we could do the radius times the angular speed so if we come back down here the angular speed I'm going to put 27,600 Pi okay and then you can go per hour let's put this down here and then times what's the radius well the radius is going to be given to us here as 15 in so 15 in Right Here If I multiply this out and let me scroll down to get some room going If I multiply this out I'm going to get that 414,000 Pi okay and then we'll say inches per one hour okay this is not what I want though because I want it in terms of miles per hour and I have it in terms of inches per hour okay so I've got to do another unit conversion and this is where these things kind of tough so if you know the relationship between inches and feet 12 in equals 1ot and then you can convert from feet to yards and then from yards to miles I'm going to shorten this up so the relationship between inches and Miles 63,360 in is equal to one mile so what I can do I want inches to go away so I'm going to put 63,000 and then it's going to be 360 in okay I want that to go away so I want that in the denominator and I put one mile in the numerator again 1 m is equal to 63,360 in so this is really a fraction that has a value of one okay so this would cancel with this so those units are gone so now this going to be in terms of miles per hour okay so what you'd want to do is really just take 44,000 Pi okay and then divide it by this number here 63,360 it's going to be an approximation because you're going to get a decimal number but if you wanted to simplify this first you could divide this by 720 okay and you could divide this by 720 and that would give you 575 and this would be 88 okay so you could really write this as 575 Pi miles okay per your 88 hours but again this is kind of useless because we want miles per hour so if you punch this up on a calculator you could say that this is about 20.5 this will be in terms of miles per per hour okay so that's our linear speed so let me just erase all of this so to answer this with a nice little sentence I'll say the bike the bike is traveling at about because again this is an approximation 20.5 this is in terms of miles per hour so the bike is traveling at about 20.5 miles hour all right let's take a look at one more problem so at a local high school Max Rides cart around a circular track at two revolutions per minute if the radius of the circular track is 150 M how fast is Max's cart traveling in meters per second okay so a similar setup I'm going to use Omega once again to say it's Theta over T right so this is my angular speed so what is Thea in this case well we have two revolutions and this is per minute so down here this would be let me put put t as 1 minute and get that out of the way and I'm just going to abbreviate it for Theta it's 2 * 2 pi again one revolution or one time completely around a circle is 2 pi radians or 360° so you have 2 pi okay times you've got two of these revolutions so two so this would be 4 Pi 4 Pi radians radians per minute okay so that's your angular speed now we want this in terms of seconds but let's convert this in a moment because it's going to be a little bit EAS for us we know that with the linear speed okay if we're trying to find out how fast Max's card is traveling in meters meters per second well what I do is I take this guy right here okay and I plug it in I plug it in and I multiply it by the radius okay so let's do that so we have 4 pi over 1 minute times the radius it tells us is 150 M so it's 150 M okay and so let me scroll down and get some room going here okay so 4 * 150 we know is 600 so this ends up being 600 Pi M okay per 1 minute so the last thing we want to do is get this in terms of seconds okay and the reason I waited here is because you have to divide by 60 and 600 divided by 60 is pretty easy to do okay so that's why I waited so you can multiply this now by I want seconds in the denom minator because I want it as per second and there's 60 seconds in 1 minute let me slide this down I kind of ran over a little bit let me slide this down and I'll put one minute up here so that it will cancel again you want to and let me abbreviate this so that's consistent you want to put these across from each other again so they cancel so this will cancel with this and I'm just left with seconds now right I'll have meters per second so 600 ID 60 is going to give me 10 so this would be 10 pi m per second or you could say it's about you could say it's about 31.4 M okay meters per second okay so to answer this little word problem I'll say that Max's cart Max's cart is traveling at about again it's an approximation 31.4 m per second okay so Max's card is traveling at about 31.4 m per second in this lesson we want to talk about graphing s and cosine all right so to start out the lesson I'm going to talk about something called a periodic function so when we think about s and cosine these guys are called a periodic function because their values repeat in a regular pattern so to see what I mean by that let's revisit the the unit circle from the last section so from our unit circle what we see here is that as we travel around our unit circle so in other words if I start here let's say I make a counterclockwise rotation okay so a positive angle measure and I do one full rotation well that's going to be an angle measure of 2 pi we already know that okay but as I continue past 2 pi so let's say I created an angle that was greater than that so I keep going around okay so let's say I make an angle that's 4 Pi so two four rotations what's happening is these values for S and cosine are going to repeat themselves okay so anything past 2 pi so when we talk about 2 pi for S and cosine this is going to be known as our period it's the smallest possible positive value for where these function values okay these values for S and cosine are going to repeat themselves okay let's think about this with a simple little example let's say I think about this angle measure here which is pi 6 or 30° I know that the sign of pi/ 6 is given by the yalue of that point right there okay so this is equal to 12 okay but also if I was to rotate around another 2 pi okay so another 2 pi I'm going to get back to that same point right there okay so I have a co-terminal angle and my sign value is going to be the same so in other words if I said the s of Pi let me use parentheses here Pi / 6 Plus another full rotation would be 2 pi let's go ahead and write that as 12 pi over 6 so we can have a common denominator this is going to be equal to a half as well okay if I add these up Pi + 12 Pi is 13 Pi so this is 13 pi over 6 okay this is 13 pi over 6 if you punch this into your calculator you'll see that you get the same value either way and you can keep going around another 2 pi or you can go the other way you can go clockwise 2 pi so basically Ally as you're adding these multiples of 2 pi you're getting these co-terminal angles and so your s values are the same and also your cosine values are the same so if we go back up to our little definition of a periodic function you see we have f ofx x is just some member of the domain in the case of s or cosine it would be any real number this is equal to F of you have that same X Value Plus n which is some integer times P which is going to be our period which in this case is going to be 2 pi okay okay so I could say something like the S of X and I could say really this is equal to the sign of I'll say that same X Value Plus some integer n times the period P which is 2 pi okay this is something that is very important to write down we're going to use this not only in this section but also when we solve trigonometric equations so then we could say the cosine of x is equal to the cosine of you have your same value for x plus your n * your 2 pi okay in our example we saw let me write this off to the side that the sign of basically the pi/ 6 was equal to the sign of we had pi over 6 again okay and then what we did was we added just 2 pi right our n was one but really you could put n times let me scoot this down because it's going to run out of room n times your 2 pi here okay so you can plug in anything for n as long as it's an integer and this statement would be true you would basically get 1/2 on the left and one half on the right after you evaluate okay before we move any further I want to clear up something that is a common source of confusion and that is what to do with this point right here okay which contains an x value and a yvalue when you're working with something like y equal the sin of x or y equals the cosine of x which we're going to get to in a moment let me just erase this okay we'll build up to what we need to do okay let me actually erase this I want to show you again where things come from even though we talked about in the last section we have a unit circle here okay so that means the radius is one okay so this r value here or this distance from right here to right here is one okay you can see that from here which is the origin 0 comma 0 to here which is 1 comma 0 that's one from here to here is one from here to here is one from here to here is one so we see the radius is one now if I go back to my definition of cosine of theta and S of theta again if I have a point on the terminal side well the cosine of theta is given as X over R and the S of theta is given as y/ R well here my r value is one so the cosine of theta is exactly equal to X or the x coordinate from that point so this is exactly equal to X for the S of theta again y/ r r is 1 so this guy is exactly equal to the y-coordinate from that point so it's really important to understand that when I grab this x value it's the cosine of theta when I grab this y value it's it's the S of theta okay now we move on and we introduced this s now instead of theta okay that further complicated things the reason we're using this is so that we can have real numbers as our domains for the trigonometric functions right we started to call these guys circular functions so s here is the arc length of the intercepted angle okay so s is exactly equal to Theta okay so now we can say that for this point right here the x value is the cosine of s okay this length here and then the sign of s is going to be y okay so that's all we really need to know when we come back to our little unit circle if we think about something like Y is equal to the S of X or let's say Y is equal to the cosine of x it's important to know what's going on what you're plugging in for x and what you expect as an output for y we are plugging in angle measures here typically in radians for x and we are getting s values out for y okay okay so in other words if I look at let's say this point right here I am not taking this as the x value and this is the yv value that is not how that works I want the angle okay so pi/ 6 in other words let me erase this for a moment if I said Y is equal to the S of pi/ 6 what's the Y value well the sign of pi over 6 is given by this y value here for this point right the y-coordinate from that point which is 1/2 so this guy right here would be 12 in other words a point would be < / 6 comma 12 another one would be let's say I did piun over 4 well the S value would be < TK of 2 over2 so another one would be pi over 4 comma < of2 over2 okay you can continue making points like this to sketch your graph of sign I'm going to show you a better way in a moment but for right now I just want to make sure that's clear for you okay if I did a few with cosine just really quickly so y equals the cosine of x again if I plug pluged in Pi / 6 so y = cosine of < / 6 well what is the cosine of piun / 6 I'm not looking at the Y value now I'm looking at the x value for this point okay it's of 3 over2 so this would bequ of 3 over2 and so the ordered pair there would be pi/ 6 comma > 3 over 2 okay so it's very important that this is clear for you in terms of what you're plugging in for x and what your output is for y now with that being said let's go through and let's think about some of the basic point that you're going to get from a sign graph okay so we're going to be looking at y equal the sin of X this is a function so you can use function notation F ofx equal sin of X it's not a one: one function but we'll talk about that later on in the course okay so if we look at these y values here we're going to be focused here here here here and here right so as we start from an angle measure of zero okay and we go to pi over 6 pi over 4 pi over 3 to Pi / 2 you'll see that the S of X or the Y value okay is going to increase from zero to a half 2 over two 3 over two to one okay then as we go from Pi / 2 to Pi this guy is going to now decrease from one down to zero okay so you're going to go to square 3 over two again 2 over two and then 1 12 and then back to zero okay then as we get into our quadrant 3 and our Quadrant 4 remember sign is going to be negative so from PI down to 3 Pi / 2 we're going to be decreasing from zero down to negative 1 okay so you go to negative a half 2 over two andg 3 over2 finally you get to Nega 1 and then lastly from 3 pi over 2 up to 2 pi so up to one full rotation we're going to increase from this ne1 back to this zero right so you're going 3 over 22 two and then finally one2 now you do not need all of these values as ordered pairs you can just start off by taking the first few guys that you get in the first quadrant along with your quadrantal angles okay and that's basically enough to get started after that I'll show you that you really only need five key X values to draw this guy over one period you're going to draw a WAV like function okay so if we start with zero that's our angle measure the Y value is zero then Pi 6 you get a half then if we do p piun over 4 you get2 over2 then if you do p piun over 3 you're going to getun 3 over2 then if you do p piun over 2 you get 1 Pi you get zero 3 pi over 2 you get negative 1 and 2 pi which is one full rotation you are back to zero all right so now that we have some basic points let's go through and draw our first graph of I'm going to go ahead and label this as f ofx is equal to the S of X I'm specifically going to say here because this is over one period so from 0o to 2 pi so one rotation I'm going to say that X is greater than or equal to0 and less than or equal to 2 pi I'll show you in a minute that this continues to the right and to the left forever okay so what we've done here we've put a little bit of detail into this more detail than you really need you'll see I have these notches that are going to be an increments of pi/ 6 we're going to start out with an angle measure of zero okay again the sign of Zer is zero so your x value is zero your y value is zero okay then as you get to Pi / 6 what's going to happen is your s of pi over 6 is a half so you can say this right here is a half okay you could label that this is pi over 6 comma 12 and I'm trying to make that a little bit better here okay then as you continue then you have your pi over 3 and the sign value for that is going to be square 3 over2 so this right here would be your pi over3 and then sare < TK of 3 over2 okay and then at Pi / 2 you get to the max maximum y value which is one okay so that's right there then you start decreasing again remember after Pi / 2 you're coming back down to a yvalue of zero so you're decreasing so now this is your maximum then you're going to come back to your of 3 over2 in terms of your y-v value you're going to come back to your 1/2 in terms of your y-value and then you come back to zero or an X intercept when you get Pi okay so then as we continue from PI to 3 pi over 2 we are decreasing from zero down to - 1 which is the minimum okay for the yvalue then as we go from 3 pi/ 2 back to 2 pi back to one full rotation we're going back to zero this just goes directly in line with what we're doing on the unit circle now when you graph these guys let me erase all of this you really do not need all of this detail okay I put some notches here in terms of pi over 6 typically if you're asked to graph something over one period you're going to make five notches for the the five key X values okay you're going to have a starting point you're going to have a quarter point let me Mark these off so you have a starting point a quarter way point a middle point a three4 way point and then an ending point okay for the basic sign graph it's really easy because these key X values are going to be your quadrantal angles right so this is if you think about it in degrees if that's easy to remember this is 0 degrees 90° 180° 270° and then 360° Okay so very easy to get those five values I'm going to talk about in a moment splitting up the period into fourths but I'll get to that in a moment so let's go down here and I want to think about what happens when we extend this to the right and to the left well again if I start off here with an x value of zero let me write this out so now this is f ofx equals sin of X okay no restriction we're going all the way through one full rotation or one cycle would be I would go up to one I would go back through my x intercept down to the minimum and then back up to my x intercept okay so this is one cycle or one rotation on the unit circle every other part of that graph is just made up of repetitions of this okay you can see this is called a sine wave okay so when you graph this guy you're going to have your five key points so 1 2 3 4 five given by the five x values which I'll talk about in a moment and then the corresponding y values obviously and then you're going to sketch that wave and just continue that wave indefinitely to the right and to the left based on what your teacher says okay you've seen that I've marked this off in increments of 2 pi so that's your period and another way we want to think about this if you're going this way okay think about what a negative angle measure means this graph is in terms of pi over 2 right those are The Notches so this is negative pi/ 2 this is piun this is -3 piun / 2 remember what it means to have a negative angle it means I'm rotating clockwise if I go back to the unit circle if I start here and I rotate this way okay I'm going to end up with netive piun / 2 which is co-terminal with 3 pi over 2 so the S value is ne1 then as I keep going I'm going to end up with a negative pi the sign value is zero right this is co-terminal with pi then as I go and I continue then I'm going to end up with3 pi over 2 this is coterminal with Pi / 2 so the S value here is going to be one and then finally as I make my for rotation going clockwise now okay I'm going to end up with -2 pi and my S value there is going to be 0 once again so when you return if you think about this again as we get to an x value of negative pi over 2 our yvalue is going to be Nega 1 as we get to an x value of negative pi our yval is going to be zero as we get to an x value of3 pi over 2 our y- value is going to be 1 as we get to an x value of -2 Pi a full rotation we're right back to zero or hitting the x intercept now here are are a few things that you absolutely need to know when you work with the sign function first off this guy is continuous across its entire domain the domain is a set of real numbers okay so that means you can plug in any real number you want and you'll get a corresponding y value it also means that when I say continuous you can graph this guy without Ever Lifting your pencil there's no holes no gaps nothing like that okay additionally when we think about the X intercepts so you think about here and here and here and here and here and here and here these are of the form let me write this out the X intercepts are of the form n * pi okay or n is some integer so you can think about this as being 0 * pi that's zero then 1 * pi is pi then you'd have 2 pi then 3 Pi then 4 Pi Works going negative so you'd have basically negative pi and negative - 2 pi so on and so forth now your range for S and also for cosine is from negative 1 to positive one so the minimum yv value would be negative 1 and that's included and the maximum y- value is positive 1 and that's included so we can say our range and let me put the E there forgot that so our range is going to be from netive 1 to positive 1 like that now the last thing I want to talk about is that this graph is symmetric with respect to the origin so when this happens we call this an odd function so let me go back to the unit circle and let me demonstrate this real quick for an odd function f ofx is equal to the negative of f ofx now this is going to come up in our section on identities we're going to need this what we call negative angle identity to rewrite things but for right now I just want you to understand this concept if I have the S of negative X let me put this in parenthesis this is equal to the negative of the S of X okay let's just go back to this pi/ 6 if I do the sign of let's say negative pi over 6 what would that be well that would be me starting here and rotating clockwise okay to right there it's co terminal with 11 pi/ 6 so the yv value for that point would be -2 so this is - 1/2 let me write that this is equal to the negative of the S of X okay I have Negative X I want just X so the negative of the S of pi/ 6 so the S of pi over 6 is given by this yvalue here so this is going to be the negative of s of Pi / 6 which is going to be negative -2 as well okay let's move on now and talk about graphing variations of the S graph I'm going to go through all the different transformation that you're going to come across we're going to start off with a vertical stretch or vertical compression okay and in order to graph these Variations by hand what you want to have are the X intercepts you want to have your maximum point and then your minimum Point okay if you're looking at one period okay so when we look at one period or one cycle you think about your X intercepts for your basic sign graph let me write this out again so f ofx is equal to the S of X or X is greater than or equal to zero and less than equal to 2 pi your X intercepts occur at the beginning the middle and the end of your period okay so 0 Pi 2 pi okay then in terms of your maximum and your minimum your maximum is going to occur at a quarter way through the period okay so 2 pi ID 4 or 2 pi * a 4 that's going to give you pi over 2 and then your minimum is going to occur at 3/4 way through your period so 3/4 * 2 pi would give you your 3 pi over two okay so what you want to do to get these X values in any situation is you want to set yourself up a little table okay or you could say just some scratch work and you want to First divide your period by four okay so with the basic sign function the period is 2 pi okay we're going to divide this by four so we cancel this with this and get a two so basically you have your pi over two okay so you're going to start off where your cycle Begins for the basic sign function it's going to start at zero okay so let me put zero here here and then what you're going to do is you're going to just keep adding pi/ 2 to the previous value so here x sub 2 or the second x value is this x sub1 which is zero plus your pi over 2 so that's pi over 2 okay and then here it's pi over 2 plus pi over2 which is pi and then plus pi over 2 again which is going to be 3 pi over 2 and then plus pi over 2 again which is going to give me back to 2 pi okay typically as a check a lot of books will say take your ending value minus your beginning value so 2 pi minus Z if that equals your period you're pretty much good to go okay now that doesn't mean you didn't make some mistake in between but at least you know that your cycle is the right length all right so let's start talking about some of the function Transformations that we're going to see the first type is very very easy this is where we have a vertical stretch or a vertical compression you'll recall from earlier in the course if we have something like a multiplied F ofx basically if this a value we'll say the absolute value of that if that's greater than one you're going to have a vertical stretch if you have the absolute value of a being greater than zero and less than one now you're going to have a vertical compression okay so all your function values or your y values are being multiplied by this guy a here okay so what we see here is we have f ofx is equal to a that's multiplying the S of X so the original function is s of x a is now multiplying that and I'm specifically going to say that a is any real number that is not zero okay so some nonzero real number so the example we start with is f ofx = 2 * the S of X so basically we know that since this is a two and the absolute value of two is two and that's greater than one this would be a vertical stretch right because all the function values all the Y values are being multiplied by two when we compare to the original function f ofx equal sin of X and you can see this so the X intercepts here here and here obviously the yv values are zero so those are going to stay the same because 0 * 2 is still zero but if you look at this maximum yvalue of one okay if you multiply that by two you're now at two okay 1 * 2 is two if you look at the minimum yvalue which is NE 1 you multiply that by two you're down here to -2 okay so this guy has been vertically stretched by a factor of two now when we think about the period for a function of this form it's still going to be twoot okay so we haven't introduced any type of shift we haven't introduced any type of horizontal stretch or compression so the period is still 2 pi we're still starting out with an x value of zero so you still have the same five key X values at 0 pi over 2 pi 3 pi over 2 and 2 pi okay so basically to get these points you really just need to think about the fact that you're multiplying the Y values from your original function by two okay that's all you really need to do now I'm going to talk about in a moment the fact that you can use this a value to find your maximum and minimum yvalue but we're going to talk about that later on when we talk about the amplitude Okay let's look at another example again I'm working with f ofx is equal to a * the S of X here we have F ofx = 12 * the S of X so what do you think's going to happen here we're going to have a vertical compression okay so basically all your y values now are multiplied by a half and that's compressing your graph okay so again you get your X values in the same way the period has not changed so it's still going to be 0 Pi / 2 pi 3 pi over 2 and 2 pi that part's really easy so the X intercepts the yv values are always going to be zero and then for these yv values here on your new graph again you can just multiply the previous ones by a half or you can evaluate the function there or again I'll talk in a moment about how to get the minimum and maximum value you might notice that when we had 2 * the S of X it was 2 and -2 when we just had s of X it was 1 and negative 1 here we have 1/2 * the S of x X and it's basically a half I can put that in this is a half and then down here this is one2 so you see the pattern is that your maximum and minimum is related to this right here we'll talk more about that in a moment all right let's take a look at one more of these in this form so now we're going to look at F ofx = -2 * the S of X so you actually have two different things going on one you have a vertical stretch of a factor of two and then you have that negative there that's going to cause a reflection across the X AIS so compared with the graph of FX equal the sin of X basically all your y values are multiplied by -2 okay so again you have the same period so you have the same X intercepts so you're can have this guy this guy and this guy those are going to be the same right y values are still going to be zero but when you look at this guy and this guy okay you can go from your basic sign graph and you can just multiply those y values by -2 so one multiplied by -2 gives you this -2 and then -1 * -2 is going to give you this positive2 okay and if you think about this in terms of your f ofx equals 2 * the sin of X again the negative there if you have the negative of f ofx whatever you're working with it's a reflection across the x-axis so if we go back and look at this guy one more time you can see that we have this peak here okay where your x value is Pi / 2 and your y value is two okay so this is going to be reflected across the x axis and so now this is going to be your minimum point right so at pi/ 2 comma -2 and this guy right here which was your minimum right where you had 3 pi/ 2 comma -2 well now this is going to come up and be your maximum it's getting reflected across the x-axis so now it would be 3 piun / 2 comma pos2 if we go back you'll see that's exactly what we have here okay so let's make a few generalizations you should notice that this is -2 here and the maximum is still two and the minimum is is still -2 so what I would say here is that if you have this F ofx is equal to a * the S of X where again we're going to say a is not zero but it could be any real number you want other than that then basically if you think about the range okay and this is going to give us our maximum Y and our minimum y then what I do I'll put some brackets here and I'll say it's from the negative of the absolute value of a to the absolute value of a and I'm using these Square brackets here because these values are included right if we look at this -2 here the absolute value of that is 2 so you'd want -2 to positive2 again both are included if you went back to a half it would be negative a half to positive a half both are included if you had positive two again it would be negative2 to positive2 both are included if this was 57 then it would be Nega 57 to positive 57 you know so on and so forth all right the next concept I want to talk about is known as the amplitude this is something you need to know when you work with the the s or cosine function and it's going to be the same for the basic sign graph and also the cosine graph which we'll talk about later on so over one period let me write this out so F ofx equal the sin of X again X is greater than or equal to Z and less than or equal to 2 pi so we're looking at this right now and basically you'll see that I've written that the amplitude is one so what does that mean and where does it come from well basically when you think about the amplitude you're thinking about something that describes the height of the graph both above and below your horizontal line that passes through the middle of the graph so for the basic sign or cosine function that's going to be your x axis okay that's going to be your x axis so it would be the line Y equals 0 okay so from this guy right here to this guy right here your minimum y-value is Nega 1 it's a distance of one or you could say from right here to up here that's a distance of one as well however you want to draw it now the formula we use is basically to graph grab the a value and take the absolute value of that and that's going to be your amplitude here you can think about your a value being one okay so the absolute value of one is one so the amplitude is one a more precise definition comes from the fact that you're taking half of the difference between the maximum and minimum value so if you take your maximum which is one you subtract away your minimum which is negative 1 minus a negative is plus a positive so this becomes 1 + 1 which is two if you took half of that you're right back to one if you want to look at another example of this again we've been working with f ofx = 2 * the S of X again just take this number out in front which in this case is a two take the absolute value of it so if it was two or negative -2 the amplitude is just going to be that number and again you can see the distance from this guy right here down to your minimum okay that's going to be a distance of two units and again you could just go 2 minus a -2 that's going to be pos4 you half that and you're back to two okay if you wanted a more generic formula again you can go f ofx = a * the sin of X where a is not zero and you could say okay well I know the range I know the range from earlier is from the negative of the absolute value of a to the absolute value of a okay so that tells me that the minimum y value would be the negative of the absolute value of a so let's just go ahead and say the maximum y value is going to be the absolute value of a so a minus a negative a is going to be 2 a okay so this is 2 a and if I half that I just get a and of course we're going to use absolute value bars just to account for the fact that this guy could be positive it could be negative so we want to account for that so we use absolute value bars so your amplitude I'm just going to put is always found as the absolute value of a this guy out in front all right what I want to do is consider when the period for s or cosine is not 2 pi I'm going to work with the S function but this works the same for the cosine function so we have something like f ofx = a * the sign of now we have BX and I'm specifically going to work with B being greater than zero in our examples today you might get examples where B is less than zero I'll talk about how to deal with that but we're not going to get to that scenario until the next chapter okay so I'm going to put a comma here I'm also going to say that a is not equal to zero so any real number that's not zero so in order to consider how this is going to change our graph and how this is going to change our period I want you to First consider that if B is one that that's what we've been working with so f ofx is equal to atili the S of X this has a default period so this has a default period of 2 pi okay now where does that come from well it comes from the fact that as you rotate on your unit circle starting at an angle measure of zero in terms of radians around one rotation going counterclockwise to complete that rotation you end up with an angle measure of 2 pi okay so that's one cycle so we would say X is going to be greater than or equal to zero and less than or equal to 2 pi okay so this is what happens as long as you have this format it does not matter what the value for a is remember this is affecting us with a vertical stretch or vertical compression this could be negative it could be positive it could be a fraction it doesn't matter it will not change your period now when you consider that you're throwing this B into the mix now something is multiplying X okay so it's changing the speed at which you're going through one cycle okay so you can throw this into your inequality okay and you can basically say that I'm going to put a BX inside of here because my function is going to complete one cycle as BX now increases from 0 to 2 pi so this might give you a horizontal stretch or horizontal compression okay it's going to change the speed at which you complete one cycle and basically to figure out what your period is you're just going to divide everything here by B okay you're going to solve this for x 0id b well we know that b is not zero so this would be zero okay so let me write zero here this is less than or equal to well the B's would just cancel out B ID B is 1 so this is just X and then this is less than or equal to your 2 pi over B okay so we don't have any horizontal or vertical shifts going on your starting point for the cycle okay would still be zero for the x value and now your ending point for the cycle would be 2 pi over B okay so we'll update the period and just put 2 pi over B okay like that and this is what you want to use as a general formula okay let me get rid of this and I want to talk about some other things first and foremost if you are in this section in your textbook and they are giving you problems where B is negative you can just use the absolute value bars for right now when we get to the next section I'll show you how to rewrite this based on the fact that s is an odd function and cosine is an even function to where B is always greater than zero okay but for right now you can use the absolute value bars to get your period now what I said earlier was that this would create a horizontal stretch or a horizontal compression now where is that coming from let's think about a few basic examples let's say I said I had something like f ofx is equal to forget about the a we'll just keep that as one each time I'll say the S of 2x well let's first think about the period so the period okay would be what well I would have 2 pi over my B value this is my B value it's the coefficient of x so 2 pi over 2 well these are going to divide out it's just Pi so if I think about this it's completing a cycle okay starting with an x value of zero and ending with an x value of pi versus normally starting with an x value of zero and ending with an x value of 2 pi so it basically did it in half the time right so this is going to compress our graph horizontally okay so basically you can think about the fact that if your B is greater than one it is increasing the speed at which this BX here is going from 0 to 2 pi okay so obviously that is giving me a horizontal compression by a factor of 1 over B okay so this is horizontally compressed by a factor of 1 overb let me erase this and let's talk about the next situation so if this was a value between zero and one so a fractional value now it's slowing it down okay so I'm going to be horizontally stretched so let's say we looked at something like f ofx is equal to let's just do I'm going to keep my a value as one so I'll do sine of 12 * X well basically this is going to stretch me out okay by a factor of one over you would take this 1/2 here okay so you're going to flip this and this is going to be two okay so it's going to stretch me out by a factor of two again to think about this with the period your period is always going to be and I probably didn't write that correctly so let me put this back in here so the period would be 2 pi over B now if I'm dividing by half I might as well just multiply by two so this is just going to be four pi okay so before it went from 0 to 2 pi now it's going from 0 to 4 Pi so that is stretching the graph horizontally right it's taking twice as long so let's look at a simple little diagram here I think this will make a little bit of sense for you so we have F ofx equal a times the S of BX again we we don't have any vertical or horizontal shifts so the starting x value for your cycle is going to be at zero and then you would add quarter periods okay to get these other notches so you have your 2 pi over B you multiply by a fourth to get your quarter periods so you can cancel this with this okay and basically it's going to be pi over 2B okay that's what you're going to be adding each time so plus pi over 2B plus pi over 2B plus pi over 2B plus pi over 2B you're still going to have your same 3 x intercepts okay and essentially you're going to have your maximum and your minimum and you can get those from your amplitude which is the absolute value of a right there okay your maximum is going to occur at the absolute value of that gu guy and then the minimum is going to occur at the negative of the absolute value of that guy all right let's look at two examples here real quick I'll give you one where there's a horizontal compression and then I'll give you one where there's a horizontal stretch so we have F ofx = 3 * the S of 2x so the first thing I'd like to do if you're going to graph this find your amplitude Okay this is going to help you to figure out what's the maximum and the minimum so this guy again if this is three it's always the absolute value of that number so it's going to be three okay okay then find your period so what is the period here it's 2 pi over B in this case your B value is 2 so 2 pi over 2 so this is just going to be Pi okay so we know because this guy right here this two is greater than one we're going to have a horizontal compression by a factor of one over that number so by a factor of 1/2 and you can see our starting x value is zero our ending value is pi so this is being completed in terms of one cycle in half the time okay so once you have those two things you're going to go through and add these quarter periods let me erase this real quick and just put Pi here okay so a quarter period would be this guy divided by four so it's pi over 4 okay so we don't have any shifts yet so we're starting with an x value of zero then you add pi over 4 you had pi over 4 you had pi over 4 you had pi over 4 right so really easy you can see that Pi minus 0 okay gets you back to Pi which is your period that's always something you check I know it's not that important here but when you get into things like a phase shift which we'll talk about here shortly it's going to be important to do that okay so you have pi over 4 you have pi over 2 3 pi over 4 and then Pi okay so you know that at the beginning the middle and the end you have your X intercepts so the Y value is already figured out for you it's going to be zero when you get to a quarter of the way through and you get to three4 of the way through you know that you're going to be at your maximum for a quarter of the way through and your minimum for 3/4s of the way through through and again from your amplitude you know that's three so the maximum is at three okay that's your y value and your minimum is atg3 okay so that's enough to get you the five key points for this sign graph here and then just connect the points with a smooth S Wave that's all you need to do all right let's look at an example now where we're going to have a horizontal stretch so we're going to have F ofx = 3 * the sign of X over 3 and you might want to rewrite this I'm going to go f ofx is equal to I'll put my three times my sign of let me make that sign a little bit better my sign of instead of X over3 I'm going to put 1/3 being multiplied by X like this that way it's Crystal Clear that the B value here is a thir okay so let's start off by thinking about our amplitude so I'm going to put a m for amplitude obviously it's just the absolute value of this number so again it's just going to be three okay so that tells me the maximum and the minimum it's going to be three and neg3 right three for the maximum y value and then -3 for the minimum y value okay now when we think about our period here so when we think about our period we're going to have 2 pi over this B so 2 Pi / 1/3 basically you're going to flip that fraction okay and you're going to have 3 over one and you're going to multiply that by 2 pi so the period here is going to be 6 Pi okay so now if we think about one cycle now starting from this x value of zero out to this x value of 6 Pi so that's going to be one cycle for us and essentially we can see that it takes longer to go through this cycle right instead of 0 to 2 pi now it's 0 to 6 Pi so what's happening is this is going to be a horizontal stretch of the graph okay by a factor it's always by a factor of 1 over B here B is 1/3 so 1 over 1/3 is equal to 3 okay so horizontally stretched by a factor of three now when we talk about sketching the graph of this guy really all you need to do is get your key X values okay so you know this first one is going to be at zero and you know the yv value there is going to be zero so this is 0 comma 0 that's pretty easy to get from there you want to add quarter periods right so you're going to go 6 pi over 4 if I divide 6 by two I get three so let me write this as a three if I divide four by two I get two so 3 pi over 2 is what I'm adding each time so basically this notot right here would be an x value of 3 2 so this point up here would be 3 Pi / 2 and then I'm going to be at the maximum which is three okay so always when this guy is positive I don't have any kind of reflection or anything like that where we saw earlier we would reflect across the xaxis if this was a negative three I know that a quarter way through I'm going to be at the maximum okay because this is positive three now halfway through right here I'm going to be back to my x intercept so to get that I'm just going to add another 3 pi over 2 to my 3 pi/ 2 that would give me 3 PI right so this point right here would be 3 Pi comma 0 okay it's another x intercept then I'm going to add another 3 pi over 2 to this so that's going to give me 9 pi over 2 okay so that's going to be my guy right here so this point right here would be 9 pi over 2 and then comma you're going to be at your minimum here that' be3 so again as long as this is positive here your minimum is going to occur 3/4 of the way through okay again if you get a negative here you got to reflect across the x-axis so you got to keep that in mind okay for right now this guy is positive so it's pretty easy then you add another 3 pi/ 2 to this 9 pi over 2 you're going to get your final x value of 6 pi and of course this is another x intercept so this is 6 Pi comma 0 okay so once you get these five key points this is enough to sketch one period of any sign function so again you're just going to connect the points with a smooth sign wave all right so we have two Transformations left that we want to talk about one one is going to be a horizontal shift so either shifting to the left or the right when we talk about this with the circular functions like s or cosine this is called a phase shift we also want to talk about a vertical shift so shifting up or down that's a lot easier to deal with so we'll deal with that in a moment okay so right now we're going to look at f ofx equals the a Time the S of you're going to have this BX minus C inside of parenthesis and for right now we're going to work with examples where B is greater than zero just to keep things simple again as we get into the next chapter I'll talk about what to do when B is less than zero and also what I'm going to do here is I'm going to say that a is not equal to zero that should be obvious but I want to list that as a restriction just so it's clear okay now when we talk about the phase shift I want to First go back to something we learned a long time ago just to jog your memory so you know if you have something like f ofx is equal to let's say the sare root of x and you compare that to G ofx which is equal to the sare root of let's go x - 2 let's go H ofx let's let's say this is the square < TK of x + 2 okay remember when things are happening inside the function it's a little bit counterintuitive right when you see X values you're thinking horizontal axis so I'm moving left or right when you see this minus two most people assume that means that you have a shift based on this original graph F ofx equal x that you're going to go two units to the left but in fact it's the opposite this guy is going to shift this would shift two units to the right okay now what causes that well you have to think about the fact that to undo what's being done to X we've got to add two right that gets me back to where I was in other words one ordered pair let's say 0 comma 0 that's on this guy to get the same yalue for this guy the x value has to be two units larger so it would have to be 2 comma 0 okay if you wanted to do another one let's say you did for this guy something like 9 comma 3 okay for this guy again the x value has to be 2 units larger so so now it would have to be 11 comma 3 okay because you've got to go 11 - 2 to get to 9 and then square of 9 would get you that three here it's just square of 9 which is three if you look at this guy again what's being done I'm adding two so to undo that I've got to subtract two so that means I'm shifting two units to the left so I'm going to let me change my marker here I'm going to shift two units left okay so you got to keep that straight in your mind it's the opposite of what you think right you got to undo what's being done to X now let me erase this and what I'm going to do is I'm going to write this in a different way okay so I'm going to write f ofx is equal to a times the S of I'm going to factor this B out now you might be saying well there's no B over here so how are you going to do that all you need to do if you want to factor that out is basically divide each term by B so in other words I can put a b out here and then inside the parentheses I can go ahead and say well BX / B is X and then minus the C / B would be C over B okay so this is going to make your shift a lot more clear you can do it from this form if you want most people do but I advise if you're having trouble with this to write it like this I can also say that a is not equal to zero and I can say that b is greater than zero for our examples and you might want to make this crystal clear and put some brackets here they're not needed but that makes it a little bit cleaner okay so if I look at it in this form I know that if C is positive and B is positive that basically I'm subtracting away a positive number okay and I'm going to be shifting to the right by C over B units okay it's very very simple if for example this was a negative C I know B is positive because I've made it that way well then you have minus a negative which is plus a positive right so I could really write this as plus C over B in this case I'm going to shift to the left okay by C over B units or you could say really to make that Crystal Clear since I said that C was negative by the absolute value of C over B units okay to make that Crystal Clear let's put this back to this original form now if you want to work with it in this form it's really the same thing okay all you would do is start off with BX minus C and you could set that equal to zero and you can say well how do I undo what's being done to X well what I could do is I could add C to both sides okay cancel that and say I have BX is equal to C divide both sides by B and I'll say have X is equal to C over B so that's going to be your phase shift okay again dealing with whether it's going left or right is just based on the sign that you have if B again is restricted to be positive well then if C is positive and you have it in this form you're going to be shifting to the right by C overb units okay if C is negative meaning you have plus a positive you're going to go to the left by the absolute value of C over B units okay this one's a little bit tricky when you think about the signs when you're in this generic form so let's let's go ahead and look at an example I think it'll be a little bit clearer okay so let's look at this F ofx = 4 * the S of we have the 2x - 2 pi over 3 so if you wanted to you could rewrite this I'll go f ofx is equal to You'll Go 4 times the S of I would factor out the two okay and I would put that out in front and then inside you're going to divide everything by two so if I divide 2x by two I get X and then minus if I divide 2 pi over 3 by two it's like multiplying by a half basically you would cancel that two out with the two in the denominator so you would have your pi over 3 okay so really when you look at it you can just go back to my simple example with the square root you have this minus Pi over3 so your phase shift is just going to be Pi over3 units to the right okay if this was a plus here I would be going pi over 3 units to the left okay that's how I would write it it's the simplest way to do it if you don't want to go through all this trouble of factoring it you can always say that I want to do my phase shift as C over B okay so my C is going to be 2 pi over three let me write this out so my C is 2 pi over 3 and then divided by B so I'm dividing by two okay so it's like multiplying by half and again this two would cancel with this two and you end up with pi over 3 and again because this was positive here what that means is that we're going to be shifting to the right by pi over3 units okay so let me write out here that my phase shift is pi over 3 and this is to the right so Pi over3 you could put units if you want and you will put right okay so you can see that by looking at the graph normally we start at 0 comma 0 now we've been shifted pi over 3 units to the right so the first guy for this cycle to start it off the x value is pi over3 and the Y value is zero so pi over 3 comma 0 okay now now when we talk about the other five key X values from this point it's the exact same procedure okay same song and dance so I'm going to find my amplitude again that's super easy to do absolute value of this number out in front so the absolute value of four is four okay and this is not a negative it's a positive so you know your maximum is going to occur at the quarter way through the period your minimum is going to occur at 34s of the way through the period so we'll get to that in a moment but you know basically the yvalue here would be four let me go ahead and write it out and the yv value here would be ne4 okay so that's pretty easy you also know that your yvalue at all your intercepts okay which is going to occur at the halfway point and at the ending point is going to be zero so that's pretty easy too so let me go ahead and put those in now how do we get these X values how do I get this one and this one and this one and this one again you're just going to add quarter periods so we first need to find the period to do that so let's go ahead and find the period And basically if we look at it we have a b value of two so we have 2 pi over 2 which is just going to be Pi okay so if you wanted quarter periods it's pi over 4 so I'm going to add increments of pi over 4 okay and you can see I already have this listed this is something you would have to do with either a calculator or by hand so basically pi over 3 plus pi over 4 is going to give me this 7 Pi over2 so this is 7 pi over 12 add pi over 4 again and I'm going to get this 5 pi over 6 so 5 piun over 6 add Pi over4 again I'm going to get this 13 pi over 12 so let me write that in there so this is going to be 13 pi over 12 and then add pi over 4 one more time and I'm going to get to 4 pi over 3 so 4 pi over 3 now I talked about this earlier and it might have not made sense at the time but when you have a phase shift you want to take this guy and this guy and you want to do a little subtraction and make sure it equals the period that way you know you didn't mess up when you were adding quarter periods okay so 4 pi over 3 minus pi over 3 does give me the period of Pi so I know I'm good to go right I have all my points and then I can just sketch My Graph by making a smooth sine wave through those points now if you wanted to think about all the Transformations that have occurred some teachers will ask you for that basically you think about okay it's been vertically stretched by a factor of four it's been horizontally compressed by a factor of a half right so remember it's one over B your B value here is two so one over two or a half okay this guy is basically running through a cycle in half the time your phase shift is going to be Pi over3 units to the right all right let's talk about the last graphing transformation which is basically where we're going to shift up or down okay so a vertical shift we have something like f ofx = a Time the S of BX minus C and then plus d now we're still going to say B is greater than zero again we're going to deal with B less than zero in the next chapter and I'm going to specifically state that a is not allowed to be zero this is just for clarity okay so here because of this plus d this is a vertical shift you're going to go up by D unit if D is positive you're going to go down by the absolute value of D units if D is negative let's go ahead and look at an example we have F ofx equal 4 let me write this out here so F ofx = 4 then multiplied by the sign of you're going to have this 3x + 3 pi over 4 I'm going to factor that three out so inside instead of 3x I'm going to divide by 3 I'm going to have X then Plus instead of 3 pi over 4 let me write this out so nobody gets lost I'm going to divide by three so that's like multiplying by a thir so you're going to see that the threes are going to cancel and you're going to have pi over 4 right here let's write that in close that down okay let me scooch this over so everything fits you also have this minus one here so we'll put that out there if you were to think about this let me make that a little bit cleaner if you were to think about this in terms of all the things that have been done based on the original sign graph that we were working with well you have a vertical stretch by a factor of four okay you have a horizontal compression by a factor of 1/3 so 1 over B in this case b is three you have a phase shift left by pi over4 units because you have X plus pi over4 inside here okay so you're going left by pi over4 units and then you're going down by one unit that's what that negative 1 is going to do okay so if we think about this we can get our amplitude again we know that that is going to come from this number right here the absolute value of four is four and normally if we don't have a vertical shift that's going to tell us the maximum and the minimum right so we would know it would be 4 and-4 so the range here is going to change because of this minus one instead of four and4 what you're going to have is everything shifted down by a unit so now instead of four for the maximum yvalue you're going to have three instead of4 for the minimum yvalue you're going to haveg five so your range is going to be from neg5 to positive3 okay so keep that in mind now when you talk about your period okay your period that's going to be 2 pi over B and B is just three so that's pretty easy and then you're also going to need the quarter periods so 2 pi over 3 * a 4th let's go ahead and cancel this two with this four and get a two so your quarter periods here are going to be pi over 6 so I'm going to put quarter quarter period is just going to be pi/ 6 okay so let's talk about how we can get this graph if you're trying to do this quick remember you start off with 0 comma 0 on the basic sign graph so think about all the things that are going to be done to this guy you're going to have a phase shift of Pi over4 units to the left so that's going to take me this way by pi over4 units so that point would be here now and then we're shifting down by one unit so that Point's going to be right there okay so I got moved from 0 comma 0 to that point which is going to beunk over4 comma the Y value instead of it being zero right normally that's the x intercept it's been shifted down by a unit so this is going to be-1 okay now as we go to the next X value so you've got this one which is at the quarter this one which is at the half this one at the 3/4 and this one at the full period we're just adding quarter periods right so we're adding pi over 6 to get to the next X value okay so when I add pi over 6 to negative pi over4 I'm going to get piun over 12 okay so this is piun over 12 how do I get my yvalue again this is positive here so I know at a quarter way through I'm at the maximum and the maximum is going to be three okay so pretty easy there then as I go back this would have been an X intercept right but it's not it's been shifted down by one unit so I know the Y value here is going to be NE 1 that's pretty obvious and then for the x value I just take my negative pi over 12 and again I'm just going to add my pi over 6 just add a quarter period and that's going to put me at pi over 12 so this one is pi over 12 right there okay okay and then at 3/4 of the period I know I'm going to be at my minimum right the minimum is5 so I'm going to put a neg five in there okay and for this guy I just take my pi over 12 and I add another quarter period so pi over 6 and that's going to give me this pi over 4 so this is pi over 4 comma -5 then to get my last Point here in this cycle what I would want to do is add pi over 4 and pi/ 6 that's going to give me 5 piun / 12 and I know the Y value that would have been an X intercept it's been shifted down by a unit so instead of zero it's going to be negative 1 okay so that's how you can get those Five Points so this one this one this one this one and this one and so that's going to be one cycle or one period of this graph just connect these guys with a smooth sine wave and you're basically good to go if you want to go to the right or to the left you're just making repetitions of that one cycle all right now that we've mastered graphing the S function we're going to move on now and talk about graphing the cosine function so the reason I spent all of the lessons so far on the sign function it's a little bit easier to start with one function learn all of the graphing Transformations and then take that knowledge and apply it to the other one okay it's the same process so if I think about f ofx is equal to the cosine of x how will this graph be different from f ofx equals the sin of X well basically you take your s graph and you do a phase shift of pi/ 2 units to the left and you're going to have your cosine graph you'll see that in a moment so all we really need to think about is how this guy is going to be different in terms of the five key X values that you pick what are the Y values there how is that going to be different once you memorize that you just take the knowledge from the sign graph that you learn those strategies those techniques and you apply to the cosine graph and you're good to go so I'm just going to think about these quadrantal angles we start with an x value of zero would produce a cosine value or a y value of one one okay then as we go to Pi / 2 we're decreasing right so we're going to end up at zero when our x value or our angle is Pi / 2 then as we continue to Pi we're decreasing from zero down to Nega 1 okay and then as we go from PI to 3 pi over 2 we are increasing from netive - 1 up to zero okay so right there and then lastly as we go from 3 pi over 2 to our 2 pi right we're making one full rotation we are increasing from 0o to 1 okay so you can refer to this table if you want to have some notes on that now when we think about this guy over one period so f ofx is equal to I'm going to put cosine of x I'll say that 0 is less than or equal to X Which is less than or equal to 2 pi so this is just one period you want to think about how these points are going to be different from the S graph if you memorize those differences you're basically good to go your period here is the same it's still going to be 2 pi we should know that from the unit circle now you'll notice that you have have a maximum here and a maximum here and this if I draw a line down the middle I want you to notice that you could take this point and reflect it across this line and you'd end up with this point remember that is an even function right so where f ofx is equal to F ofx we've seen this with the squaring function right if you saw something like f ofx = x^2 doesn't matter if I plug in a two there 2^ squ is four or a negative -2 -22 is also four you get the same y value okay so cosine is going to be an even function we'll talk more about that in a moment okay so another thing you want to think about is the fact that with the S graph remember you had three x intercepts you had an X intercept at the beginning the middle and the end okay here you only have two x intercepts and it's going to be at the quarter of a period and 3/4s of a period okay and then your minimum occurs at halfway okay so you want to remember that you start at the maximum you end at the maximum halfway through you're at the minimum okay and a quarter way through and 3/4 way through you're at the x intercept that's really all you need to memorize you just take the other information from the sign graph and you're good to go now let's extend this to the left and right so this is f ofx equals the cosine of x so here's going to be one period from here to here okay so one period again this is just the sign graph that's been phase shifted to the left by pi over 2 units okay that's all it really is it's the same type of graph now a couple of key things this f ofx equals the cosine of x it's continuous across its entire domain so just like with f ofx equals the S of X again the domain for each is going to be all real numbers the period for each is going to be 2 pi so that stuff's the same the range is negative 1 to positive 1 that's the same the X intercepts this is going to be different so the X intercepts is going to be of the form you're going to have 2 n + 1 * pi over two okay I should probably move this so it's a little bit more clean let me move this down there and basically n is any integer let me make that a little bit better because that's a little bit bad in terms of handwriting let's say n was Zero 2 * 0 is z so you'd have 1 * pi over 2 so that's this x intercept here okay if you put in a one you'd have 2+ 1 is 3 3 * pi over 2 is 3 Pi over2 so on and so forth now the last thing I want to do is talk about the fact that this guy is an even function we already saw that right when I basically drew a vertical line at xals Pi but you should know that basically with cosine of x you could say that the cosine ofx is equal to the cosine of x we can go back to the unit circle and you can grab a value real quick let's say you grab pi/ 6 and you can say the cosine of pi/ 6 is equal to the cosine of pi/ 6 really easy to prove that cosine of pi 6 isk 3 over2 and the cosine ofk 6 I'm just rotating clockwise now so I'm co-terminal with 11 piun 6 so this would be 32 as well okay so cosine is an even function okay let's take a look at one example with graphing cosine I'm going to put all the different Transformations into this guy again if you can do this for sign you can already do it for cosine now what I'm going to do is rewrite this to start you have F ofx equal 3 * the cosine of 3x + 2 piun over 3 then you have this plus two so let me write f X is equal to I've got 3 * the cosine of I'm going to go ahead and Factor this three out again I think this is the simplest way to identify the phase shift but you don't need it so it's up to you so three times inside the parentheses divide everything by three 3x ID 3 is X then plus if I had 2 pi over 3 and I multiplied by a thir same thing as dividing by 3 this would be 2 pi over 9 okay close that down and then don't forget your plus two at the end okay so now that we have this set up we can identify what's going on here we know that this three out in front the absolute value of that guy which is just three is the amplitude Okay so that's going to be three now with the normal cosine graph just like a normal sign graph it's going to tell you your minimum is atg3 and your maximum is at positive three but if you look at the graph that's already drawn you see the minimum is at negative 1 and the maximum is at five so what's going on you have this plus two at the end okay and that's giving you a vertical shift up so instead of -3 for the minimum you add two units and you get to 1 so if I think about my range okay the minimum is now going to be Nega 1 the maximum again it would have been three add two units now it's going to be five okay we're going to use this in a little while when we start getting these points now additionally when I think about the period it's going to be 2 pi ID B now B here is three so just 2 pi over 3 if you want the quarter periods again this is for us to get those X values well it's just 2 piun over 3 multiplied by a 4th so basically this cancels with this and gives you a two so this is pi over 6 okay so we're going to have pi over 6 for the quarter periods and then when we think about the phase shift so let me put the phase shift in here you have X plus 2 pi over 9 so again if you have plus 2 pi over 9 I'm shifting to the left by 2 pi over9 units so that means I'm going to shift to the left by 2 pi over 9 units now the last thing I want to put in here and I'm running out of R so let me just put it off to the side my vertical shift as we talked about earlier I have a plus two here so that's up two units okay so now that we've identified all these things from just looking at our function we can go through and think about how our points would have moved based on the original cosine graph so it's really really easy okay okay now what I would say is that this guy based on the original cosine graph has been stretched vertically by a factor of three from that guy it's been compressed horizontally by a factor of 1 over 3 it's always 1 over b b is three so it's been compressed horizontally by a factor of 1/3 and then we've had a phase shift of 2 pi over9 units left and a vertical shift up by two units okay so that's all our different Transformations so if we go to our original cosine graph remember the cycle starts at the maximum the x value is zero and the Y value is one okay so based on this function here I know I'm going to start at the maximum the maximum is five so I know the yvalue for this first point is going to be five what's the x value going to be well I'm just going to consider this phase shift which is 2 pi over 9 units left so 2 pi over 9 units left so that would give me a - 2 pi over9 okay and then comma five so that's going to be this point right up here okay that's going to be the first point we get so this is negative let me put my 2 pi over 9 comma five okay so pretty easy to get that point now when you go to get the other points what I would suggest is just getting your yv values first they're really easy to figure out remember at the beginning of the period and at the end of the period you're at the maximum so here's your beginning and here's your end okay so you're going to be at the maximum so let me put a comma here and then put a five okay so I know the yv value There is five I know that halfway through I'm at the minimum the minimum is netive 1 we already figured that out so that's going to be right here let me put a comma in the ne 1 at a quarter way through the period and three quarters way through the period you would have had an X intercept but because we have this plus two or this vertical shift up by two units now the y-coordinate is going to be two okay so for right here the y-coordinate is going to be two and for right here the y coordinate is going to be two again if you find this confusing what I'm doing you can always take your X values plug them into the function and evaluate and you can get your yv values that way this is just quicker for me and for most people as they sketch the graph now to get these four missing X values you already have the first one you're just going to add quarter periods okay so I'm going to add piun over 6 to the existing x value so -2 piun over 9 + piun over 6 is going to give me piun over 18 so piun over 18 okay so that's going to be the x value for this point right here okay so 18 comma 2 then as I get to my minimum I add another pi/ 6 to thisk over 18 and what I'm going to get is pi over 9 so this point would be pi over 9 comma 1 then I add another pi over 6 to this pi over 9 and I'm going to get to 5 piun over 18 so 5 pi over 18 comma 2 and then lastly I add another pi over 6 and I'm going to get to the x value for this point which is going to be 4 pi over 9 let me make that a little bit better okay let me make that a little bit better so 4 pi over 9 comma 5 Okay so this would be one cycle okay for our graph and again if you want to keep going to the left to the right you're just making repetitions of that one cycle in this lesson we're going to talk about graphing tangent and cotangent all right so in the last lesson we talked about how to sketch the graphs for S and cosine and we found that the process is really not that bad the main thing to understand when you're in this section is that you are not expected to make a perfect graph you're just expected to know the key information and how you could sketch the graph if you had to generally speaking we're always going to use computer software either something that's online or graphing calculator like a ti3 to sketch our graphs so with that being said let's go through how to sketch the graph for tangent and cotangent we're going to start with a different period so basically we know that the period for S and cosine is 2 pi but when you work with Tangent and cotangent you need to know that the period is now going to be Pi that's extremely important especially when we get to solving trigonometric equations so something like the tangent of X is going to be equal to the tangent of X plus pi n where n is any integer and we would also say that X is a member of the domain remember that when you work with Tangent you have places where it's undefined then we have cotangent of X is equal to the cotangent of this X plus pi n again where n is an integer and X is a member of the domain so let me just quickly demonstrate that the period for tangent and also cotangent is going to be Pi you can show this with identities but we haven't gotten that far yet first off I'm just going to pick something like pi/ 6 for the angle so this guy right there and you know that if you want the tangent of pi/ 6 you can find it as the S of pi/ 6 / the cosine of pi over 6 in other words we should know that the tangent of X is equal to your s of x X over your cosine of x so the tangent of pi/ 6 is going to be equal to the S of Pi 6 and then this would be divided by the cosine of pi/ 6 so what would this be equal to we know that the S of Pi / 6 again that's going to be given by the y coordinate right there so that's 1 12 so let's write that how do we divide fractions we multiply by the reciprocal so this is times you would have the cosine of 6 that's 32 but you're going to find the reciprocal of that so this is going to be time 2 over the < TK of 3 and then basically the twos are going to cancel and let's say that this is 1 over the < TK of 3 we want to rationalize the denominator here so times the < TK of 3 over the < TK of 3 so this equals theare < TK of 3 over 3 so this is the square < TK of 3 over3 now I want you to notice that if you add Pi or you could say if you rotate by pi so I'm going to come over here into quadrant 3 three and we're going to be at 7 piun 6 now I want you to think about if I found the tangent so the tangent of 7 pi/ 6 what would I get well you're going to find that you're going to get that same square of 3 over3 and the reason for that is you're dividing in terms of absolute values the same numbers it's just that now you're going to be dividing a negative by a negative which is also going to give you a positive so in other words this is going to turn into your s value is -2 and then divided by your cosine value which is3 32 we're going to again multiply by the reciprocal so let's go negative we're going to go 2 over the < TK of 3 and so the negative * the negative will give you a positive the twos will cancel so this would end up being 1 over the < TK 3 which we know if we rationalize the denominator that's going to be the square 3 over 3 so you can see that the tangent of piun / 6 and the tangent of 7 piun / 6 that that's the same so that's one way you can demonstrate this I'll show you with some other ones just so you can see the stuff in quadrant 2 and four so that's where tangent is negative so let's say you pick something like let's say 2 pi over 3 so that guy right there and that color doesn't show up so well so let me do that like that and again if you add pi to this so if you add Pi you're going to come into Quadrant 4 and you're going to be at this 5 piun over 3 so it should be true that the tangent of 2 piun 3 is equal to the tangent of 5 5 Pi 3 because again I added Pi every time I add Pi I'm coming back to the same tangent value so let's just try that real quick so we have the tangent of 2 piun over 3 that would be equal to the S of 2 piun over 3 which is the square < TK of 3 over 2 and then divid by the cosine of 2 piun over 3 which is -2 so I'm going to multiply by the reciprocal so times 2 over -1 and of course the twos are going to cancel and you have 3 over1 Let's just say that's the negative of thek of 3 then if I come down here to 5 piun over 3 again if I look for the tangent of 5 piun over 3 it's going to be the same thing I'm going to be dividing opposites here it was basically a positive s value divided by a negative cosine value but here it's going to be a negative s value divided by a positive cosine value well we know if we have a positive divided by a negative or a negative divided by positive I'm going to get a negative so that's what's happening here so this is just going to be my Nega < TK of 3/2 that's the S value here divided by the cosine value which is pos2 so again I'm going to multiply by the reciprocal so time 2 over 1 and let's go ahead and cancel this with this and so you get < TK of 3 so hopefully that gives you a little bit of insight into why the period is going to be Pi all right so what we're going to do now is just look at the basic graph for tangent we're going to do this by grabbing some ordered pairs and just sketching this guy from negative pi p 2 to Pi / 2 with both of those guys being excluded before I show you the graph I want to cover a few things that basically cause a lot of confusion when you first see these graphs the first thing is we should remember that when we see a negative angle we're rotating clockwise in other words something like NE pi/ 2 that means I'm starting here and I'm rotating clockwise so it's going to be this guy right here it would be co-terminal with 3 piun / 2 so this is piun / 2 now why does it say that it's undefined here we know that the tangent of X is equal to we have the S of X over the cosine of x well if I want the tangent of < / 2 that would be equal to the S of piun 2 over the cosine of piun / 2 since this piun over 2 is coterminal with 3 piun over 2 we can just take the value for S and cosine from this guy right here and so basically you could say that this is equal to the S of ofun 2 well that's 1 and then the cosine of piun 2 well that's 0 what do we know about division by 0 it's undefined so basically you stop at this point and you would say that this is undefined undefined so this would always be true whenever the cosine value is zero so basically this is going to happen at Pi / 2 then 3 Pi / 2 then 5 piun over 2 7 piun over 2 9 pi over 2 so on and so forth but also you have to consider that would happen at piun / 23 piun / 2 5 piun over 2 so on and so forth so a lot of books will cover this by saying that this guy is going to be undefined at odd multiples of Pi / 2 so that's a good way to remember it you could also write it as if you want to do it this way you could say that it's undefined at Pi / 2 so this guy right here plus pi n where n is an integer so basically if it's a positive integer then I'm rotating counterclockwise so let's say I add Pi that's going to put me right here at 3 pi/ 2 and if I add Pi again then I'd be at 5 pi over 2 then 7 pi over 2 so on and so forth if n was a negative integer well then basically now I have a clockwise rotation so let's say I subtract pi and I rotate clockwise like this well now I'm not at 3 pi/ 2 I'm going to be at negative pi over 2 and if I do that again and subtract away another pi well now I'm going to be AT3 piun / 2 and so on and so forth so all of those places it's going to be undefined because we're going to have a cosine value of zero and division by zero is undefined now the other thing you need to think about is that this guy is going to be zero when the sign value is zero and the cosine value is not zero so something like let's say add zero you have a sign value of 0 and the cosine value is one then at Pi you have a s value of zero and a cosine value of negative 1 so basically for this you could say that the tangent value is going to be zero at Pi n where n is any integer in other words you could start here and you could add pi and you would get to Pi you could add Pi again and you would get to 2 pi and then 3 pi and then four Pi so on and so forth and also if you wanted to start here and rotate clockwise so if you wanted to subtract Pi away well now I'm going to be at netive Pi then if I do that again I'm going to be at -2 Pi then -3 Pi so on and so forth so basically that's where you're going to have a tangent value of zero or you could say where your x intercept is going to be and you can see on the table that we've made that we do have this 0 comma 0 okay so let's look at a basic sketch for yal tangent of X so just using those points that we just had in our little table so you have something like piun over 3 comma < 3 so that would be this point right here then you have piun over 4 comma 1 so that' be this point right here so this point is actually one we're going to focus on it'll be important for of course later on then we have piun / 6 comma 3 over3 so that's this point right here then we have our 0 comma 0 so that's our x intercept and we'll focus on that one later on then you have your P Pi 6 comma 3 over 3 then you have your P Pi 4 comma 1 again this is one we're going to focus on and then we have our pi over 3 comma 3 so that's going to be this guy right there so one of the things to focus on here is you do have a vertical ASM toote at this pi/ 2 and then also Pi / 2 and that's being created because again at Pi / 2 and Pi / 2 you have division by zero which is undefined so we've seen this previously in our course when we look at something like yal 1 /x we know that if you plugged in a zero for X there this is undefined so if you look at the graph of that you're going to end up with a vertical ASM toote at x equals 0 so here when you have y equals the tangent of X the tangent of X which you could really write as the S of X so the S of X over the cosine of x to make it more clear well again where this is zero at NE Pi / 2 or Pi / 2 something like that 3 pi over 2 5 piun over 2 7 Pi / 2 so on and so forth well you're getting division by zero so you're getting these vertical ASM tootes now when you think about the behavior as this guy right here approaches pi over two coming from the left so coming from the left basically as we approach this x value of Pi / 2 this guy is going to increase without bound as they say so approach positive Infinity so I have another little table for you a lot of books will this so basically at an x value of 0 we get a yv value of 0 at pi/ 6 you get about 0.58 that's just approximating 3 over 3 at pi over 4 you get 1 at pi over 3 you get about 1.73 that's approximating the square root of three and then we have 1.5 is about 14.10 1.53 is about 24.50 1.57 is about 1,255 77 and then this 1.57 7 is about 10,381 33 and then notice that as you get really close to pi/ 2 so 1.57 796 you get 3,60 2331 so you can get as close to this as you want you just can't touch it when you actually get to pi/ 2 you're going to get undefined but as you get closer and closer to it this number is going to get larger and larger so again it's going to approach positive infinity or what they'll say is increase without bound now similarly as you approach negative pi/ 2 coming from the right this guy is going to decrease without bound so basically it's going to approach negative Infinity all right so just like we saw with graphing s or cosine after we did one cycle of the graph we were able to continue that to the left or to the right as much as we wanted so here I have y equals tangent of X so I've extended this to the left and to the right a little bit just so you can see a little bit more so basically this is what we graphed already so from piun / 2 to piun / 2 we see that we have a vertical asmp toote at piun 2 and then also another one at Pi / 2 those are what we call consecutive vertical ASM tootes so when you graph a variation of yal tangent of X you're going to first find those guys I'll tell you how to do that in a moment and then basically you have these three points so this one right here is at a quarter of the way through and let me just label this as a for right now so a let's just say that's PK over 4 comma 1 that's definitely something you want to memorize then let's call this B this is going to be the half way point or the Midway as a lot of books call it that's your x intercept so that's 0 comma 0 and then let's call this point C so that's 3/4 of the way through so basically that point is pi over 4 comma 1 so you can use these points to basically graph any variation of yal tangent of x if you want these points here or here or wherever it is if you just want this one right here remember the period for tangent is pi so I could take the x value here let's just call this point d let's call this point e and let's call this point F so if I wanted those let's just go ahead and say that for Point D well the y- Val is going to be the same all I have to do is add pi to the x value so pi over 4 pi over 4 plus pi so 4 pi over 4 to have a common denominator and that's going to be 3 pi over 4 so this would be 3 piun over 4 and then common 1 then for E I know that basically I would just take zero and add Pi so that would be pi comma 0 y coordinate staying the same and then this is going to be for f again I'm just going to take this pi over 4 so pi over 4 I'm going to add Pi so 4 pi over 4 so that's 5 pi over 4 and then basically you're going to have the same y-coordinate so it's still going to be one and you can basically use that process to extend the graph to the left or to the right as much as you want all right so just some information that you need to know so for the tangent function the domain contains all real numbers except odd multiples of Pi / 2 some books will write that as Pi / 2 plus pi n where n is in the integer so basically it's just telling you things like Pi / 2 3 pi over 2 5 pi/ 2 7 pi over 2 you could also have negative Pi / 23 pi/ 2 so on and so forth we have f ofx is equal to tangent of X you can also write yal tangent of X is discontinuous at odd multiples of Pi / 2 so basically we've seen that the graph contains vertical ASM tootes at those values the X intercepts are of the form xal Pi n where n is an integer we know that that's because that's where the sign values are zero and 0 divided by something that's not zero gives you zero so that's why that occurs the x intercept occurs at half of the way between consecutive vertical asmp tootes so if we're using yal tangent of X over that interval that we talked about so from piun / 2 to pi/ 2 with both being excluded that's going to occur at 0 comma 0 then we have our point on the graph that's a fourth of the way between consecutive vertical asmp tootes this is going to have a y-coordinate of ne1 so again using that example yal tangent of X over the interval from piun 2 to piun / 2 both excluded we have that piun 4 comma 1 and then the point of the graph 3/4s of the way between consecutive vertical asmp tootes has a y-coordinate of one so again using yal tangent of X over that interval pi2 to Pi 2 both being excluded we have that Pi / 4 comma 1 so we already know that the period is pi and then there are no minimum or maximum values so that's important we saw that the range is from negative Infinity to positive Infinity so it tells us that the graph has no amplitude so that's important because you might get that on your test it might ask you for the amplitude remember that when you think about the amplitude we learned in the last lesson when we're graphing s and cosine that's half the difference between the maximum and minimum values well here again there are no minimum or maximum values so you're not going to have an amplitude here then we can also say the graph is symmetric with respect to the origin remember that tells us we have an odd function so the tangent ofx is equal to the negative of the tangent of X for all X in the domain again f ofx equals the tangent of X is an odd function all right let's talk about the steps involved with graphing the tangent function so what I'm going to do is give you a simple model here and then if we get something that's a little bit more challenging we can just modify our steps so the first thing is we have y = a * the tangent of BX - c and we're specifically going to say here that b which is the coefficient of x is greater than zero of course we'll see scenarios where B is negative and I'll show you how to deal with that with identities for right now let's go ahead and talk about the period so what is the period again when we think about this from our s and cosine lesson we know that if we had something like y equals the S of BX well the period here so the period here was the standard period for S which is 2 pi divid b well it's a similar thought process here when you have y equals the tangent of X the standard period is now Pi not 2 pi but Pi so let me put that off to the side so y equals the tangent of X the period is pi so we know that well here the period for this guy right here because you have B that's multiplying X it's going to be pi over B now I have a little sample illustration here to go through the procedure so basically the first thing you're going to do you're going to find these two consecutive vertical ASM tootes and the way you do that is you go back to again y equal the tangent of X we know that we generally graph one cycle from piun / 2 to Pi / 2 with both being excluded so you have xal piun 2 and you have xal Pi / 2 so you could say that one cycle occurs where X is greater G thank 2 and less thank / 2 so thisk 2 andk / 2 those are consecutive vertical ASM tootes now we don't have an argument of X we have an argument of BX minus C so we just have to adjust this let me get rid of this real quick we're going to use the same thought process though we're going to say that we're going to do one cycle where bx- C is greater than pi/ 2 and less than Pi / 2 and we can find the vertical asmp tootes by saying that BX - c is equal Tok / 2 so that would be one of them we'll put and BX - c is equal to piun over two so that would give you this guy and this guy so the two consecutive vertical asmp tootes that you're looking for once you have that you need to find these three points so you're going to have one that's a quarter of the way halfway and 3/4 of the way the way you do that is you find your quarter period just like we did with s and cosine and so basically if the period is pi over B well you can multiply that by a 4th and you can say that the quarter period is going to be < over 4B so you take your leftmost vertical ASM toote so that's x equals some number you take that value and you add pi over 4 B to that and you're going to get the x value for the quarter of the way point then you add the quarter period you get the halfway point then you add the quarter period you get the three quarters of the way point and you could add a quarter period to get the next vertical ASM toote if you wanted to but again you should already know what that is because you already set up those two equations so the main thing is that you just have to find the X values there the Y values are very easy to find you can just use the fact that at a quarter of the way through the y-coordinate is negative a so whatever this is It's the negative of it halfway through it's going to be zero and then 3/4 of the way through is just going to be a again going back to what we saw earlier when we did y equals the tangent of X remember I told you to memorize these three points you have PK over4 comma -1 well you could put a one here if you want for clarity well it's the negative of one that's how you get that y-coordinate then you have another point which is 0 comma 0 that's just your x intercept and you have another point which is pi over 4 comma positive one okay so positive one there so I would just memorize those three points from this basic tangent function if you do that you're going to be good to go with any graph that you come across all right let's take a look at an example with the tangent function so we have f ofx = 2 and then tangent of 2x + 2 piun 3 and then + 1 so what I'm going to do is just use the procedure that we just talked about there's alternative approaches when I get done with this problem I'll show you another way that you could do it I'll just do a few points using graphing Transformations only the first thing is to think back to what we talked about a moment ago if we're graphing something like y equal the tangent of X generally speaking if you're graphing that over one period or if you say you want to graph one cycle of this guy well we know that we're going to do X is greater thunk / 2 and less than Pi / 2 we know there's a vertical asmp toote at piun / 2 and also at Pi / 2 because again the tangent of piun / 2 and the tangent of Pi / 2 is going to be undefined because we'd have division by zero so what you want to do is set up an inequality here the argument is not X like we have here the argument is 2x + 2i over 3 so this guy right here just gets plugged in for this right here so you're going to set up the inequality let me get rid of this where you say 2x + 2 piun over 3 is greater than PK / 2 and less than piun / 2 and I'm just leaving myself some space because we're going to multiply them all so we're going to solve this inequality for X so this is the goal right there so we do have some fractions involved I would personally like to clear those some people don't mind them but I find that working with fractions is very tedious so what I would do is look at the denominator here of two then you have a denominator of three then a denominator of two so the least common denominator is going to be 3 * 2 or six so just multiply each part here by six and then you can get rid of the fractions and you won't have to deal with them so basically 6 ID two would be three and so you'd have3 Pi there so this is less than this gets distributed to each guy so 6 * 2x is 12x and then plus you have 6 * 2 piun over 3 so this six would cancel with this three and give you a 2 2 * 2 pi is 4 Pi so let me write that there and then basically this is last than six would cancel with the two and give me a three so 3 * pi is 3 Pi at this point again if you're trying to isolate x what I would do is subtract 4 Pi away from each part so let me do that over here this is going to cancel on the left you have -4 piun - 3 piun that's -7 Pi this is less than in the middle you're going to have 12x Which is less than on the right you're can to have 3 piun - 4 piun that's piun again if you want to isolate X in the middle just divide each part by 12 so divide by 12 divide by 12 and divide by 12 so this right here is going to cancel let me slide down here a little bit and you're going to have -7 pi/2 on the left and this is less than x Which is less than pi over 12 so this is telling me that my function is going to go through one cycle from 7 piun 12 to piun / 12 where 7 piun 12 and piun 12 are my two consecutive vertical ASM tootes and let me come down to this page right here let me go back up and let's talk about the period so what is my period and then I'm also going to need to know what is my quarter period so the period again if you're looking at y equal the tangent of X well the period is just going to be Pi if I change this to yal the tangent of BX well now now provided that b is positive again if you don't have a positive B you can just use absolute value bars but provided that b is positive you'd say the period is pi over B so you're just dividing by whatever the coefficient of x is so here we have a coefficient of x that's two so the period is pi/ 2 and then the quarter period is < / 2 * a 4 so that's going to be P pi/ 8 so this is pi over 8 if you want to know if you got the right period let's come down here and paste this then and just put a little border around it you can actually subtract so you're going to go larger minus smaller and you should get Pi / 2 so < over 12 so that's for this guy right here and then you're going to go minus this 7 pi/ 12 convert this into plus minus a negative is plus a positive so this becomes you could write negative 1 there so -1 piun + 7 piun that's going to be POS 6 piun / 12 and of course this simplifies to Pi / 2 so that confirms the period right there so you have two consecutive vertical asmp tootes let me put vertical ASM tootes here you have one at x = -7 piun over 12 and then the other I'll just put and here x is equal to piun over 12 if you come to the graph and I have a link for you on Desmos if you want to get something that has more detail I have all the points labeled for you if you want to click on that this is what we've found so far so this is the vertical ASM toote and this is the leftmost one that we found so X = 7K 12 again if you plug that in for X there you're going to get undefined so that's why you're getting a vertical ASM toote there then this guy right here is going to be our X = piun /2 again if you plug that in for X there you're going to get undefined so you're getting this vertical ASM toote here our next goal is to figure out what is this point this point and this point so this is the quarter of the way through this is halfway or Midway and this is 3/4 of the way through so let me come back and let's think about how we can use this information so the first thing is you can find your x value vales by just adding quarter periods so you can start with this one x = -7k 12 add the quarter period of piun 8 and you just keep doing that to get the next X value and the next X value and the next X value the yv values you can either plug into the function or you can use graphing Transformations so I'll show you both let me go ahead and grab this real quick so you have -7 piun over2 and then plus piun over 8 I'm not going to have enough room to do this on the sheet so I'm actually going to go to another page if you're trying to find your x value you for the first quarter point let's multiply this by 3 over3 and let me slide this down I'm going to multiply this by 2 over2 just so we can have a common denominator if you notice you have a denominator of 12 and a denominator of 8 so this is basically 2^2 * 3 and this right here is 2 cubed so basically the LCD is going to be 24 it's going to be 8 * 3 so this right here would turn into -14 piun over 24 just multiplying 2 * 7 piun that's 14 piun pi and then 2 * 12 is 24 and then plus over here 3 * piun is 3 pi and this is over 8 * 3 that's 24 so if you do this - 14 Pi + 3 Pi that's 11 pi and this is over 24 okay let me write this right here so for my first quarter point the x value is -1 pi over 24 the quarter period we have that listed as pi/ 8 but again we could see that we could write that as 3 pi/ 24 so we have a common denominator so go ahead and add here so1 piun / 24 + 3 piun 24 that would be8 pi over 24 you can simplify that but I wouldn't do it yet let's come down here so this is for the halfway or Midway so basically this is going to be 8 piun 24 and if you simplify that 24 ID 8 is going to be 3 and 8id 8 is 1 so this becomes piun over three okay then we want our 3/4 of the way point so start with this and again just add a quarter period so this is plus 3 pi over 24 and let me slide this down over here so this should be -5 piun over 24 just doing 8 Pi + 3 Pi that's 5 pi over the common denominator of 24 so 5 pi over 24 now coming to the graph we can see that for the first quarter point the yv value is going to NE 1 for the halfway point it's going to be postive 1 and then for the 3/4 of the way point it's going to be three now how could we get that well essentially you could plug into the function or you could use graphing Transformations let me just show you with this one right here that you could do it that way and then I'll do the graphing Transformations because that's a little bit faster so coming back up here you would want to say what is f of -1 pi over 24 so you're just plugging in so this equals 2 * the tangent of you have 2 * -1 pi over 24 you're just plugging this right here in for X right there so that's all you're doing and then we'd say + 2 piun over 3 and then + 1 you can think about this canceling with this and giving me a 12 down here and then I could multiply this let me sco this down a little bit I could multiply this by 4 over 4 so I have a common denominator so this would be 11 pi over 12 and then plus this would be 8 pi over 12 so let me write that in and then you would have1 piun + 8 piun that's -3 piun so you have -3 piun over 123 piun over2 and of course that's going to simplify to piun over 4 so piun over 4 we know that the tangent ofun over 4 is 1 so basically this becomes 2 * -1 and then + 1 so 2 * -1 is -2 and then -2 + 1 is1 so so this equals -1 again if you go to the graph you can clearly see the y coordinate here is1 so let me put1 in right there now what's the other way to do this because that's pretty slow well you could have used graphing transformations in other words you know that on y equals the tangent of X you have these points which are for the quarter of the way through again it's piun over 4 comma 1 and then for halfway through it's going to be 0 comma 0 that's your x intercept and then for 3/4 of the way you're going to have < over 4A 1 well what's happening in terms of the vertical Transformations well you have this two out in front that's giving me a vertical stretch by a factor of two and then you have this plus one right here which is going to give me a shift up by one this is done first and this is done second so in other words if I take this let me move this down here a little bit so it fits so if I take this right here this y coordinate of -1 and I multiply by two first and then I add 1 second well that's -1 * 2 that's -2 and then -2 + 1 that's -1 if I go through this one and I go 0 * 2 + 1 well 0 * 2 is 0 + 1 is 1 positive 1 and then if I look at this one I have 1 * 2 + 1 so 1 * 2 is 2 2 + 1 is going to be 3 and that's exactly what we saw if you look at for the halfway point well basically the y coordinate is 1 if you look at the 3/4 of the way point the y-coordinate is three once you know those Y coordinates when you go to the right or to the left to extend the graph those are always going to be the same so for the first quarter point it's always going to be NE 1 for the halfway point it's always going to be positive 1 for the 3/4 of the way point it's always going to be positive 3 so you only need to figure out the x coordinates for the vertical asmp tootes and for the points that you're getting and you can just do that by continually adding quarter periods in other words in terms of getting the X values you could start here add a quarter period add a quarter period add a quarter period add a quarter period add a quarter period and you would just keep getting to the next guy so let's come through here and just put that this is one and then put that this is three so those are your points again you can get more points if you want that's going to be up to you I left you a very detailed graph on Desmos where all the points are filled in but basically I can come through here and try to fill this in it's going to be really tight this point right here let me just Arrow to it and say it's -1 pi over 24 and then comma -1 then you have your NE pi over 3 comma 1 so this right here is piun over3 comma postive 1 and then this one right here is5 piun over 24 comma 3 so again you can click that link and go to Desmos and see all of these points I've labeled them for you if you want to get more detail all right before I move on to the next problem I want to show you this using only graphing Transformations this is something you should be aware of in some cases it's faster you may prefer to do it this way but generally speaking it's very error because a lot of students have trouble with horizontal Transformations so things that affect the x coordinate so let's just look at F ofx equal 2 tangent of 2x + 2 piun over 3 and then + 1 and compare it to your f ofx is equal to your tangent of X so with this one let me actually grab this real quick and let me come down here and paste this in you have your basic points which we already know at this point let me put a little border here the first quarter so basically that's going to beunk over 4 comma -1 you have your halfway that's going to be 0 comma 0 and then you have your 3/4 and basically that's going to be pi over 4 comma 1 so you think about the transformation so you're going to have a new x coordinate and a new y-coordinate of course you also have the asmp tootes to consider so for that you just have to think about the X so You' have X is equal to PK / 2 and so you'd have X is equal to piun / 2 and these right here these are the vertical asmp tootes so let's come back up and the first thing I would do is I would Factor this so I would say f ofx is equal to 2 * the tangent of pull out that two there it's just going to make the phase shift a lot easier to find so pull that out and then you're going to have x + < over 3 and basically close that down and say this is + one I always like to check this so 2 * X is 2X and then plus 2 * piun 3 is 2 piun over 3 so that's good to go so here is where it starts to get a little bit confusing you have Transformations that are happening to the Y those are the vertical Transformations and you have Transformations that are happening to the X those are the horizontal Transformations the vertical Transformations are very straightforward if you have something like a * let's say f ofx plus b so let's say G of X is equal to this we already know that basically this right here is a vertical stretch if a is greater than one and it's a vertical compression or sometimes you'll say a vertical shrink if a is between Z and one then for this one right here this B this is giving me a shift up if B is positive and a shift down if B is negative so hopefully you know that already and in terms of the order we follow just the order of operations you do multiplication before you do addition in the order of operations so this is done first and this is done second so if I look at this this is done first and this is done second so we know that for the y-coordinates again we saw this earlier you're going to multiply by two first and then you're going to add one so if we think about those points and let me slide this over so this fits when I go over here and I do my transformation actually let me move this to another page because I don't think it's going to fit the way that it is basically you can get the Y values again you multiply by two first so 1 * 2 is -2 then you add one so -2 + 1 is -1 and then for this one 0 * 2 is still zero then you add one so that's one and then for this one 1 * 2 is 2 + 1 so that's three so that's very straightforward it's done almost instantly now let's come up and let's think about the harder guy which is going to be what's happening to the x coordinates basically what are the horizontal Transformations we know that if we have something like let's say G ofx is equal to F of let's go ax here this is going to be if a is greater than one you have a horizontal compression or you could say a horizontal shrink and then if a is between Z and one now you're going to have that horizontal stretch so it's always the opposite of what you think so basically here when you think about that two well that's something that's greater than one so that is going to give me a horizontal compression or you could say a horizontal shrink however you want to say that now this is the confusing part because there's different terminology some people will say that there's a horizontal compression by a factor of two if you want to say that that's fine typically I'll say it that way other people will say that it's been horizontally shrunk by a factor of 1/2 whatever you want to use in terms of the terminology just realize that you're going to either be dividing by two or multiplying by half whatever you want to do there and whatever you want to say just realize what's happening now this part right here this is your phase shift and again it's the opposite of what you think if you have X plus pi over 3 that means that I want to go Pi over3 units to the left I want to go minus Pi over3 it's always the opposite of what you see so what you would want to do here is actually do this one first and then this one second so why is that the case if we were think about the order of operations then you would want to do what's inside the parenthesis first and then what's outside which in this case would be multiplication second but again on the inside you have to reverse everything this looks like I'm multiplying by two well I have to do the opposite of that I have to divide by two this looks like I'm adding pi over 3 I have to do the opposite of that I have to subtract Pi 3 and then the same thing happens with the order of operations it has to go in reverse now so basically it has to be this one first and then this one second so if we think about all the X values the first thing you're going to do is you're going to multiply by2 or again you could divide by two and that's because you have to undo this part right here you undo multiplication by two by dividing by two or multiplying by half then the second guy here that you're going to do you're going to subtract away pi over 3 so you can just write subtract piun over 3 or you could put add piun over 3 so this is what we need to do to each X to get those x coordinates so let me come down here we'll do it to the asmp tootes first so if you start with pk/ 2 and you multiply by 1/2 first that's going to be PK over 4 and then in step two you're going to take PK over 4 you're going to need a common denominator so let's multiply this by 3 over3 now so that would be -3 piun /2 so you have 3 piun / 12 and you're going to subtract away piun over 3 let's write that as 4 piun over 12 just getting a common denominator if you do the subtraction you're going to get -7 piun over 12 notice that matches exactly what we got right here when we solve the problem in a different way so you get x = 7 piun over2 the other guy is going to be X = piun over2 so let me erase this and then for this one again all you're going to do you're going to multiply by a half so pi p 2 * a half that would give me let me actually show this so Pi / 2 * 12 this would be < over 4 so that's done first and then once you have that you subtract away pi over 3 so you have piun over 4 and then minus piun over 3 again get a common denominator this would be multiplied by 3 over 3 so this would be 3 piun over 12 multiply this by 4 over 4 so this would be 4 piun over2 so 3 piun - 4 piun would be piun over the common denominator of 12 so P 12 again as I just told you X = pk2 that's your other vertical ASM toote that we found earlier using the other approach and we'll Arrow to this so the vertical asmp tootes here for this guy you would have X is equal to7 pi2 and then X is equal to let me put and in there piun 12 so that's how you can find that using the graphing Transformations only let me go ahead and copy this real quick and let me come down here and paste this in for getting the x coordinates for these three points again you just follow the same procedure multiply by a half then subtract away piun over 3 piun over 4 * a half piun over 4 * 12 that's going to be < over 8 so once you have that you're then going to subtract away your pi over 3 so minus piun over 3 we need a common denominator here that's what makes it take so long so let's say this is 3 pi over 24 just multiplying this by 3 over 3 and then minus let's multiply this by 8 over 8 so 8 < over 24 and we get -1 piun over 24 and we know that's the case so1 piun over 24 again if you check with what we did earlier here's 11 piun 24 comma 1 the next one's a little bit easier because you're working with zero so 0 * a half is still zero so then you just need to subtract away piun over 3 so this becomes piun over 3 and so piun over 3 comma 1 again here's your halfway point so piun over 3 comma 1 and then for the last one you have pi over 4 * 1 12 so that's done first so this would be < / 8 so then we have to subtract away pi over 3 from that so you have pi over 8 and then minus < over 3 again just get a common denominator I'm going to multiply this by 3 over 3 so this is 3 pi over 24 and then minus you multiply this by 8 or 8 so this is 8 piun over 24 and this gives me5 pi over 24 so this is5 Pi let me make this five better here over 24 comma 3 and let me show you that real quick here's your 5 piun over 24 comma 3 that we found earlier so of course you can use this approach many students like it because it's a little bit faster but you really have to understand all the different graphing Transformations particularly what's being done to X and what is the correct order so if you have that down then you could use this method in some cases it's a little bit faster all right let's take a look at another example with Tangent so here we have f ofx equals the tangent of x /2 - 5 piun 4 and then + 1 so again you can use the book procedure the one that I started out with or you could use graphing Transformations it's really up to you I'll do this one both ways again and then for the ones that we're going to do with cotangent we'll have two examples I'll just pretty much stick to the book method for those so let's start off with our book method again you're going to take this right here the argument for the tangent function and you're going to say x / 2 - 5 piun / 4 and you're going to first say that this is greater thank / 2 and less than < / 2 so your goal is to solve for x so we want to solve for this guy right here so what you could do again if you have fractions involved is just clear them first it just makes your life a lot easier and so what you see is you have a denominator of two 2 four and two so the least common denominator would be four so just multiply everything by four here and let me do that like that and like that so this four we cancel with this and give me a 2 2 * piun is -2 piun so this is less than four we cancel with 2 and give me 2 2 * X is 2X and then minus these fours would cancel so you have 5 pi and this is less than four we cancel with two and give me 2 2 * pi is 2 pi again if I'm trying to solve for x what do I do well first I would add 5 pi to each part so let's go ahead and do that and let me move this in a little bit more like that so so on the left 5 piun - 2 pi is going to be 3 Pi so this is less than 2x let me show this as cancelling Which is less than 2 Pi + 5 Pi that's going to be 7 Pi again if you want to isolate X here just divide each part by two so let's go ahead and say this is divided by two this is divided by two and this is divided by two so this would cancel me slide down here a little bit and so now we have X is greater than 3 piun / 2 and less than 7 piun / two and let me come down here to a fresh page so again just getting the information for the consecutive vertical asmp tootes so vertical asmp tootes you're going to have X is equal to 3 Pi / 2 and then and you're going to have X is equal to 7 Pi / 2 so if you want to come to the graph again I gave you a link for a more detailed version on Desmos this is your x = 3 piun /2 and then this is your x = 7 piun /2 so so so far we have that but we need these three points again your quarter of the way halfway 3/4 of the way the first thing is we do need the period so what is the period let's go back up so this right here is X over2 you could really write that as 12 * X so you could say this is f ofx is equal to the tangent of let me Factor this because we're going to do this with all graphing transformations in a little while so that'll already be done let me pull out a 1/2 so inside you'd have X xus so then for this part right here if you pull out a 1/2 it would be 5 piun / 2 close that down close this down and then plus 1 again if you want to check this 1 12 * X is 12 X or X over2 if you want to write that way then you have minus and minus and then 1/2 * 5 pi over 2 would give me 5 pi over 4 so that checks all right so if we look at it now you have this 1/2 here so the period is going to be Pi /2 so Pi /2 so you can go pi over 1 1 * 2 over 1 which is 2 pi okay so now we know that the period is 2 pi and then the quarter period again you just divide this by four or multiply it by a 4th we know that 2 pi over 4 would be pi over 2 that's your quarter period so if you want to get your X values you're going to start with this 3 pi/ 2 and then you're going to add quarter periods let me come down here so let me start with 3 pi over two and I'm going to add pi over 2 and then basically that's going to give me 4 pi/ 2 so let me write this as the quarter point so the first quarter you're going to have 4 pi/ 2 if you want to do that now is going to be 2 pi if you simplify but keep it like this for adding so for the next guy you're going to go 4 pi over 2 plus pi over 2 that way you don't have to worry about getting a common denominator this would give me 5 pi over 2 so for the halfway point or the Midway as a lot of books say you have 5 pi over two for the x coordinate and then we'll do this one more time so 5 piun / 2 and then plus piun / 2 this is going to be 6 piun / 2 which of course is 3 Pi so let's say that this is the 3/4 point and basically the x value there would be 3 Pi just doing 6 pi divided by two okay we don't need this anymore we have what we want and basically all you need to do at this point is get your y-coordinates and if we go back to the graph you can see that this right here is your first quarter and it's going to be at zero so you're on the x- axis then the Midway or the halfway point you're going to be at postive 1 for the y-coordinate and then for the 3/4 of the way you're going to have a y-coordinate of two so it's going to be 0o 1 and then two now where is that coming from well basically you just have this plus one here that's the only thing you're doing in terms of a vertical transformation this right here is just the one if you want to write it so basically all you have to do is take your normal y values which you get - 1 0 and one and just move them up by one so in other words instead of negative 1 you get zero instead of Z you get one and instead of one you get two this is going to be zero this is going to be postive 1 and then this is going to be positive2 again that's coming from those basic points from the tangent graph so piun over 4 comma 1 your y coordinate of netive 1 you add one you get to zero then your halfway point is 0 comma 0 your y coordinate of zero you add one you get to one then your pi over 4 comma 1 you take that y coordinate of one you add one you get to two so those are your points for the quarter half and 3/4 so you can have 2 pi comma 0 let me label that so this right here is 2 pi comma 0 then this one right here is 5 pi over 2 comma 1 so let me label that one this one is 5 piun over 2 comma 1 and then this one is 3 Pi comma 2 so this one is going to be 3 Pi comma 2 now in terms of continuing this graph to the left or to the right remember that for these points you're always going to have the same y value so this one the first quarter point you're always going to have a y-value of zero for the halfway you're always going to have a y coordinate of one for the 3/4s way you're always going to have a y coordinate of two so that's not going to change the only thing that's going to change is the x value and the way you can do this is you can just constantly add if you're going to the right or subtract if you're going to the left quarter periods that's all you really need to do to find your additional points and your additional vertical asmp tootes okay so one more time I'll do this with with Transformations only I know a lot of people want to see this because you don't get a chance to really practice this in the book because they don't cover it in most cases so again when I think about this I can put a one right here so really in terms of a stretch or compression vertically there's none right so basically you have 1 time the tangent of whatever this is and then plus one so there's no vertical stretch or compression this is just a one but you do have this plus one on the outside here which is giving me a shift up by one unit so in terms of the vertical Transformations I know that I'm just going up by one again if you want to take these right here and let's slide this down and compare them to the original points from y equals the tangent of X let me do this like this you have PK over 4 comma 1 you have 0 comma 0 and then you have pi over 4 comma 1 well again all I'm doing is I'm just adding one to each yvalue so negative 1 goes to zero 0o goes to one one goes to two so we did that already but basically that's all we're doing for the vertical Transformations okay let me put this right here and now let's talk about the horizontal again this is the one that's a little bit more confusing and basically now we have the case where we have this 1/2 here so remember if it's between zero and one now you're getting a horizontal stretch again it's always the opposite of what you think how do you UND do multiplication by a half or dividing by two if you want to think about it that way well you need a multiply by two so the first step here so step number one you're going to multiply by two this is going to be horizontally stretched by a factor of two and remember this is going to be what you do first then this part right here this is minus 5 piun / 2 so this is going to be done second again it's the reverse of what you think so since it's minus 5 pi over 2 I've got to add 5 Pi over2 so I want to shift to the right by 5 pi over 2 so let me put for number 2 this is going to be add 5 pi/ 2 so again if you always remember that things on the inside are the reverse so basically this is 1/2 so to undo that I need to multiply by two this on the inside is- 5i over 2 to undo that I need to add 5 pi/ 2 and then you even go in the reverse order so you reverse the order of operations normally you do what's inside a parentheses first and then you would multiply second you reverse that order so you would do this one first do the multiplication first and then you do this part right here here what's inside the parentheses second so you have to keep everything straight with the horizontal Transformations I know it gets super confusing so if this is not something that's straight in your mind I would just go with the other procedure you could start off with the vertical ASM tootes so just start with your x equals piun / 2 and x equals you could do Pi / 2 like this so these are your vertical ASM tootes when you think about the standard y equal tangent of x that you start with so basically I'm going to multiply by first so let's go piun 2 * 2 that's going to give me then once I've done that I want to add 5 piun 2 so let me start now with pi like this and I'm going to add 5 piun / 2 well let's write this as - 2 piun / 2 and what is that going to give me -2 Pi + 5 Pi that's 3 pi and then over the common denominator of two so this would go to x is equal to 3 piun / 2 then for this one again you're going to multiply by 2 first piun over 2 * 2 let me show this so nobody's lost that would give you pi and then you're going to add 5 pi over 2 so write this as 2 pi over 2 you add that together you're going to get 7 pi/ 2 so this right here is going to become X is equal to 7 Pi / 2 let me put and there and again if you go back and check we found from the other method that X is equal to 3i / 2 and X is equal to 7 pi/ 2 those are your vertical asmp tootes now for the other X values again from that first quarter point from yal tangent of X you have an x value of piun over 4 so you take piun over 4 you multiply by two first this would cancel that gives you two down there so this would be equal to piun over 2 and then from there you're going to add 5 piun over 2 so this would give me 4 Pi / 2 and 4 pi over 2 is 2 pi again if you look here for the first quarter point you have an x coordinate of 2 pi again we already figured out that the y-coordinate was Zero now for the next one you have an x coordinate of zero so you multiply that by two I don't even need to show that that's still zero and then you add 5 pi over 2 that's going to give you 5 pi over 2 so see our x coordinate here is 5 pi over 2 for the Midway point so we just need the last one so you have pi over 4 for that 3/4 of the way point for the x value and again I'm just going to multiply by two so this would cancel give me two down there so now I have pi/ two and then once I've done that I'm going to add 5 piun / 2 and this becomes 6 piun / 2 which is 3 Pi so again you come over here and confirm with what we already did for the 3/4 Point you're going to get 3 Pi for the x value so that's an alternative approach if you just want to use graphing Transformations as I told you in the last example I know a lot of students struggle with it particularly with the horizontal Transformations because everything is the reverse or you could say counterintuitive to what you think all right let's talk about the coent function now so most of you know that the coent of x is equal to the cosine of x over the S of X so again just like we saw with our tangent function the period is pi you could take this point right here which is pi over 4 and the cotangent value would be one you're dividing 2 over2 by2 over2 this is your cosine value this is your s value the same nonzero number divided by itself would be one if you swing around by pi so if you add pi and you go to this 5 pi over 4 which is right here here again you're dividing the same nonzero number by itself here you have A2 over2 and here you have A2 over2 so you're still going to get one and the same thing would be true if you looked at let's say 3 pi over 4 and then you looked at 7 pi over 4 well here the cotangent value is negative 1 because you're dividing opposites so if you swing around you're going to see the cotangent value is negative 1 there as well at 7 pi over 4 so the period here is going to be Pi again just like with Tangent when we think about the fact that cosine of x is now in the numerator well anywhere where the cosine value is zero so again we know that's at piun / 2 3 piun 2 5 piun 2 or you could go netive piun over 23 piun over 2 basically odd multiples of Pi / 2 or you can say piun / 2 + < n where n is any integer that's going to be where the X intercepts are now so basically that's where this is going to be zero and this will not be zero so 0 divided by something that's not zero will give you zero then in terms of where it's going to be undefined well that's going to be where the sign values are zero because again that gives you division by zero so that's going to be right here at zero then at Pi then at 2 pi 3 Pi or you could have Pi -2 Pi so basically you would say Pi n where n is an integer now I have a few basic points here so just like we saw before so we have at zero this guy's undefined again because your sign value is zero then at pi over 6 you're going to get square > of 3 at pi over 4 you're going to get one at pi over 3 you're going to get 3 over3 at pi over 2 you're going to get zero at 2 pi over 3 you're going to get get 3 3 at 3 piun 4 you're going to get1 at 5 piun 6 you're going to get 3 and then at Pi you're going to get undefined so we're going to graph this over the interval from 0 to Pi with both of those being excluded because again at zero it's undefined and at Pi it's undefined so again we're going to get vertical asmp tootes at those values all right let's take a look at a basic sketch of y equals the cotangent of X just using these points right here again when you first do the graph of this you're just going to work from0 to Pi with both being excluded so you can see that at an x value of0 the Y value is undefined because if you have cotangent of 0 you have the cosine of 0 which is 1 over the S of 0 which is 0 and 1 divid 0 is undefined so that's why you get undefined again as this guy approaches an x value of zero so coming from the right as we do this this guy's going to increase without bound so it's going to approach positive Infinity then similarly for an x value of pi again the cosine of pi is negative 1 and the S of Pi is 0 so if I say I want the cotangent of Pi you're setting up the fraction 1 over 0 that's going to be undefined so as your x value approaches Pi coming from the left so coming from the left going this way this is going to decrease without bound we're going approach negative Infinity now in terms of this stuff in between you have Pi / 6 comma 3 so here's your pi over 6 for the x value this would be 3 for the Y value so this point right here is p piun 6 comma 3 then you have your pi 4 comma 1 so this is one that you want to memorize that's going to be your first quarter point we're going to use that then you have your P Pi / 3 comma 33 so it's going to be this guy right there then you have your pi over 2 comma 0 so this is your halfway point so you need to memorize that one then you have your 2 pi over 3 comma 3 over3 so that's this point right here then you have here 3 piun over 4 comma 1 so that's going to be this point right there that's going to be 3/4 of the way through so that's one that you need to memorize and then lastly you have this 5 piun over 6 comma three so that guy right there so that's your basic points usually you just put this point this point and this point you make a vertical ASM toote here you don't need to make one here you can if you want but typically you leave that off because this is the y- axis so you don't draw anything and basically you just draw this to where it approaches the y-axis and doesn't touch it and this guy is going to approach this vertical asmp toote which is x equals Pi but not touch it so that's all you really need to do so just like with the graph for y equals the tangent of x when you expand this to the right or to the left all you're doing is just copying what you previously did in other words we graph this right here so from 0 to Pi with both being excluded but basically the same thing is going to happen from PI to 2 pi 2 pi to 3 Pi or if you went this way you can say from negative pi to 0 you could say from -2 pi to negative pi so on and so forth so you're going to have a vertical ASM toote you're going to have three points vertical ASM toote so it's just that pattern over and over and over again and these three points again the benefit to these guys right here is that basically all you need to know is what is the vertical ASM toote here and what is the vertical ASM toote here so I'll show you how to get that in a moment once you have this one right here on the left in this case it would be zero you just add a quarter period Well a quarter of the normal period the period for cotangent is pi so a quarter of that is pi over 4 so I know the x value right here is pi over 4 and of course the Y value is one we can see that then this one right here again you just add a quarter period so p piun over 4 + p piun over 4 that's 2 piun over 4 or piun / 2 and the Y value is Z then for this one again I'm just adding a quarter period let me make that Arrow better so if you have Let's Go 2 pi over 4 so we have a common denominator plus pi over 4 that's going to be 3 pi over 4 comma 1 for the y- value so the yv values are things that you want to memorize because typically you're going to do this with graphing Transformations so if you know those yv values it's going to save you a lot of time you can always plug in if you want to but I find that to be a very tedious process for the X values again all you're going to have to do is find the vertical asmp tootes figure out the period and the quarter period period and then just go through the process of adding quarter periods to get to the next X value so I start here add a quarter period I get to this x value add a quarter period add a quarter period add an ASM toote add a quarter period add a quarter period add a quarter period add an ASM toote so on and so forth all right let's talk a little bit about the cotangent function in general so the domain contains all real numbers except for pi n where n is any integer so 0 Pi 2 pi 3 Pi or even pi2 Pi that's where the sign value is zero and again division by Z is undefined so f ofx equals c of X is discontinuous at Pi n where n is any integer and the graph contains vertical asmp tootes at those values the X intercepts so again this is where the numerator would be zero and the denominator is not zero so these are of the form xal piun / 2 remember that cosine of pi over two is zero and then plus pi n where n is any integer the x intercept occurs at half of the way between consecutive vertical asmp tootes so if we use that basic yal cang of X over the interval from 0 to Pi both being excluded this is going to happen at pi/ 2 comma 0 so you want to memorize that point then the point on the graph 1/4 of the way between consecutive vertical asmp tootes has a y-coordinate of 1 so we saw that already so if you're using yal cang of X over the interval from 0 to Pi with both being excluded well you're going to have pi over 4 comma 1 lastly the point on the graph 3/4 of the way between consecutive vertical asmp tootes has a y-coordinate of negative 1 again using yal cent of X over the interval from 0 to Pi with both being excluded you saw that we had 3 Pi 4 Comm 1 so again the period is pi there are no minimum or maximum values the range is from negative Infinity to positive infinity and the graph has no amplitude so the graph is symmetric with respect to the origin so if you did Cent ofx it's equal to the negative of cent of X for all X in the domain so f ofx equals the cotangent of X is an odd function so just like we saw with Tangent I gave you a little model you can use it's not going to fit perfectly with every scenario but we can just adjust it so we have yal a cent of BX - C where B is greater than zero again if B is negative which you'll see in the future you can rewrite it using identities if you need the period you can just use absolute value bars it's really not that big of a deal it won't cause that much concern so what we would do is say that you're going to take this BX minus C like this and you're going to find an interval over which you're going to graph it by saying it's greater than zero and less than pi and that's coming from the fact that if you have y equals the cent of X we we said that we're going to go X is greater than Z and less than y so again all you're doing is you're taking this argument right here which is the bx- C and you're just putting it in that place right there and you're going to solve for x that's all you have to do so basically once you've done that you're going to find these two vertical asmp tootes so you could do it as an equation BX - c is equal to Z so that's for this one and then BX - c is equal to Pi for this one and then once you have that just like with the tangent function you can figure out what the period is so it's just going to be pi over B let me write that out so the period is just pi over B so whatever that is that coefficient of x if you end up factoring that out so if you have something like y equals aent of let's say we pull the B out because we want to use graphing Transformations let me move this up so it fits so then inside you would have x minus C over B like that well the period is still pi over B so in this form it's the coefficient of x in this form it's outside of the parenthesis so just keep that in mind now once you have that period you can just figure out the quarter period so the quarter period again that is just pi over 4 B so just take this and multiply by a fourth and once you have that you take the x value here you add a quarter period you're going to get your x value here so the y-coordinate would just be given by a in this form but again if you have a shift you have to consider that as well you do this multiplication here first by your typical y-coordinate for yal cang of X which in that case is going to be positive 1 and then once you've done that you need to apply your shift either up or down this 1/4 point the y-coordinate is a because again 1 time a would be a then at the halfway point the x intercept well again I could just add a quarter period to get the x value there and then for the Y value in this form is just going to be zero but again if you have a shift up or down you have to consider that then you add another quarter period to get the x value and then you're going to have your 34s point and the y-coordinate is now negative a because normally if you have yal coang of x 3/4 of the way through you have that 3 pi over 4 comma 1 so if you take that1 you multiply by a you get Negative a that's where that's coming from so just keep this in mind as a general model and with that being said let's just go ahead and look at an example all right let's take a look at an example now with cotangent so we have F ofx equal the cotangent of x / 2 + < / 3 to start things off I'm just going to go x / 2 + pi over 3 so that's going to be the argument for the cotangent function there this guy and basically you're going to say that this is greater than Z and less than Pi so I'm going to get rid of these fractions here you have a denominator at of Two and a denominator of three so what I would do is just multiply both sides by the LCD which would be six so 0 * 6 would still be zero let me just show it for right now multiply this by six multiply this by six let me slide this over here and then multiply this by six let me drop down a little bit so as I just said 0 * 6 is still 0o this is less than 6 / 2 would be 3 and 3 * X is 3x and then plus 6 ID 3 is 2 2 * piun is 2 pi and then this is less than 6 * piun is 6 Pi if you want to isolate X in the middle then what you're going to do is subtract 2 pi away from each part here let me just put minus 2 pi here and so this is going to cancel we're going to have our 3x in the middle and this is greater than -2 piun and it's less than 6 piun - 2 pi would be 4 pi as a final step here let me divide each part by three just to finish things up so this is going to cancel and what you're going to have is that X is greater than -2 piun over 3 and less than 4 piun over 3 let me come down here and just paste this in and I'll put a little border and I'm going to say that my consecutive vertical asmp tootes so vertical asmp tootes you're going to have X is equal to you have this -2 piun over 3 and then and let me write that better there and let me fix this three here we're going to say x is equal to 4 piun over 3 let me show you this graphically first so coming down to the graph you see that you have a vertical ASM toote here and this is our x = -2 piun 3 and then you have a vertical ASM toote here this is going to be our xal 4 pi/ 3 so that we know we need to figure out this point this point and this point so let's go back up so what you want to do is figure out what is the period and for this again let's go up here when you look at this you have X over 2 so you can think about that as 1/ 12 * X so the stand standard period is pi again all you're going to do is go pi over 1 /2 or times the reciprocal of that which is 2 over 1 so this is 2 pi for the period let's write the period as 2 pi and then the quarter period this is going to be 2 pi over 4 which is pi/ 2 so again all you have to do at this point is just go through and say well I'm going to go -2 piun over 3 plus pi/ 2 so just starting at that leftmost vertical Asm tote that we found and you need a common denominator here so let me actually write this I'm going to multiply by 3 over 3 and I'm going to put this as let me scooch this down a little bit so we have some room this would be 3 pi over 6 so I have a common denominator and this guy right here I'm going to write as -2 piun over 3 * 2 over 2 so this would be -4 piun over 6 so I'm just going to start with that so -4 pi/ 6 and then plus the quarter period so that's 3 pi/ 6 6 that's going to give you < over 6 so let me write this on another page for this one your first quarter point you're going to have pi over 6 we'll give the Y values in a second I think all of you can see that there's no vertical Transformations that are applied so you can just put one now if you want but I'll just do that separately so then we have the 1/2 and then we have the 3/4 and let's go up and get those so I'm going to start now with the pi/ 6 and I'm just going to keep adding quarter periods so I'm going to add 3 piun / 6 so this gives me 2 piun over 6 3 Pi minus Pi that's 2 pi and 2 pi over 6 you can write that as pi over 3 let's come down here and say this is pi over 3 and then let's do that one more time so we would want to do 2 pi over 6 so 2 pi over 6 plus the quarter period we'll write that as 3 pi over 6 so that gives me 5 pi over 6 let's come down and put 5 pi over 6 now for the Y values here again when you look at these points you see the yv value here is one the y-value here is zero and the yvalue here is negative 1 that perfectly matches what we see with y equals the cotangent of X and the reason for that is you don't have any vertical Transformations here this you could really put as a one so there's no vertical stretch or compression and out here you don't have anything so you can put plus Z to make that clear there's no shift up or down so you would match those y values perfectly so for the first quarter point you have a y- value of one for the Midway point you have a yv value of zero and for the 3/4s of the way point you have a y value of negative 1 so let's put one here let's put zero here and then netive 1 here so we're ready to go in terms of labeling those points you would have piun over 6 comma 1 so that's going to be this first point right here so this is piun over 6 comma 1 that 6 could be a lot better then we have pi over 3 comma 0 so this one right here let's say that's pi over 3 comma 0 and then this one the 3/4 of the way it's 5 piun 6A 1 so this one we'll say that this is 5 piun over six let me make that six a lot better there and then comma netive 1 so those are your three key points again if you want to go to the right or to the left you just keep adding quarter periods so in other words if I take this 5 pi over 6 and I add a quarter period I'm going to be at 4 pi over 3 and you could just keep doing that to get the next point the next point the next Point vertical asmode the next point the next point the next Point vertical asmodee the yv values are always going to be the same in other words for the first quarter point you're always going to have a yvalue of one for the halfway point or the Midway Point you're always going to have a y value of zero and then for the 3/4s of the way point you're always going to have a yval of Nega 1 all right let's take a look at another one so here we have F ofx = 12 and then cent of we have 2X + 2 piun over 3 and then plus 2 all right so again I'm just going to take the inside part here so the argument for the cotangent function and I'm going to start by saying 2x + 2 pi over 3 is greater than Z and it's less than Pi so we'll solve this first and so let me slide down here just a little bit I would get rid of this denominator of three so just multiply everything by three so you would have 0 * 3 let me show this so 0 * 3 is less than 2x * 3 + 2 piun over 3 * 3 and this is going to be less than < * 3 so 0 * anything is 0 so this is still 0o this is less than 2x * 3 is 6X and then plus these threes would cancel so you have 2 pi so which is less than 3 * pi is 3 Pi so at this point I'm going to subtract 2 pi away from each part let me write that like that and so this is going to cancel you're going to have 6X in the middle so this is greater than you're going to have -2 pi and less than 3 Pi minus 2 Pi is Pi to finish things up let's divide everything by six and so that's going to give me this cancelling in the middle so X is greater than if you think about 2 and 6 well 2 ID 2 is 1 and 6 ID 2 is 3 so let's say this is piun over 3 and it's less than piun over 6 so let's grab this okay let me put that in there like that so again your two consecutive vertical asmp tootes so vertical asmp tootes you're going to have X = Tok over 3 and then and you're can to have X is equal to piun over 6 all right so this one's a little bit busy so again I've given you a link on Desmos if you want to click that to get more detail this is your vertical ASM toote which is going to be x = piun 3 so that guy right there then this is the other one so this is the vertical ASM toote it's xal piun over 6 so we're going to be finding these three points right here so let's go back to right here if you want your period again it's Pi which is your standard period divided by 2 the coefficient of x so it would be pi/ 2 so let me write that the period is going to be pi over 2 and then your quarter period so this is pi/ 2 * a 4 so that's pi over 8 so again all you have to do is take this piun over 3 and you're going to add a quarter period so you're going to add pi over 8 now again you need a common denominator so let's just go ahead and write now multiply this by 3 over3 just so that we're not struggling the whole time to get common denominators so this right here would give me 3 pi over 24 okay so let me adjust this I'm going to write piun over 3 as -8 piun over 24 and then I'm going to add 3 pi over 24 so then for the first quarter point the x value would be8 piun + 3 Pi that's -5 pi and this is over 24 and let me put for the first quarter point you want -5 pi over 24 okay so we're going to have halfway and then 34 we're going to figure those out right now so let me take this5 piun over 24 and all I'm going to do is add a quarter period so let me rewrite that it didn't take it with it so + 3 piun over 24 so5 piun + 3 piun that's going to be -2 piun over 24 so if you think about that you can simplify 24 ID 2 is 12 and 2id 2 is 1 so this would be piun over 12 so this is pi over 12 and let me get rid of that and we only have one more to do so let me move this over so let me add 3 piun over 24 one more time and so -2 piun + 3 Pi that's going to be pi and this is over 24 so this would be pi over 24 okay in terms of the Y values you do have to do a little bit of work here because you do have two Transformations that are vertical so let me start off with these basic points here so on the yal cent of X you're going to have your pi over 4 comma 1 then you're going to have < / 2 comma 0 and then you're going to have 3 piun over 4 comma -1 so we're going to think about how we're going to transform here again in terms of vertical you have this guy that's done first this is a multiplication by half so that's giving me you could say a vertical compression by a factor of two or again a lot of people like to say it as a vertical shrink by a factor of a half it's just understanding that you're either going to divide each yvalue by two or you're going to multiply by 1/2 is the same thing either way just a matter of how you want to say it this right here this plus two that's going to shift me up by two units so this is done first this is done second so the order does matter here so let me write this out we're going to for part one multiply by2 or again you could divide by two if you want and then for part two we're going to add two so again this is a vertical compression by a factor of two then you're going to add two that shifting me up by two units okay so we're first going to multiply one by a half so 1 time 1/2 would be 1/2 then I'm going to add two so you would have2 Plus for two you could write that as four over two so this would be five halves so this is five halves all right for this one right here you have 0 times a half that's still zero then you add two that's going to be two and then for this one you have -1 * a half that's - one2 and then you add two so let's add 4 over 2 so this would be 3es so that's basically it just a matter of understanding those graphing Transformations again if you get confused by that you can always plug in so you can always take5 piun over 24 plug into the original function you'll get 5 Hales you could plug in negative piun over 12 you'll get two you could plug in pi over 24 you'll get three halves all right let me get rid of this and let's just label things and we'll be done so for this one right here you have 5 piun over 24 comma 5es so that's going to be this one right here so this is -5 piun over 24 comma 5es and if you look at this it's hard to see but this Notch right here is two so this Notch right here would be 2.5 because this Notch right here is three and then if you look at this one right here that's going to beunk over2 comma 2 so that isk over 12 comma 2 and again this Notch right here is two it's covered up by that point and then if you look at this one right here we're going to have pi over 24 comma 3es so this is pi over 24 comma 3 Hales and again this Notch right here is one this Notch is two so this is halfway so that's 1.5 or three Hales in terms of your yv value so you see that does match up again if you want to click the link on Desmos you can see I've labeled all of these points for you so you can go through there and you can get some additional practice by basically just adding quarter periods to this guy so to the X Valu see here you'll get your vertical ASM toote and then you're going to get the x coordinates for this guy this guy this guy vertical Asm toote this guy this guy this guy vertical ASM toote so on and so forth you know the Y values are always going to be the same so for the first quarter point you're going to have five halves then for the Midway Point you're going to have two then for the 3/4 of way point you're going to have three halves so on and so forth so if you want to play around with it and get some additional points going on your graph going to the right and to the left it'll be good to get some practice in this lesson we want to learn about graphing secant and cosecant all right before we get started with the lesson I want to remind you that when you're in this section and you're graphing trigonometric functions we want to understand the concept so how could I graph the function if I wanted to but generally speaking if we want to get a perfect graph we're going to use graphing software and that's something we're going to rely on as we get out of this chapter so don't worry too much about trying to make some perfect graph just do the best that you can so when we look at this we're going to start out by saying that in our lesson on cosine and sign we learned that the period was 2 pi when we think about something like the secant function so the secant of X this is equal to 1/ the cosine of x so what would the period for the secant function be well it's going to be 2 pi as well the secant of X is equal to the secant of you have this x + 2 pi n where X is a member of the domain and N is any integer in other words if I was to come to the unit circle and let's say that I took the secant of let's just do pi over 3 that's an easy one so the SEC of piun 3 this would be equal to by definition 1/ the cosine of < / 3 and so the cosine value for this pi over 3 again you just look at the x coordinate so that's 12 so this would be equal to 1 over 12 which would be two if you added integer multiples of 2 pi to this angle pi over 3 so in other words you could add 2 pi to it 4 Pi 6 Pi or you could subtract away 2 pi subtract away 4 Pi something like that you're going to end up with a co-terminal angle so you're going to have the same cosine value and also the same secant value so just to keep it simple let's say we have the secant of let's go < / 3 + 2 piun and I would write that 2 pi as 6 piun over 3 so we have a common denominator so 6 piun over 3 so this would be the secant of 7 piun over 3 and of course that's going to be co-terminal with pi over 3 so we know that the cosine value would be the same so this is 1 over the cosine of 7 piun over 3 which would be equal to again 1 over this 12 which would be 2 so the period for secant is going to be 2 pi now let's talk about cose so using a similar thought process when we think about the cose of X this is equal to 1/ the S of X and of course the period of the sign function is 2 pi so the period for the cosecant function would be 2 pi as well so the cosecant of X is equal to the cosecant of x + 2 pi n where X is a member of the domain and N is any integer all right so we're going to start out by looking at the secant function and we're going to do that with a table of values and we're going to look at this over one period so y = the see of X we're going to say where X is greater than or equal to piun and less than or equal to piun now we're not going to use this table of values approach when we graph secant in general we'll have a different procedure for that we're just using this to get some key points here now with the other graphs for S and cosine and then also with Tangent cotangent there were points that you really needed to memorize you're not going to do that here because you're going to end up using the cosine function as a guide and we'll get to that later on for right now I'm just going to focus on some things that we want to know and and so the first thing would be that you have a vertical ASM toote here atunk / 2 and then also at Pi / 2 so we want to think about where that comes from remember that the secant function so the secant of X is equal to 1 over the cosine of x so just like we saw with a tangent function you have your cosine function in the denominator there so wherever the cosine values are zero you're going to get division by zero which is undefined and that's giving you these vertical ASM tootes so that's going to happen at odd multiples of Pi / 2 so piun over 2 piun over 2 3 pi/ 2 5 pi/ 2 so on and so forth now another thing I want to call your attention to is that when you look at the behavior of this graph so let's say as this guy right here goes from an x value of zero to an x value of pi/ 2 now it's not actually going to get to an x value of Pi / 2 but you want to think about what's happening as it approaches an x value of Pi / 2 coming from the left so we would think about something like let's say secant of let's just do piun 3 again that's an easy one this is equal to 1 over the cosine of piun / 3 which is 1 over again the cosine of pi 3 is a half so notice that this value right here of one when I divide by a half I get something bigger I get two when you take one and you divide by something between zero and one you're going to get something that's larger than one for example if I did let's say 1/10th I would get 10 or let's say as another example I did 1 / 1 over 100 well that would be 100 so what's going to happen is as the x value here gets closer and closer to pi/ 2 the cosine value gets closer and closer to zero so you can get as close to zero as you'd like but you can't actually touch it so what's happening is the value is going to get really large for secant because you have that one divided by that number that's really really close to zero so this guy right here is going to increase without bound oras they say approach positive Infinity so let me get rid of this and we also want to think about the other side of that so at Pi so the secant of Pi is equal to 1 over the cosine of pi so we know that the cosine of pi is 1 1 over1 would be 1 so here this point would be Pi comma 1 now what happens as we approach pi over 2 coming from the right in other words what's going on right here and You' see that we're decreasing without bound so we're approaching negative Infinity basically what's happening is if you think about something like let's say the secant of 2 pi over 3 to make it simple so there's a pi over 3 reference angle so we know that we're in quadrant 2 cosine is negative so this would be 1/ the cosine of 2 piun / 3 and this equals 1 now you'd have -2 and so this would be -2 so it's the same thing here if you look at that point it's going to be right there so this would be your point which would be 2 pi over 3 comma -2 again as this x value gets closer and closer to Pi / 2 again coming from the right now the cosine value is getting closer and closer to zero but it's on the other side of zero so it's going to be something between 0 and negative 1 now so this guy is now going to approach negative Infinity so that's just something you really need to understand about the behavior of this graph that's why you have this going this way that's why you have this going this way and that's why you have this shape right here so once you understand that it's basically good to go in terms of why the graph looks the way that it does and now what I want to do is just erase some things and give ourselves a little Clean Slate here and want to go through some of these basic points so first off you should know this point right here which is 0 comma 1 that's one that's going to be on the graph of your cosine function and your secant function then this one right here this is your Pi comma 1 again that's going to be on both and then this one right here that's going to be piun comma -1 now in terms of these other points if you want to go through them in more detail I left you a link on Desmos so you can stop the video here click that link and you can look at all these points and think about where they're coming from I'll just read through them real quick if you have Plus or Min - pi/ 2 again that's where we're going to be undefined because the cosine value is zero you have 0 comma 1 so that's this point right here then this plus orus piun over 6 you get 2 * 3 over 3 let me actually show you that real quick so these other guys make sense coming back to the unit circle again if I asked you for the secant of Pi / 6 or I asked you for the secant of piun / 6 you're going to get the same thing because again when you look at piun / 6 the cosine value is 32 when I look atunk 6 again that's rotating clockwise so that's co-terminal with 11 piun 6 let me write out that this is piun 6 here so the cosine value is again sare 32 so whether I take the see of piun 6 or the see of NE piun / 6 again I'm going to be in either quadrant 1 or quadrant 4 where cosine and see are positive and I have the same reference angle so I'm going to get the same result let me just do this one real quick you can get rid of this one I'll just show you this real quick so this would be equal to 1 over the cosine of < / 6 and of course this is 1 over the 3 /2 and so this would be equal to you're just going to flip this and say this is 2 over the 3 and then multiply by the 3 over the 3 to rationalize the denominator so you get 2 * 3 over 3 and again this would be for < / 6 or piun 6 so coming back down again Plus or - piun 6 you're going to get 2 * 3 3 plus or - piun 4 you're going to get2 Plus or - piun 3 you're going to get 2 plus or - 2 piun over 3 you're going to get -2 Plus or - 3 piun over 4 you're going to get2 and then Plus or - 5 piun 6 you're going to get -2 * 3 3 and then plus or minus Pi you're going to get Nega 1 you can look at that link again on Desmos if you want to go through those points again we're not going to focus too much on these points we're going to use the cosine function as a guide so don't worry about this too much okay now let's actually get into the procedure so how do you sketch the graph of the secant function where you're going to use the cosine function as a guide so you know how to graph the cosine function at this point if you do it over one cycle you would do this part right here and I'm just going to trace over this real quick so from 0 to 2 pi so this is what your basic cosine function looks like and we've graphed that a million times at this point so you would graph that guy right there and then what you're going to notice is that you have a shared Point here so at0 comma 1 and then also here at Pi comma 1 and then also at 2 pi comma 1 and I'll talk about how those come into play in a moment but you'll notice that where you have an X intercept so that's right here at pi/ 2 and right here at 3 pi/ 2 because again when you have the cosine of Pi / 2 you get zero when you have the cosine of 3 Pi / 2 you get zero so those are X intercepts and therefore that's going to make your secant function undefined because you can't divide by zero so you're going to get those vertical ASM tootes at those values so this is what you want to focus on here now that's going to change a little bit when we get to Shifting the graph up or down and I'll explain how to compensate for that for right now we're just going to learn it this way so when you think about this all together just going over the whole graph that we have here you would say that wherever the Y value is one you're going to have a shared point so that's going to happen at an x value of zero and then an x value of 2 pi and then 4 Pi 6 Pi 8 Pi so on and so forth and then going this way you'd have -2 pi4 Pi 6 Pi so on and so forth so you could say at 2 pi n where n is in the integer the yv value would be one and you're going to have a shared point between your secant graph and your cosine graph then similarly when you think about having a yvalue of negative 1 again that's going to have a shared point so something like let's say Pi comma 1 and if you add 2 pi to that you'd have 3 Pi comma 1 if you subtracted away 2 pi from this Pi here you'd have Nega pi and then comma 1 and then here's your -3 Pi comma 1 so this would be at Pi + 2 pi n where n is any integer so 5 Pi comma 1 would be there 7 Pi comma 1 would be there something like5 Pi comma 1 would be there so on and so forth the idea here is you want to think about that as being being shared points because when you see that there's a shared Point you're going to what we call interchange The Hills And The Valleys so what I mean by that is when you think about the graph of cosine you know that you have a maximum Point anywhere where the Y value is one and a minimum Point anywhere where the yvalue is negative 1 because the range of cosine is from negative 1 to positive 1 with both being included so here you'd have a maximum point and then again here and then here so when you think about it this maximum point right here corresponds to what we call a local minimum point on the graph of the seein function so this right here is a hill and this right here is a valley so the hills and the valleys are going to be interchanged so this right here is a hill this is a Valley Hill Valley then when we get down here let me use a different color you think about the fact that okay now when we look at our cosine graph this is going to be a valley or a minimum and then here this would be a local maximum or Hill and then when you come here it's the same thing so this is a valley this is a Hill Valley and then this is a hill you have a valley and then this is a hill so that's all you really need need to memorize the heels in The Valleys will be interchanged that's how pretty much every book says it and that's how I remember it now the last thing again like we talked about a moment ago is how to get these vertical ASM tootes so without Shifting the graph up or down you're going to think about where are the X intercepts for the guide function the cosine function again that's going to happen at odd multiples of Pi / 2 so at this5 piun over 2-3 piun over 2 piun over 2 pi/ 2 3 pi/ 2 and then 5 piun over 2 so this would be a vertical ASM toote here for your seein function again you'd have one here and then here and then here and then here and then here and again this continues forever so you would have 7 pi over 2 9 pi/ 2 so on and so forth all right let's just go through the characteristics of the secant curve so the domain consists of all real numbers except odd multiples of Pi / 2 again this is because at odd multiples of Pi / 2 you're going to get a cosine value of 0 division by 0 is undefined so f ofx equals the secant of X is discontinuous at odd multiples of Pi / 2 the graph contains vertical Asm to at those values there are no X intercepts you will only see those if you start Shifting the graph up or down and then the period is going to be 2 pi so there are no minimum or maximum values we saw that the range is from negative Infinity going up to1 and including Nega 1 and then the union width you have from one and including one out to positive Infinity you will not see any y values between negative 1 and positive 1 again because of the range of the cosine function which is from Nega 1 to POS 1 and the definition of the secant function which is 1 over the cosine of x again something we talked about earlier when we were describing the behavior of the graph so the graph has no amplitude again because there are no minimum or maximum values and the graph is symmetric with respect to the y- AIS so the see ofx is equal to the see of X for all X in the domain so we can say f ofx equals the see of X that is an even function just like cosine all right so real quick we have a graphing procedure and you can write this down if you want it's something you can just memorize if you don't want to write it down it's not a big deal so graphing variations of the seant function the first thing you want to do is replace the secant with cosine and sketch the graph using a dashed curve so cosine is used as a guide then you want to sketch the vertical asmp tootes so this is the one that you really might have trouble with this step right here and the reason for that is is that if you get a vertical shift involved then you're not going to look for the X intercepts of the cosine function or the guide function it's going to be moved around and I'll show you that when we get to an example the way I remembered is these urr at a quarter of the way and 3/4 of the way through the cosine cycle again just going back to this here is the cosine graph if you look at it over one cycle so if we look at it over one cycle it looks like that so at the start of the cycle your x value is zero when you get a quarter of the way through your x value is pi/ 2 so you have an X intercept when you get to the middle your x value is pi 3/4 of the way through your x value is 3 pi over 2 so you have an X intercept and then you finish up at 2 pi for the x value so 0 Pi / 2 pi 3 Pi / 2 and 2 pi so just keep that in mind when we look at these variations of this this guy we're going to use the same thought process so a quarter of the way through and 3/4 of the way through that's where we're going to have our vertical asmp tootes for the secant function all right then we have sketch the graph by drawing the u-shaped branches between consecutive vertical asmp tootes so you have that the hills and the valleys are interchanged so all you're going to do is you're going to take those shared points you're going to start there you're going to make your u-shaped branches and you're going to approach your asmp tootes but not touch them then we have optional here use the period to find additional Cycles to the left or the right as needed and again if you're self studying I would just do one period and call it a day usually if you're given an assignment your teacher will say hey I want you to sketch it from this place to this place so just follow that okay let's take a look at an example now so we have F ofx = 12 secant of 2x + 5 piun / 6 and then + 1 so instead of just going straight into replacing see with cosine let's just get some information so let's just go ahead and say that we have f ofx is equal to 12 and then the secant of I'm going to factor out the two just makes it easier to find the a shift so pull that out you're going to have X Plus how do I undo multiplication by two I would divide by two or multiply by half so this would be 5 pi/ 6 * half which is 5 piun over 12 and so when we look at it now finding the phase shift is very easy you just want to say inside the parenthesis here how do I undo what's being done to X we have this plus 5i / 12 to undo that you want to subtract away 5 pi over 12 so for the phase shift you could say minus 5i over 12 or most people we'll just say left 5 pi over 12 now in terms of the vertical shift that's very straightforward so you have this plus one here so the vertical shift that's just going to be up one now thinking about the period so we'll need that so what is the period for this guy again the standard period is 2 pi when you're looking at either s cosine secant or cosecant and we're just going to divide that by the number outside the parentheses here if it's in unfactored form it's the coefficient of x so 2 piun / 2 which would would be pi and then for the quarter period depending on the method you're going to use to find your points for the cosine function we're going to get in a minute you would want to take this period which is pi and divide it by four so the quarter period is pi over 4 now let me put a little border here I'm going to talk about the graphing Transformations and you can use this method if you want or you can use a different method it's really going to be up to you so in terms of X remember this is the one that's really hard for students that's why a lot of people just turn to the quarter period method because this becomes a little bit challenging to remember you need to undo what's being done to X and so to undo something you have to go in the reverse order think about writing a Word document or something like that you hit control Z you're undoing the latest thing you did you don't undo the first thing you did so if you look at the order of operations here again you have parentheses here and I know you could distribute that two in but we're not doing that here so from the order of operations we know we would work inside the parenthesis first and then do the multiplication second so if I write this out now I have to go in the reverse order so this would be one and this would be two and each thing has to be undone so how how do I undo multiplication by two I would divide by two or multiply by half so you would multiply by 1/2 so that's giving me a horizontal shrink by a factor of a half I know a lot of teachers will also say this as a horizontal compression by a factor of two again I don't want to get caught up on the terminology a lot of people get confused by those two ways of saying the same thing just realize that you're having your x coordinate either divided by two or multiplied by half then the second thing is you want to undo this plus 5i 12 well want to subtract 5 piun over 12 again we saw that earlier when we did the phase shift that's going to shift me to the left 5 pi over 12 then when you think about the Transformations for y these are the ones that are very straightforward you have this 1/2 here that's multiplying this guy so the first thing is you would just follow that so multiply by2 and the second thing is you're adding one so you're going to add one again this is straightforward it follows the order of operations you multiply before you add so with this one you're getting a you could say vertical shrink by a factor of a half or you could say vertical compression by a factor of two again I know that confuses people when you have two different ways to say the same thing but just realize you're taking the y-coordinate and you're either dividing it by two or you're multiplying it by a half and then you're going to add one that's shifting me up by one unit okay so coming to a new sheet here I'm going to erase the secant and just put in cosine so remember we're going to graph this as a guide so how do you graph this cosine function again lots of ways to do it you can use all graphing trans you can use no graphing Transformations it's really up to you so I'm going to do both the quarter period method and all graphing Transformations method and then on the next ones I'll just use the quarter period method because it's a little bit easier for students to understand so when we think about using the quarter period method you start off by thinking about your points for the basic cosine function so y equals cosine of x you have a point you start off at the maximum so 0 comma 1 and then you're going to go to your x intercept so this is Pi / 2 comma Z and then you're going to go to your minimum so this would be at Pi comma -1 you go back to an X intercept so this is 3 pi/ 2 comma 0 and then you're going to finish up at the maximum so 2 pi comma 1 so these are points you should have memorized at this point and all we're going to do is think about for the first x value here you just use the phase shift in other words you have this zero right here and I'm going to think about what is the phase shift so we know it's to left by 5i 12 or you can say I'm subtracting away 5i 12 again I've got to undo what's being done to X there so this x coordinate would be - 5 piun over 12 right there to get the additional x coordinates here all you're going to do is you're going to take this value right here don't worry about these this value right here and you're going to keep adding quarter periods so the quarter period again was pi over 4 you took your period which is piun / 4 here I'm going to multiply this by 3 over 3 so we have a common denominator so it's 3 piun / 12 so you're going to keep adding 3 piun over2 again start here 5 piun 12 + 3 piun 12 that would give me -2 piun over2 I know you could simplify that but let's just hold off for a moment so we have a common denominator so -2 piun over2 + 3 piun over2 that would give me piun over 12 then < 12 + 3 piun over 12 that's 4 piun / 12 again don't simplify it yet 4 piun 12 + 3 piun 12 that's 7 piun over 12 now this right here this 4 piun over 12 you could write that as piun over 3 so this is < over 3 and then -2 < 12 you could write that as piun over 6 so piun over 6 let me make this better here before we go on I want to show you that you can get the same values here just using those graphing Transformations that we talked about a moment ago so coming back up here let me grab this real quick and come down here and paste this in and I'll just do the first two of them because this is a little bit timec consuming so for this one you stick with these guys right here so for zero you multiply by half you get zero then you subtract 5 piun over 12 you'd have 5 piun 12 then you take Pi / 2 and multiply by a half so you do that first and then you're going to subtract away 5 pi over 12 so this is very tedious because you have to get these common denominators so this becomes piun over 4 minus 5 piun / 12 so we'll say this is let's multiply this by 3 over 3 so 3 piun over 12 - 5 piun over 12 which becomes -2 piun over 12 which is going to simplify to piun over 6 so you see you have netive pi over 6 there let's just do one more I don't think we have to do all of them you can do them all on your own this is just to show you that you have different techniques to use so we have multiply by a half so let's start with pi and we multiply by half and then we subtract away 5 pi over 12 so this becomes < / 2 minus 5 piun / 12 multiply this by 6 over 6 so you have 6 piun / 12 - 5 piun over 12 so this would give you pi over 12 so pi over 12 there so again if you follow this you'll get the same result as using the quarter period method I know a lot of students do not like using graphing Transformations for X because the procedure itself is hard to get and then in this case it would be a little bit more tedious than what we did with adding quarter periods so we have our x coordinates so that's the first thing let me put a comma a comma a comma a comma and a comma and now we want to find the y-coordinates now this is one where you can use graphing Transformations very easily and that's because of the fact that when you plug in 5 piun over 12 so if I plug that in there 5i 12 + 5 piun 12 well that's going to be 0o 2 * 0 is 0 so you would get the cosine of 0 which is one so notice that this right here when I go through the process I get this y-coordinate right there so I could have shortcutted that and not plugged in and just said okay well it's going to be 1/2 times whatever this y value is which is one and then plus one again if we go back that's what we came up with here multiply by half and then add one that's all you're doing when you look at this one here again if you plugged in and I'll just do it for this onega piun over 6 if you plug that in there and you go through the steps you're going to end up with this guy right here being zero because when you plug in a negative piun 6 and you go through the steps you're going to end up with a cosine of pi 2 and let me just show you that real quick just so that's Crystal Clear if you did the cosine of you have 2 * let's Plug Ink over 6 and then plus 5 piun over 12 okay so we just need this part right here I'm going to multiply this by 2 over2 so this would be -2 piun 12 and what would that give me well -2 piun over2 + 5 piun over2 that's 3 piun over 12 and 3 piun 12 you can say is pi over 4 well 2 * pi over 4 that's gonna be pi over two so the inside part here this part right here is going to be Pi / 2 and the cosine of Pi / 2 is 0 well notice that again the y-coordinate there is 0o so all I had to do was sit there and say that I know that the result of this will be zero so I can just say well it's going to be 1/2 * 0 + 1 and then if I was to plug in Pi / 12 the result of this part right here would be ne- 1 so 12 * -1 + 1 again so on and so forth these are just duplicates so you don't have to do this so I'm just going to say that this is 1 half time 1 versus 1 12 plus let's write 1 is 2 over 2 so this is 3es then this one anything * 0 is 0 so this is just 1 and then - 1 * a half is NE one2 and then plus this will be 2 over two so this is 1/2 so this would be three halves 1 and 1/2 let me erase this and write those in so this is three halves one and 1/2 and again for this one it's going to be one again and then for this one it's going to be three halves again when we think about graphing this guy let me grab this real quick I have given you a link for all of these guys that we're going to look at today on Desmos so just click on that if you want something that has a little bit more detail or is easier for you to follow along with so you can just pause the video and check these points out but you see that this point right here again at the beginning of the cycle you have 5 Pi over2 comma 3 then this right here is a quarter of the way through you have piun / 6 comma 1 so you want to focus on this one right here because this x value is where you're going to have your vertical ASM toote it's always going to be a quarter of the way through and then you're going to have Midway of piun 12a 1/2 and then this one right here is 3/4 the way through again that's where you're going to take this x coordinate of Pi 3 you're going have a vertical ASM toote there your 3/4s of the way through so P pi over 3 comma 1 and then you're going to the end of the cycle so 7 piun over 12 comma 3s what a lot of teachers will do is they will draw a horizontal line which in this case would be yal 1 they're showing you that the midline of this guy has been shifted up by one unit so this right here would be Y is equal to 1 so a lot of students will do that instead of remembering that it's at the quarter and 3/4 and just say okay well anywhere where there's a point that's on that line well this has been shifted up like that so I'll look at the x coordinate in this case is NE Pi 6 it's covered up a little bit in this case it's pi over 3 and that's where I'm going to have the vertical asmp tootes drawn so go to the next graph that's what we have so here's your x = piun 6 here's your X = < over 3 that's all your focused on this is a quarter of the way through and 3/4 of the way through that's how you want to draw this in terms of getting the vertical Asm tootes for the seein function now once you have that you're ready to put everything together so let me just highlight this again so we have this vertical ASM toote and this vertical ASM toote so this is extended to the left and to the right a little bit but again if you're doing this on your own you're going to have this point right here that we found this point right here that we found and this point right here that we found again this point right here and this point right here that's not really relevant anymore because we use it to make the vertical ASM tootes so we don't need that the idea is that again when you see a hill here it becomes a valley here so that means that this right here would be drawn to where it approaches that vertical ASM toote and then this right here would be drawn to where it approaches the vertical ASM toote same thing for this part right here and again the way you know that this is going down like that this right here is going to be a valley so this right here becomes a hill this minimum here becomes a local maximum here this maximum here on the cosine guide is going to become a local minimum here then when we look at this part right here so again this is a hill so this is will be a valley so this maximum here on the cosine guide is going to be a local minimum on the secant graph so this guy right here will be drawn to where it approaches the vertical ASM toote and doesn't touch it all right let's take a look at another example with secant so here we have F ofx equal the secant of X over 3 - 2 piun 3 + 1 so again let me just Factor this guy so we can get the phase shift more easily let me make that a little bit better so this is the secant of I'm going to pull out a 1/3 and this one's very straightforward let me slide this down so I have room so inside you'd have x - 2 pi close that down close this down then plus one okay so looking at this just getting some information if you want the phase shift which we're going to use with the quarter period method you have to undo what's being done X so you have this minus 2 pi here so I'm going to add 2 pi to undo that so that means I'm going to the right so I'm going to the right by 2 pi then for the vertical shift again you have this plus one here so that means I'm going up one for the period again you're going to take the standard period which is 2 pi and you're going to divide by this number outside the parentheses in factored form or if it's not in factored form you have that X over 3 so the coefficient of x is 1/3 so it's just 2 pi ID A3 which is 2 piun * 3 which is 6 Pi so your period here is going to be 6 pi and then the quarter period that's going to be 6 pi over 4 me slide down a little bit so 6 piun over 4 6 ID 2 is 3 and then 4id 2 is 2 so it's 3 piun over 2 okay so if you want to do this using graphing Transformations again typically we always do that for y but for X it just depends on the student I'll just give you the information if you want to do it that way that's up to you so let me put out X and Y here so for X again you have to undo what's being done to it and to do that you go in the reverse order so as we talked about in the first problem this is done first and this is done second because you're reversing the order of operations you go inside the parentheses first so now this would be second and you multiply second so now this would be first and you have to undo what's being done so multiplying by a third to undo that you need to multiply by three so multiply by three and then to undo this subtraction of 2 pi you need to add 2 pi so add 2 pi and so what's happening is this multiplying by three that's giving you a horizontal stretch by a factor of three and then adding two that's going to shift me to the right by 2 pi then for y again that's very straightforward you don't have anything outside here you can just put a one if you want so really I put one and two but you just need one thing here so you wouldn't even consider that because multiplying by one is not going to change anything so really all you have to do here is just add one and so that just shifts me up by one unit so again if you want to do it this way it's up to you I'm going to use the quarter period method because most students find that easier so let me go ahead and grab this real quick and I'm just going to erase this and just write in a cosine okay so going back to the basic points we had so we're going to say that we have 0 comma 1 again starting at the maximum then you have pi/ 2 comma 0 that's your x intercept then you have Pi comma 1 so down to the minimum then you have 3 Pi / 2 comma 0 so that's your x intercept me slide down a little bit and then you'd have 2 pi comma 1 so you're at the maximum for the end of the cycle so we're going to again you can use the quarter period method or you can use the graphing Transformations that we wrote on the other page we're going to find the x coordinates first so again you would start with a shift because this guy right here is zero so you're just going to add to Pi because we're shifting 2 pi to the right so this guy is going to occur at 2 pi now and then from that point on you're just going to add quarter periods so the period was 6 pi and the quarter period was 3 pi over 2 so if I add that to 2 pi what that's going to give me let me multiply this by 2 over2 so you could say this is 4 Pi / 2 so this would give me 7 piun / 2 so this is 7 piun / 2 and then from this point on you're just going to keep adding 3 piun over 2 so 7 piun 2 + 3 piun over 2 that's 10 piun / 2 again you could simplify that but we'll do it in a moment 10 piun over 2 + 3 piun over 2 that's 13 piun / 2 and then 13 piun over 2 + 3 pi/ 2 that's 16 pi over 2 so this right here would be 5 Pi if you simplified and then this right here would be 8 Pi if you simplified okay so now that we have the x coordinates again for the Y coordin you can just use graphing Transformations you could plug in if you want but again it's the same thing so what I would do is I would just say that whatever the y coordinate is here I'm just going to add one because we're shifting up by one unit so 1 + 1 is 2 0 + 1 is 1 negative 1 + 1 is 0 and again for this one and this one you can just copy so for zero it goes to one and then for one it goes to two and let me paste this in once again I'll just mention this I have a link for you on Desmos if you want to follow along more closely maybe you're having trouble because of the scale of the graph or using a smaller device or something like that but you see the beginning of the cycle we have this 2 pi comma 2 then we have 7 pi over 2 comma 1 then we have 5 Pi comma 0 you have 13 pi over 2 comma 1 and then 8 Pi comma 2 the common mistake that students make is they're not paying attention and they say okay I have an X intercept here so they draw a vertical ASM toote that's wrong you want to look at what happens at the quarter so at the quarter and 3/4 okay because this guy has been shifted up by one unit again a lot of teachers we'll take and draw that new midline so this is yal 1 normally it's yal 0 or the xaxis but because of the plus one it shifted up by one unit so whenever you have a point that's on that line you're going to take the x value there and you're going to say that's where my vertical ASM toote is going to be for the secant function so xal 7 pi/ 2 would be a vertical ASM toote xal 13 pi2 would be a vertical ASM toote so on and so forth again if you want to continue this to the left of the right that's going to be up to you again you can keep adding quarter periods to your X values to keep going to the right remember the Y values are going to cycle so for the beginning it's two then for the quarter it's one halfway it's zero then for the 3/4s it's going to be one and then finish up at two all right so continuing again at X = 7 piun over 2 you can have a vertical ASM toote and then xal 13 pi/ 2 you can have a vertical ASM toote okay so coming to the graph let me just highlight what we just did so this is your x = 7 piun 2 this is your x = 13 piun / 2 so those are the two vertical ASM tootes we found for the secant function and again all you want to do is you want to take that point and let's just say it's about right there and you're going to sketch this to where it approaches the vertical ASM toote when you look at this right here for the cosine this is a hill so here it's going to be a valley so this maximum becomes a local minimum here and then again similarly let's say we make a point about right there and we would say that you could go this guy approaching the vertical ASM toote and then this way you would approach the vertical ASM toote so just like that again it does not have to be perfect just get it as good as you can again this Valley right here on the cosine graph becomes a local maximum on the seeking graph let's say you have this point about right there that you found so this guy right here would be drawn to approach the vertical ASM toote but not touch it again if you want to extend this to the right or to the left you can click the link on Desmos and just practice that way but generally just drawing one cycle of this guy particularly if you're self-studying that's good enough all right now let's change things up and look at the cosecant fun function so with this one you're going to see that a lot of things we learned with the secant function is just going to carry over so let's do yal the cose of X and again I'm going to do this over one period now negative pi and Pi will not be included here because that's going to give you undefined so let's say that X is strictly greater than negative pi and strictly less than Pi so let's think about this using the unit circle me come back up here for a moment so remember that the cosecant of X by definition is 1 over the S of x so a moment ago I said that it would be undefined at pi and Pi but just thinking about where the S value is zero so that's right here at zero right here at Pi so just using the standard unit circle but again this is going to repeat every Pi so you have 0 Pi 2 pi 3 Pi 4 Pi 5 Pi so on and so forth or if you wanted to rotate clockwise you could say let me just show you this real quick if you rotated like this this would be Pi so that is pi that is co-terminal with pi so it's the same sign value so you could do PI -2 pi3 pi4 Pi stuff like that you're still going to get an undefined value for cosecant because the sign value is zero so going back you will see that anywhere where the sign value is zero you're going to get these vertical asmp tootes so again at negative pi you're going to get one you get one at zero but you typically do not put a dashed line right there because this is the Y AIS so you can put one in if you want just know that that is a vertical ASM toote so what would would be xal 0 or again the Y AIS and then here you have Pi but if you keep going you'd have 2 pi 3 Pi 4 Pi if you kept going this way you'd have -2 pi3 pi4 Pi so on and so forth so you can basically say this is going to happen at Pi n where n is an integer when you look at this point right here so this point right here you have a y value of one so this will be a shared point for cosecant and also for sign and then here you have a yvalue of NE - 1 that'll be a shared point for cosecant and for s and I'll show you that when we get to the combined graph so just remember this point right here this is going to be pi/ 2 comma 1 and then this point right here this is going to be piun / 2 remember that's coterminal with 3 piun over 2 and then comma -1 so again just thinking about the behavior of this guy as you go from Pi / 2 for that x value where the S value is one to an x value of pi where the S value is zero well again because the cosecant of X let me write this out because the cosecant of X is equal to 1 over the s of X well the value here as you go this way and you get closer to an x value of pi it's getting closer and closer to zero but it's never going to touch it so one over something that's really really close to zero again this is going to blow up and we say this increases without bound or it's approaching positive Infinity then similarly over here if you follow this part right here you're going to decrease without bound or approach negative Infinity so just like we saw with the secant function same process is happening it's just happening at different places on the graph now we can go through a few basic points this is just something you can read through for reference so if the x value is zero again Y is undefined this is also going to happen at plus or minus Pi you would get undefined again this is pi n where n is an integer because you have a signed value of zero then when you think about something like pi/ 6 you get two again just to show you that one just so this is Crystal Clear if you look at this right here this is Pi / 6 well the S value is a half so the cosecant of pi/ 6 is 1 over the S of pi/ 6 which equals 1 over 12 which again is 2 if you looked atunk over 6 I'm rotating clockwise so we saw that earlier and so this is coterminal with 11 pi/ 6 but the thing is cosecant and S they are negative in Quadrant 4 so that means I'm going to get the opposite of this so I would want -2 so the cosecant of PK / 6 is equal to 1/ the S ofun / 6 which is equal 2 1 over -2 again you can take that y coordinate right there and so this is equal to -2 so that's why these guys are opposites so coming back down let's think about you have pi over 4 that's squ two then negative pi over 4 would be negative two again they're opposites pi over 3 you have 2 * 3 over 3 negative piun over 3 you have -2 * 3 over 3 then pi over 2 you get 1 neg piun 2 you get ne 1 then you have 2 piun over 3 you get 2 * 3 3 -2 piun over 3 you get 2 * 3 3K 4 you get > 2 3un 4 you get2 and then 5un 6 you get 2 and 5 pi/ 6 you get -2 so just like we saw with secant and cosine we're going to use the S function as a guide when we graph cosecant again where there's a yvalue of one you're going to have a shared point where there's a y-value of negative 1 you're going to have a shared point so you're going to use those shared points again when you have a hill here you're getting a valley here so this maximum here again the range for sign is the same as the range for cosine it's from negative 1 to positive one with both being included so whenever you have a y value of one you're at a maximum on your s graph so a maximum here gives me a local minimum here again there is no minimum or maximum for the cosecant function or the seant function but you do have those local minimums and local maximums some books call that the relative minimum and the relative maximum if you're used to that terminology but this hill right here becomes a valley here or you could say this maximum here becomes a local minimum minum here then this Valley right here becomes a hill here so in other words this minimum here becomes a local maximum here and so on and so forth then again the other thing I would call your attention to is that in this form because there's no vertical shift when you look at where you have the X intercepts on your sign graph that's where you're going to have your vertical asmp tootes for the cosecant graph so something like Pi you would also have zero again we typically do not draw that in for the y- axis if you want to it's not wrong but you would have Zer Pi 2 pi 3 Pi again going over here negative pi 2 pi 3 Pi so on and so forth so just like we saw with the secant function the range here is going to be from negative Infinity to1 where that's included and the union with one going out to positive Infinity so you don't have any y values between negative 1 and positive 1 all right let's spend a little bit of time looking at the characteristics of the cosecant curve so we have here that the domain consists of all real numbers except Pi n where n is any integer again that's because the S value would be zero there and so your cosecant function would be undefined we have f ofx equal cosecant of X is discontinuous again at this Pi n where n is any integer and the graph contains vertical asmp tootes at those values there are no X intercepts you're only going to get those when you start Shifting the graph up or down but on the standard y equals cosecant of X there are no X intercepts the period is 2 pi there are no minimum or maximum values of course we have those local minimums and local maximums that we use to sketch the graph the range is from negative Infinity to1 with this guy being included and the union with you have one to positive Infinity again where one is included but you don't have any y values between negative 1 and postive 1 so keep that in mind the graph has no amplitude because again there are no minimum or maximum values you're just going to see amplitude when you work with cosine and sign now the graph is symmetric with respect to the origin so the cosecant ofx is equal to the negative of the cosecant of X for all X in the domain so f ofx equals the cosecant of X is an odd function so just like the sign function all right so now let's talk talk about graphing variations of the cosecant function so this is pretty much the same thing as we saw with the secant function directions so we're going to replace cosecant with s and sketch the graph using a dash curve so that's your guide function s is used as a guide sketch the vertical asmp tootes again when you think about this because you have vertical shifts you cannot just say that the vertical asmp tootes will occur wherever the X intercepts are on the sign graph that you're using as a guide that will only work if you don't have any vertical shifts so what I always teach people is they occur at the beginning middle and end of the sign cycle because remember on your normal sign graph you have let me just come back to this for a moment so your cycle is going to go from right here to right here so from 0 to 2 pi you're going to start out at an X intercept you're coming up to the maximum and then you're coming back down to an X intercept so right there let me do that in a different color and actually let me Mark this one and this one and this one in a different color and go back to this one so let me come down so you're going down to the minimum and then you're coming back up to the end of the cycle which is an X intercept so the beginning the middle and the end if you remember that this midline is y equal 0 so you're on the x- axis if you had something like a vertical shift up by one well then that midline is going to be yal 1 if it was let's say a shift down by four well now it be y4 so a lot of people will draw that midline in you can do that but what I always do is just say okay what point am I going to have that's first third and fifth so at the beginning the middle and the end that's where I'm going to take those X values and draw my vertical asmp tootes for the cosecant function I'm working with all right then I have sketched the graph by drawing the u-shaped branches between consecutive vertical asmp tootes so these guys are going to approach the vertical ASM tootes but not touch them and we have here the Hills and Valleys are interchanged and then optional I have used the period to find additional Cycles to the left or right as needed if you're self-studying I recommend just doing one period you can of course just look on Desmos and see it to the left or right as much as you want but if you're getting this from a teacher as an assignment they'll generally tell you well I want you to sketch it over this interval so just follow that all right let's take a look at an example so we have F ofx = 12 cose of x +un / 3 and then plus 2 so normally I would Factor this so I can get the phase shift but here the coefficient on X is one so you don't really need to Factor Anything You Can immediately just get the phase shift and say that we're going to go left by pi over 3 again you've got to undo what's being done to X you have plus pi over 3 so that means you need to do minus piun over 3 so that tells me I'm going left piun over 3 and then in terms of your vertical shift you're going to have plus two here so that means I'm going up two so if you wanted to graph a horizontal line through the middle of that guy when you're graphing the sign function as a guide you would sketch yal 2 for that because normally it's y equal 0 and this has shifted me up by two units so for the period for the period it's just the standard 2 pi so you take 2 pi and you divide by the coefficient of x and so in this case that's one so it's just 2 pi and then the quarter period if you're using that method well it's just going to be 2 pi over 4 which would be pi/ 2 so if you want to use graphing Transformations then let's just put down for X here you only have one thing you'd have to do again you're just shifting to the left by pi over 3 so you would just subtract away pi over 3 and that's from each x value from your standard points for the and we're going to work with the sign function a moment when I SW this out so for the sign function and then in terms of the Y you have two things you need to do let me put a little border here the first thing is you multiply by half you multiply by2 and the second thing is you add two so we're going to use this if you want to use this that's up to you in this case I think it's probably equal time to using the quarter period method but a lot of people do not like using graphing Transformations for X for whatever reason so thinking about this this gives me a shift to the left by pi over 3 this right here gives me a vertical compression by a factor of two or again you might say a vertical shrink by a factor of a half again I will not get into a debate about that just understand that you're either dividing by two or you're multiplying by half for each y-coordinate and then you're going to add to that shifting me up by two units and come down here and paste this in so I have some room so I'm just going to rewrite F ofx equal 12 s of x + < over 3 and then plus 2 okay let me get rid of this move this up and we just need to get our points so just start with your standard points for sign so you have 0 comma 0 you have piun / 2 comma 1 then you have PK comma 0 you have 3 piun / 2 comma 1 and then you're going to finish up at 2 pi comma 0 so all you have to do is start things off with your phase shift so that is going to be minus piun over 3 so you take 0o subtract away pi over 3 so that gives me piun over 3 once I've done that I'm just going to add quarter periods that's all I need to do so again if you're using this method you're not going to touch any of these X values you're only going to do that if you're using the graphing Transformations method and I know that gets confusing but essentially you're just saying okay well the quarter period here is going to be pi/ 2 I need a common denominator so let me multiply this by 2 over two let me multiply this by 3 over 3 so that would give me 3 pi over 6 so 3 pi over 6 and this would be -2 Pi so -2 pi over 6 you can eras this now so I'm going to add 3 pi/ 6 so that gives me pi over 6 so this is pi/ 6 and again you're just going to keep doing this so you're just going to keep adding 3 piun over 6 p piun over 6 + 3 piun over 6 that is 4 piun over 6 don't simplify yet 4 piun 6 + 3 piun over 6 7 piun over 6 7 piun 6 + 3 piun over 6 10 piun over 6 okay so now you can simplify you just want to make sure you're not simplifying then you got to get a common denominator again so 4 pi over 6 4id 2 is 2 and 6id 2 is 3 so it's 2 piun over 3 and then 10 piun over 6 10 ID 2 is 5 and 6 ID 2 is 3 so that's 5 piun 3 the other way you could have done this we only have one thing we're doing to X we need to subtract away pi over 3 so if you're using that method you're working off of these values only so 0 - piun over 3ga pi over 3 pi over 2 let me show you this real quick so pi over 2 minus piun over 3 let me make that better here that's going to give you pi over six so if I multiply this by 3 over3 multiply this by 2 over two I would have 3 piun - 2 piun and this is over 6 and of course that is piun over 6 same thing as I got right there and you can keep doing this if you want you could do PI minus pi over 3 let me just do one more of them just so this is Crystal Clear how you can use the different methods so let's write 3 pi over 3 minus pi over 3 and of course that gives you 2 pi over 3 so you're getting the same thing so it's the same result either way you just have to be completely clear on which method you're using and what steps work with each method I don't want anybody to get confused with that being said let's go through and get our y values again you can always plug in but you know that if you plug in negative pi over 3 right here this part right here is going to become this right here again piun 3 + piun over 3 is 0 the sign of 0 is 0 so this becomes this right here so the shortcut is again just to use graphing Transformations you're just going to say multiply by half times whatever this is so that's zero then plus two so then this would be 1/2 times this guy right here this one and then plus two so on and so forth so this right here would be2 time your zero again and then plus your two and of course this is a duplicate of this so you don't have to do this one we can just erase this one you could do this one so 12 * -1 and then plus 2 so this would be two we know that 1 12 * 0 is 0 and you add two so you get two so this one's two and of course this one's two and of course this one's two for this one 1 12 * 1 is 1/2 and then plus two you could say this is 1 12 plus 4 over2 so this is 5 over2 so five halves is what you're looking for there okay here this would be 1 12 * 1 that's -2 and then plus 2 so let's say + 4 over two and that would be three halves so that would be for that one so not too bad overall again getting the Y values is always much easier let me go ahead and grab this real quick and let me come down here and paste this in and really I don't need these I just need these so let me move this over so you can see everything again I gave you a link for Desmos if you want to click on that but essentially when we graph this sign function as a guide you have this piun 3 comma 2 you have this piun 6 comma 5es you have 2 piun 3 comma 2 you have 7 piun 6 comma 3es and then you have 5 piun over 3 comma 2 remember when you think about this at the beginning the middle and the end again that's where you have your X intercepts on that basic sign function this guy right here has been shifted up by two units again this is what a lot of teachers do to make this crystal clear so this is yal 2 again it was y equals 0 but you have this plus two here that shifted everything up so anywhere where you're hitting that line yal 2 2 or your y coordinate is 2 so that means that the x value there that's going to be a place where the cosecant function is undefined so you'd have a vertical ASM toote in other words if I plugged in a negative pi over 3ga piun over 3 plus pi over 3 that's 0 the sign of 0 is zero but again if you had the cosecant of 0 you would get undefined because the S value is zero same thing would happen with 2 pi over 3 and this is 5 piun over 3 hard to see it because it's circled let me recircle it here in a smaller way so the idea is you'd have a vertical ASM toote here at xal 3 and then here at x = 2 piun 3 and then here at x = 5 piun 3 so just showing this to you so we draw these in as X = piun over 3 and then you can have x = 2 piun over 3 and then x = 5 piun over 3 all right so let's put everything together again here's your vertical ASM toote x = piun 3 here's your x = 2 piun 3 so another vertical asmt toe and then here's your x = 5 piun over 3 of course you can use your period to get additional vertical ASM tootes going to the right or to the left just depending on how far you want to extend this what I'm going to do right now is just concentrate on what we worked on so we know that we had a point that was let's just say about right there and then one that was about let's just say right there so again you're going to Interchange your Hills and your valleys so this right here is going to be a hill for the sign function the guide function so that would mean that it's going to be a valley here in other words this maximum point on your sign function here becomes a local minimum on the cosecant function here and then with this guy you have a valley on the S function so that becomes a hill on the cosecant function or another way to say that is you have a minimum on your sign function and then that's going to turn into a local maximum on your cosecant function so what we're going to do is draw this to where it approaches the vertical asmp toote but doesn't touch it so there's that one and then I'll do that over here as well and then I'll come down here same thing so approaches the vertical ASM toote but doesn't touch it and then let me do this part right here so again that would be good enough for me if I'm self studying you've done one period so if you want to continue this to the left or the right again you have that link on Desmos you can calculate all the things you need and then you can just use that link to check it all right let's take a look at another example here with cosecant so f ofx = 2 cosecant of x /2 minus 5 piun over 6 and then minus one so again I'm going to factor this just to get things started so F ofx equal 2 cosecant of pull out a 1/2 and so inside you'd have x minus if I'm pulling out a 1/2 how do I undo multiplication by half I need to multiply by two so this one you might need to write it out so 5 pi over 6 * 2 and so this would cancel with this and give me a three down there so this is 5 pi over 3 and I know this gets a little bit tricky when you're factoring like that but you can always check it a half * X is 12 X or X over2 then minus and minus 1/2 * 5 piun over 3 that is 5 piun 6 so that checks so then we have minus one so at this point we can just get the phase shift right away that's why I factor so for the phase shift you're going to get again how do I undo this minus 5 pi over 3 I need to add 5 pi over 3 so I'm going to the right 5 pi over 3 this part right here is a vertical shift I'm going down by one unit so vertical shift so it's down one unit then for the period so for the period you look at this 1/2 here that again is the coefficient of x in unfactored form in factored form it's outside the parentheses but essentially you take your 2 pi which is your standard period and you divide by half well if I divide by half it's like multiplying by two so this gives me 4 Pi for the period if you want the quarter period because you're using that method so the quarter period Well you just take four Pi you divide by four you get Pi that's all it is if you want to do graphing Transformations again it was more simple on the last one this one you have two things that are happening on the inside so again you've got to undo things and to undo something you have to go in the reverse order essentially if you look at this I'm going to go inside of the parentheses first if I'm using the order of operations I know you can distribute this but just the way it is I would go inside of this first and then I would multiply second so I've got to undo this in the reverse order so now this is one and this is two and I'm undoing things so for X how do I undo multiplication by half or division by two well I'm going to multiply by two so I'm going to multiply by two so in other words I have a horizontal stretch by a factor of two then this guy right here I need to undo the subtraction of 5 pi over 3 well that means I need to add 5 pi over 3 so I'm going to add 5 pi over 3 and that's shied me to the right by 5 pi over 3 then again for the Y this is always much easier and very straightforward you just follow the order of operations so you have that two that's multiplying this so we're going to multiply by two and that's going to give me a vertical stretch by a factor of two and then the second thing is you have this minus one so we're going to subtract one and then that is just going to shift me down by one unit so if you want to use this that's fine you just take those original points for your sign graph and then you take the x coordinates you're going to take each one multiply by two and then add 5 pi over 3 that's going to be slower in this case because of getting common denominators it's going to be faster to use the quarter period method so I'm just going to stick with that maybe I'll show you one or two points with this just for clarity so let me go ahead and grab this and let's paste this in and of course what we're going to do is get rid of that cosecant and we're going to catch the graph of the guide function so we're going to put in a sign and make the N better there all right so once we look at this again you think about your basic points for the sign graph so you have 0 comma 0 you have pi/ 2 comma POS 1 you have Pi comma 0 you have 3 pi over 2 comma -1 and then you're going to have 2 pi comma 0 all right so let me start out by doing the quarter period method and then once we've done that I'll do a few points points using the graphing Transformations so the quarter period here is going to be Pi now your phe shift is plus 5 pi over 3 so what you'd want to do is write this with a common denominator and you could say this is 3 pi over 3 that three could be a lot better and so the first thing here is going to be I'm just going to shift to the right by 5 pi over 3 so I'm going to say instead of zero now I have 5 pi over 3 and then to get the additional guys here you're just going to keep adding add in quarter periods so don't worry about these X values here you're done with that you just want to add this guy right here to this guy right here and they just keep going so 5 pi over 3 + 3 piun over 3 that is 8 piun over 3 then 8 piun over 3 + 3 piun over 3 that is 11 piun over 3 then 11 pi over 3+ 3 pi over 3 that is 14 pi over 3 that one could be better and then 14 pi over 3 + 3 pi over 3 that is 17 pi over 3 so it's pretty quick in a lot of cases once you get a common denominator all right so the next thing is if you wanted to try this with graphing Transformations again we were told to multiply by two and then add 5 pi over 3 so with this one you are working with these X values this one you can just skip because you're multiplying by zero first and then you're adding 5 pi over 3 so of course you're going to get 5 pi over 3 but for this one let's just try it so pi/ 2 you're going to multiply by two first and then add 5 pi over 3 so let's see if we do get 8 pi over 3 so this right here we cancel and what do you get you get 3 piun over 3 plus 5 piun over 3 let me make this five better and of course you do get 8 pi over 3 so let me get rid of this and you can see that works if you want to do the rest of them that way that's up to you all right so for the other guy we need our y-coordinate and for that we're going to use graphing Transformations again if you plugged in 5 pi over 3 this part right here would turn into zero so to shortcut that all you have to do is say well I have this guy two right here times whatever this would be in this case it would be zero and then you have minus one and then for this one it would be 2 * this would be one and then minus one and this is going to be a repeat so we just have to do this one so you would have two times again this right here would end up being negative 1 and then Min - one so 2 * 0 is 0 and then you have - 1 which is just negative 1 so - one 2 * 1 is 2 - 1 is 1 so you can put that there and then this one's just going to be a copy so it's 1 and then 2 * 1 is -2 -2 - 1 is -3 so this is -3 and then this one you can copy it's NE 1 okay let me paste this in here real quick I'm just going to erase this and so at the beginning of the cycle you have 5 pi over 3 comma 1 then at a quarter of the way through you have 8 Pi 3 comma 1 then at the middle part you have 11 pi over 3 comma 1 then 3/4 of the way you have 14 piun over 3a3 and then finishing up the cycle at the end we have 17 piun over 3 comma 1 so you can get rid of these again where you're looking for where to put your vertical asmp tootes again at the beginning the middle and the end so just remember that beginning middle and end the horizontal line that passes through the middle of this guy is going to be at yal 1 now because you've shifted down by one unit you have that minus one there so this is yal ne1 there a lot of teachers do this to say hey if you have a point that hits that midline there well you want to take the x value there and you're going to use that to create a vertical ASM toote for your cosecant function that's all you have to remember so we'll have one at 5 piun over 3 11 piun over 3 and then 17 piun over 3 so there's your vertical ASM tootes again here's one at 5 piun over 3 here's one at 11 piun over 3 and then here's one at 17 piun over 3 so putting everything together let's do again the x = 5 pi over 3 a vertical ASM toote and then the other one that we saw x = 11 pi over 3 and then the last one that we found was x = 17 piun over 3 so we've got those again you can always get ones to the left or to the right going as far out as you want and then in this cycle here we know that we found a point let's just say it's about right there and we had a point again let's just say that one was about right there so the idea here is that this hill on your sign graph the guide graph is going to become a valley on the cosecant graph or again you can always say that this maximum here on the S graph will turn into a local minimum here on the cosecant graph then when we get to this point here again this is a valley on the S graph so it becomes a hill on the cosecant graph or again another way to say that is that this minimum here on your s graph is going to become a local maximum on the cosecant graph so once you have that down you're just going to sketch this guy again to where this is going to approach the vertical ASM toote but not touch it and again just approach the vertical ASM toote and not touch it and then this guy will do the same thing here so approach the vertical ASM toote and not touch it and then approach the vertical ASM toote and not touch it once again I gave you a link on Desmos if you want to do some additional practice you can go to the left or to the right by as much as you want and just use Desmos to check your work in this lesson we want to talk about fundamental identities in trigonometry so at this point we've already talked about the reciprocal identities we've talked about the quotient identities and the Pythagorean identities if you forgot all that stuff because it's been a while since we covered it it's okay I'm going to pull up a quick little cheat sheet in a moment before we get into what we can do with the fundamental identities I've got to give you one other piece of this puzzle okay and that's going to be the negative angle identities which some books call the even odd identities okay so it just depends on your textbook which name you're going to use now when we look at this little graph here we're going to consider two angles one is going to be this angle Theta okay so here's your initial side here's your terminal side and you can see this point here on the terminal side is just marked with X comma y just generically if I had a negative Theta so the negative of this guy so now we're moving clockwise well this point here on the terminal side notice how it's X comma negative y right so this is the Y up here and this y down here is the negative of that notice how the x value is the same okay so we're going to use that to build this little relationship here we're going to first start with s of theta okay so s of this guy right here this would be equal to what it's always y over R but when I think about s of netive theta now how do things change remember now I'm dealing with the negative of Y okay so the negative of Y over it's still going to be R okay so what I can see here is that these guys are going to be negatives of each other this guy is y over R and this guy is going to be the negative of Y over R so I can use that to build a little relationship that's always going to be true and I can just say that the sign of negative Theta is always going to be equal to the negative of the S of theta okay so this is your first negative angle identity now you might notice that this looks very familiar from earlier when we worked with functions and we talked about even and odd functions right this property here is from odd functions right so this is an odd function if you had something like f ofx and you said this was equal to the negative of f ofx remember this is an odd function that's how you check for it so let's erase this and think about cosine now so with the cosine of let's say Theta okay what is it it's X over R so it's X over R but now when I look at the cosine of negative Theta okay it's the same X right the x is the same you have X comma y you have X comma negative y so X is not changing okay so we can say that the cosine of theta is equal to the cosine of negative Theta remember this is a property of even functions so we'll say that the cosine of theta is equal to cine of theta okay all right so lastly I'm going to take the relationships here and I'm going to build something with Tangent okay so I know that the tangent of theta is equal to the S of theta over the cosine of theta now let's think about this if I had a negative Theta then all of these would just be negative Theta right so if I erase this and put a negative Theta and I erase this and this and I just put a negative Theta in here and a negative Theta in here what I want to do is think about how can I take what I just found here plug it in there and simplify and come up with something well basically we know that the cosine of negative Theta right here tells me that's the same as the cosine of theta so I can erase this and go back to the cosine of theta and I can take this guy right here the S of negative Theta and I can basically pull the negative out in front so let's get rid of that okay and I'll pull that out in front so what does this tell me if I compare this one to the one I just had so the tangent of theta where it's the S of theta over the cosine of theta where they're negatives of each other right so just like this first relationship so we can say that the tangent of negative Theta is equal to the negative of the tangent of theta all right so let me erase this using a similar thought process we can get the ones for the reciprocals okay so basically the reciprocal of sign is cosecant okay so when you look at this the signs are going to be the same right so basically it's the same property so with this guy let me put this down here in a different color if I deal with cosecant so cosecant of negative Theta it's equal to the negative of cosecant of theta okay with cosine you have the reciprocal that's going to be secant right so I can do here I'll drag this down here let me change my color the secant of let's say negative Theta will be equal to the secant of theta okay so those match then lastly this guy right here let me change the color one more time I think I'm running out of ink so let me use this one so here with the tangent again it's going to be if you flip it you get cotangent so I'm going to say that the coent of negative Theta is equal to the negative of the coent of theta okay so these are the negative angle identities or some books again call it the even odd identities because of when we're talking about even functions now I want to pull up this little PDF which is attached at the beginning of this section for you okay so I say keep this PDF handy it's going to be very useful as a reference in this section you want to start by just going back and forth and reading things and writing it out okay as you get really good at working with these identities you're going to commit these to memory okay you're really going to need to do this before you get to calculus so you have here the reciprocal identities okay we have here the quotient identities we have here the the Pythagorean identities and then the negative angle identities which we just talked about so we're going to use this in this section together these are all called the fundamental identities and we're going to start by just solving a problem that we've already seen how to solve before okay we're just going to do it in a different way so we have the tangent of theta equals 34s and S of theta is positive okay so let's stop for a minute this tells us we're where remember the all students take calculus so if I draw a little coordinate plane here the all students take calculus okay everything is positive in the first quadrant in the second quadrant it's sign and cosecant in the third quadrant it's tangent and cotangent and in the fourth quadrant it's cosine and secant okay in terms of what's positive so here we have that tangent is positive and so is s so that means I have to be in quadrant one okay so before you do anything let's just figure that out and we're just going to put here that we're in quadrant one okay so I want to find find the secant of the of and the ofet so let's start out with the secant of how can we figure this out using identities let me pop this back open and I want to look for a relationship between tangent and secant okay and I see that I have one here tangent sare Theta + 1al secant Square thet so let's go ahead and write this out so we have that the tangent squar Theta then we have + 1 is going to be equal to the secant squ okay so remember and this is going to come up later on when we're simplifying trigonometric Expressions this is the same as if I had tangent Theta squared okay this is just a different way to write this so what I'm going to do is I'm going to plug in for what I have so I'm going to plug in the 34s there this is 34s and it's being squared okay that's why I just showed you that so let's erase this and put 34s here and I'm going to square this guy okay and so if I Square three fours I get 916 let me scroll down to get some room going we'll come back up and answer this so this will be 916 + one now to do anything here I know I need to get a common denominator so I might as well just write this as 16 over 16 and just go ahead and add 9 + 16 is 25 so this would be 25 over 16 okay so this equals secant squar Theta now I want to get this guy this secant of theta by itself because that's what I'm told to find right I want to find secant of theta now SEC square of so to do this I want to take the square root okay but I need to know which one to take and I need to think about okay well if I'm in Quadrant One everything's going to be positive okay so I just want the principal square root so I'm going to take the principal square root over here and this guy over here will now be just secant of theta okay so here I can simplify and say and I'm just going to flip this around and say the secant of theta is equal to the < TK of 25 is 5 and the < TK of 16 is 4 so this guy is going to be 54s okay so that's one way we can do it obviously we saw earlier in the course that we could have done this a different way whichever way you want to do I'm just showing you different ways you can attack the same prop so let's put this up here let's erase all this and let's just drag this up here and we'll attack the next part of that problem so we have here if we go come up here we want to find the S of theta okay so let's think about this and there's a lot of different ways to do this but again if I go back to my little cheat sheet I want should have see this quotient identity here where the tangent of theta again is the S of theta over the cosine of theta now you might think well we don't have cosine of theta but we do because of the reciprocal identities we just found the secant of theta which is one over the cosine of theta okay so let's go ahead and put this down here I'm going to first just write the relationship that the tangent of theta is equal to the S of theta over the cosine of theta okay and the cosine of theta is one over so it's 1 over the secant of theta okay so you know you can flip this and you would basically have the secant of theta okay times the S of theta but I want to solve for this guy so I'm going to divide both sides by the secant of theta and that's going to give me the S of theta by itself now all I have to do is plug in the information I'm given so I have for the tangent of theta 3/4s so I have 3/4s for the secant of theta I just found that was 54 okay so basically what I'm going to do here is say let me scroll down and get some room going I'll come back up I have 34s times I'm going to flip this because I'm dividing fractions so four fths we can see that this is going to cancel and I'm left with 3 fifths so that's going to be my S of theta so let's go ahead and put this over here and say that my S of theta is equal to 35s okay for the last part of this problem we're asked to find the cotangent of negative Theta so the coent of Nega Theta okay let's just pop open this guy again we can see the coent of theta is one over the tangent of theta okay so I could say this is one over the tangent in this case I have a negative Theta so negative Theta okay and then from here I'm given that the tangent of theta is 34s but I want the negative of that remember you can pull this out we saw that earlier so you might as well say that this guy is going to be equal to 1 over the negative of 34s okay so to finish this up I'm just going to flip this and say that the coent of negative Theta is going to be equal to the negative of again flip this guy it would be 4/3 so those are going to be your answers here okay you're going to have the secant of theta is 54s the S of theta is 3 fths and the cotangent of negative Theta is 4/3 all right so the next thing we want to do is we want to write one trigonometric function in terms of another so you're going to see this in your book let's say you see something like the cosine of theta and you're asked to write this in terms of okay so in terms of the tangent of theta okay so how can we do this well the first thing you want to do is you want to find an identity that's going to work where it's going to relate your trigonometric function which in this case is the cosine of theta to the tangent of theta so if we go back here I'm going to pick this Pythagorean identity here so I'm going to pick this guy which involves the tangent of theta okay and the secant of theta so what we're going to do let's go ahead and minimize this I'm going to go to the next page I've already written it down so we have 1 plus the tangent squ of theta is equal to secant square of theta remember when you think about the secant of theta it's 1 over the cosine of theta okay so what I'm going to do first is just take the reciprocal of each side so I'm going to say this is 1/ 1 + tangent 2 thet is equal to 1^ 2 okay so now I'm going to replace I know that this guy right here again because of the reciprocal identities is going to be cosine squ Theta okay so I'm going to replace this on the left I'm going to say 1 over 1 + tangent 2 thet is equal to on the right I'm just going to put cosine 2ar Theta now I want this in terms of cosine of theta okay so I'm going to take the square root of these sides here on this side I want plus or minus because I'm not given a quadrant okay so I don't know any information about what quadrant I'm in so I'm going to write this let me change the color here I'm going to put cine of theta is equal to plus or minus the square root of this whole thing so one over you've got 1 plus your tangent squar Theta okay so from here we know that we can take this and make it a one so let's go ahead and say cosine of theta is equal to I'm going to go plus or minus this one up here over down here I'll have the square Ro T of 1 + the tangent squar Theta okay so we want to always rationalize our denominator so to do that let me scroll down and get some more room going I'm going to multiply this by the square root of this guy so 1 plus your tangent SAR Theta over Square < TK of 1 + your tangent SAR thet okay so that's just going to rationalize the denominator so I'm going to have cosine of theta is equal to so this guy right here I'm just going to have plus or minus remember 1 * anything is just itself so I have the square Ro T of 1 + tangent squar Theta over this times this would give me the radican there so 1 + tangent squar Theta okay and we're done so that's the exercise here it's just meant to give you some practice writing one trigonometric function in terms of another okay so now let's move into actually simplifying okay and I think you'll find that once you get started started on simplifying and writing things in terms of other things at first it's a bit challenging but really once you practice enough this becomes really really easy so we have the secant of theta time the cotangent of theta time the S of theta so when you get a problem like this you want to think about putting your expression in terms of s and cosine okay and if you have S or cosine in your expression that you're trying to simplify leave it alone don't try to mess with that okay you want to mess with everything else so I'm going to leave this as it is and I'm going to change this one and this one so because there's multiplication here when I go back I'm going to be thinking about my reciprocal identities okay so I'm going to be using these guys here to simplify so basically the secant of theta is 1 over the cosine of theta okay the coang of theta I know is found by flipping the tangent of theta so it's going to be the cosine of theta over the S of theta okay over the S of theta and then this is times this s of theta so you see that when you plug these things in a lot of stuff's going to cancel so this cancels with this this cancels with this and I'm basically just left with one okay so this is pretty common when you just see multiplication like this if you have S or cosine leave it alone okay the other stuff is what you want to plug in for and try to see what you can get to cancel let's take a look at another example where we're simplifying a trigonometric expression this one is going to be problematic for a lot of beginners because of the notation here so I want to make sure you understand the notation so when we work with again something like sin squar Theta remember this is the same thing as if I said s of theta being squared like this it's very important you understand that when you look at this guy right here the Temptation because of s of negative Theta okay is equal to the negative of the S of theta the Temptation here is to do what it's to go the negative sin 2 thet okay that's not actually accurate okay you have to think about it like this like this guy right here okay so the way I want to write this is I want to make it Crystal Clear what's going on and I want to say that I have the S of negative Theta okay this guy being squared okay the whole thing being squared then plus I'm going to do the same thing here so I'm going to have the tangent of this negative Theta okay this guy right here being squared then plus I'm going to have the cosine of this negative Theta being squared okay the whole thing being squared so now it's Crystal Clear okay when I pull this negative out what's going to happen we're going to have the negative of s of theta but the thing is squared plus pull this out okay I'm going to have the negative of the tangent of theta and I don't need parentheses in that case Okay and then this guys can be squared and then Plus in this case you don't need to pull anything out because the cosine of theta is equal to cosine of thet okay so I can just say this is the cosine OFA being squared okay so now I want you to see what I'm getting at here basically if you had incorrectly pulled that out and said you had the negative of sin squ Theta you would get the wrong answer okay because this guy right here because you're squaring the negative it becomes positive right so this is going to go away so you get rid of that and you basically go back to having sin squar Theta not the negative of sin Square Theta sin Square Theta okay because this negative got squared same thing here this is going to get squared so plus tangent SAR thet then plus you have your cosine squ thet okay so from here that's the first step just to understand where you go in there but from here you want to realize that if you're dealing with something that's squares here and you've got pluses and minuses involved not in every case but usually you want to look at the Pythagorean identities so if we pop these open you see that sin Square theta plus cosine Square Theta is one so that's the first thing I'm going to do you can consider this plus this to be one you can reorder this if you want to make it crystal clear so we'll say sin^2 theta plus your cine 2ar theta plus your tangent squar Theta okay this guy right here is equal to one okay we just saw that so I'm just going to substitute so I'm going to put in here a one plus the tangent squar Theta now can we make this simpler let's go back and see if there's anything that involves tangent Square Theta and one yes right here tangent sare th + 1 is equal to secant sare thet so I can replace this with just secant squar Theta okay and that's going to be my answer and you can see how we really simplify this down to just one of these guys in this lesson we want to talk about verifying trigonometric identities all right so in this particular section you might struggle a little bit when you first get to it it's a topic that you really need a lot of practice with it's not as straightforward as other topics where you have a set list of things that you need to do okay if you were solving basic equations I could give you the steps and you could perform those steps and basically you would get the right answer all the time with this there are just hints or strategies you can use to arrive at the correct answer but it's not so clear in terms of what you need to do and there's also multiple ways to do the same thing okay so I have some little tips here for you and they're standard from any textbook okay so you'll get this in any textbook the first thing is to memorize the fundamental identities so again in the last lesson we talked about these fundamental identities we talked about simplifying trigonometric expressions using the fundamental identities again we looked at a little handout here so hopefully you have that again and you want to memorize these so you have the reciprocal identities you have the quotient identities the Pythagorean identities and then you have your negative angle or again some books call this even odd identities so all of these are going to play a crucial role when you're verify Ying these trigonometric identities okay so here are some other tips you always want to try to work on the more complicated side okay so when you're verifying trigonometric identities you're not going to be doing the same thing as when you solved equations when you work with equations you might multiply both sides by the same number or add the same term to both sides something like that here you're not going to do that you're going to work on one side generally speaking the more complicated side and try to make it match the other side the simpler side okay another tip is sometimes you'll want to express everything in terms of s and cosine that doesn't always work but it's just a general tip okay you want to perform any indicated algebraic operations such as factoring or expanding here you want to think about all those special product formulas okay and all the special factoring formulas those are going of come up a lot and then we have use the formula for multiplying conjugates so remember if you have one + sin of x * 1 - sin of X this is going to be 1^ 2ar which is 1 and then minus sin^2 X so basically 1 - sin^2 x and we know from our pythagorean identities this is going to be exactly equal to cosine 2 x similarly here I could make this 1 - cine 2 x okay and that's going to be exactly equal to sin^2 X okay so let's take a look at the first example and the first few of these are going to be really really easy so go ahead and pause the video and try them on your own once I kind of get going here and then see the solution so the first thing is I'm going to think about which side is simpler so to me this side is simpler okay so I'm going to try to work on this side on the left now pretty much every book that you're going to see has different formatting generally speaking they sometimes will take this guy right here and just work on the more complicated side and the end they'll show you that it matches what I'm going to do here is I'm just going to take this guy and flip it so I'm just going to say I have 1 + sin of X is equal to over here I'm going to have cosine of x plus cotangent of X okay over the cotangent of X I'm flipping it because what I'm going to do is modify the right and leave the left alone okay so that's generally the formatting you're going to see if your teacher tells you to do something else then follow that okay so what I'm going to do is just put equals over here I'm not going to mess with this my goal is to get this side right here to match this 1 plus s of x so the first thing I'm going to do is I'm going to write things in terms of s and cosine so I have cosine here okay so I'm not going to touch that but cotangent of X again if you think about the reciprocal identities it's cosine of x over s of X so I'm going to start by saying I have cosine of x plus cosine of x over sin of X and this is over again if I have cotangent of X I'm going to write as cosine of x over sin of X okay let's scroll down and get some r we'll come back up when we get to a stopping point okay so what I want to do here is think about the fact that I have basically a complex fraction right so how do you simplify that you look at the denominator so I have S of X and S of X so if I multiply this by sin of X over sin of X basically I'm going to distribute this here and here okay and I'm going to distribute this here so I'm just going to keep my equals over here okay we might need more room so let me just kind of slide this down a little bit sin of x * cosine of x is just s of x time cosine of x then plus s of X here would cancel with s of X here so I just have cosine of x here okay then down here this s of X here would cancel with this s of X here so I'm going to put this over just the cosine of x okay so let's go down a little bit more and let's think about something here we have the cosine of x and the cosine of x what operation could I do there well I could Factor right so I'm going to put equals here and I'm going to pull that guy out okay so I'm going to say I have the cosine of x inside the parentheses I'll have the S of X okay plus if I pull this out I have a one okay then this is over the cosine of x and now what I'm going to do is I'm going to cancel this with this and what I'm left with is going to be the sine of x + 1 okay if we go all the way back up we see here that we have 1 + sin of X and here we have S of x + 1 right so I could write this as 1 + sin of X okay and then I'm just going to put a big check mark here to show that this right side now here matches the left side okay so you could have gone about this a different way there's multiple ways to verify a trigonometric identity the main thing is again start with the more complicated side work your way down okay and try to make it match the other side okay so once you found that it matches you're done put your check mark and move on try to make your steps as clear as possible so that whoever is grading it can give you full credit okay let's take a look at another example this is another kind of easy one so we have the cang of theta over sin s Theta equal cose s Theta over tangent of theta now in this particular case both sides are equally complex so I'm just going to leave the left one alone because it already contains sign okay if you wanted to do it a different way again that's fine so I'm just going to basically put equals here and I'm going to manipulate this in terms of and cosine so cose okay if I think about the reciprocal identities is 1 over s okay so basically I can say cosecant squ of theta is 1 over sin 2ar of theta okay and then over tangent of theta is going to be my S of theta over my cosine of theta okay so what I want to do here think about the fact that you are basically going to flip this guy okay basically if you're going to divide fractions you multiply the first fraction by the reciprocal of the second guy okay so let's scroll down a little bit and I'm going to go 1 over sin^2 Theta times this is going to flip so it's going to be cosine of theta over s of theta okay and can anything cancel here well no you might be expecting to cancel s of theta so at this point you might stop and say well what's going on so let's go back up here okay let's think about what we have we have cotangent of theta over sin squ thet well I have sin Square Theta here this right here if you really think about your reciprocal identities remember cosine of theta over s of theta is cotangent of theta so I can replace that okay I can replace that with coent of theta okay so I can put 1 over sin s thet like this multiply by cent of theta okay and basically we're done right all we have to do is multiply this through and say I have the cotangent of theta over sin^2 Theta okay so now your left side and your right side are going to match so you put your check mark there again just make sure you show your steps so your teacher can give you full credit all right for the next one we have the negative of cosecant s Theta time cosine s Theta is equal to 1 - cose s thet so for this guy you might be thinking that the right side is a bit more complicated it looks kind of like the difference of squares right so it is actually you could factor that if you wanted to but I'm not going to do that I'm going to leave this side alone whenever I see something that looks like this I'm thinking about the Pythagorean identities I have a minus or a plus with a one here and this is squared if I look at my little hand out here I want you to think about this one right here so 1 plus cang squ Theta is equal to cosecant sare Theta okay so we'll come back to that in a moment and I'm just going to work through this I'm going to flip the sides here I'm going to put 1 minus cosecant SAR Theta okay is equal to I'm going to put cose 2 thet time the cine 2 thet okay so what I'm going to do first is just take this side right here and express it in terms of s and cosine again for a lot of these problems that ends up being all you really need to do to end up with a solution so I'm going to put equals over here I'm going to change this I'm going to put my negative out in front cosecant Square Theta again with the reciprocal identities I can put 1/ sin^2 thet okay this is time cosine 2 thet and then over here I have the negative of I'll say cosine 2 thet over sin SAR thet can I make a replacement here yes because this guy right here is the coent of theta right so basically what I have let me scroll down a little bit is the negative of the coent squar Theta now can I do anything with this remember I talked about this Pythagorean identity here okay let's assume I moved this over here and this over here so basically I would have 1us the cose S of theta is equal to the negative of cotangent squ Theta so right now I have the negative of cotangent Square Theta and I can replace it with exactly this right here 1 minus cosecant s Theta okay so that's what I'm going to do so I'm going to say this guy right here is 1 minus the cose SAR thet okay so now my left side side exactly matches my right side I'm going to put my check mark there and I'm done all right let's take a look at a harder example so here we have two * the tangent of theta * the secant of theta is equal to 1 over 1 - the S of theta - 1 over 1 + the S of theta so when you see this guy right here or this guy right here you got to be thinking about conjugates okay so I'm definitely going to leave the left side alone and I'm just going to play with this right side before I do anything I just want to do this off to the side so I have a lot lot of room to work I know that I would want to get a common denominator here okay so how would I do that again I'm thinking about conjugates so here I would multiply by 1+ the S of theta okay here 1 plus the S of theta to make that legal again this is one I can multiply these together so you see how this formula is going to work if you had a minus B that quantity times a plus b this quantity remember this is a squ first guy squar minus b^ 2qu minus the second guy squar okay in this case it's going to be no different this is going to end up being one the first guy squared then minus you have sin Square Theta so sin squ Theta this second guy squar okay so up here this would just be 1 1 + S of theta okay you can get rid of this and then down here this is going to be one you don't need 1 squ minus the sin^2 Theta okay so that's going to be this part right here for this part right here I have a plus here so basically this part is going to be a minus so I'm going to put a minus here and I'm going to end up with what 1 minus s of theta in the numerator because again I'm going to multiply by 1us S of theta here 1- s of theta here and then down here it's the same thing right it's the same thing so it's going to be 1 - sin squ Theta okay so let me erase this and let me slide this over so I get my formatting right slide this over just a little bit put equals there and again I can write this with a common denominator so you can even do this real quick right here so we save a little bit of time time and a little bit of effort so let me just put 1 - sin^2 thet like this and then I'll put this says one numerator with a minus here but be very very careful when you're subtracting this away You're subtracting away the whole thing so go ahead and wrap that in parentheses to remind you that you need to change your signs because you're subtracting the whole thing away okay so let's keep going and see what's going to happen here so I'm going to have 1 + sine of theta again you're Distributing this negative so minus1 distribute the negative here so plus s of theta okay and this is over 1 - s^ 2 Theta so what can we do here well we see immediately that this is going to cancel okay so now I basically have 2 * s of theta so 2 * sin of thet over 1 - sin s Theta now remember in the hints we saw that 1 - sin^2 thet was cosine Square Theta again if you go back to this if you have this guy right here and you just subtract this away so 1 - sin square of theta would give you cosine Square Theta so if we come back here I can replace this with cosine Square Theta so let's go ahead and do that so I'm going to put equal 2 * sin of thet over the cine 2ar thet okay now is this getting close to what we want let's go back up and see where we are we have 2 * the tangent of time the secant of th so are we close to that well I see I have the two I need the tangent of theta and the secant of theta so to get this I'm going to break this up remember cosine s thet can be written as cosine of theta time cosine of theta okay it's just this times this gives me cosine squ thet now up here I have 2 * s of theta now think about it if I have S of theta over cosine of theta that's the tangent of theta okay and if I have 1/ cosine of theta that's the secant of theta so I already have what I need okay you just need to break things up so you realize what's going on so I'm going to put this as 2 * sin of theta over cosine of theta * 1/ cosine of theta and then I'm just going to again replace this with what I'm looking for so in this particular case it's going to be what two times this is going to be the tangent of theta times in this case this is going to be the secant of theta and now the left side and the right side perfectly match so I'm going to go ahead and put my check mark there and again I'm done all right let's take a look at one final problem and this guy is going to be pretty challenging overall so I encourage you to pause the video and try this on your own but before you do so I want to just give you a little hint that you basically want to get both of these to a common expression okay so you want to work on one side get it down to in terms of s and cosine this side is pretty hard if you are not used to working with these this side's pretty easy okay so you want to get it to this common third expression so go ahead and pause the video and try it and hopefully you did give that a try and I'm just going to start with the left hand side here okay so I'm going to start by just taking this guy and trying to write it in terms of s and cosine so how can I do that well let's start out by just putting that we're working on the left hand side okay some teachers want you to do this others don't care so just get with your teacher and I'm going to first copy the left hand side so I have the secant of X plus the tangent of x over the secant of x minus the tangent of X now lots of ways to do this I'm going to go through one that is probably not obvious if you haven't worked a lot of these problems remember the secant of X by definition is equal to 1 over the cosine of x okay so if I multiplied this secant of X which is one over the cosine of x times the cosine of x I would get one okay so in other words if I did this guy right here let me just copy this real quick so secant of X plus the tangent of X over the secant of xus the tangent of X If I multiply this guy by the cosine of x over the cosine of x when I distribute this here this is going to be one when I distribute this here this is going to be one another benefit is when I distribute this here and this here think about the fact that the tangent of X is the S of X over the cosine of x so the cosine of x is going to cancel in each case so I'm going to end up with 1 plus the S of X and 1 minus the S of X so that's pretty simple okay so that's how I'm choosing to do this again you could arrive at this a different way so let me just put equals over here this multiplied by this is going to give me one okay and then plus let me just write this out so the tangent of x times the cosine of x we know we can cancel that further and then we have again cosine of x * SEC of X is 1 minus we'll have the tangent of x X time the cosine of x okay so from here I'm going to again keep simplifying let's put 1 plus I'm going to do sin of X over cosine of x times the cosine of x again this is going to cancel so this cancels with this and this is over we get some more room here I think I'm going to run out so we have 1 minus again I'm going to write this as the S of X over the cosine of x times the cosine of x okay so this is going to cancel and I have the S of X there okay so let's just write this one more time time we'll put 1 + the sin of X over 1us the S of X so that's pretty simple it's in terms of sign okay and basically now what I want to do is try to get the right hand side to match this and you're going to find that that's pretty simple so let me start a new sheet here and I'm going to say the right hand side okay let me make that a little bit better so let me put the right and I'm going to try carefully to write this so right hand side okay penmanship is hard for me so let's come back up and let's think about this guy so we have 1 plus 2 * the sin of X Plus sin s of X over cosine s of X so let's come down so we have 1 + 2 * the sin of X Plus sin 2ar of X okay over we're going to have the cine squar of X so the first thing is you should be thinking about factoring when you see this pattern again this is basically where you have x + Yanti squar right it becomes X squar so first guy squared plus 2 * the first guy times the second guy so 2 * XY plus the last guy squar okay that is exactly the pattern you have here this is going to be basically one plus s okay of xan squar so I'm going to take this guy and let me do this properly so let me just kind of slide this down a little bit okay I'm going to put equals so I'm just going to say this is 1 + the sin of x^ 2 over your cine 2 of X okay so I'm going to put equals over here so what am I going to do with this now that I've factored it well again if you see something like cosine s of X and you see something in terms of s of X I want you to think about the fact that this can be replaced with 1us sin s of X okay so I'm going to write this as 1 + sin of x^ 2 over this is going to be 1 - - sin^2 x so at this point some of you are thinking that there's something weird going on this doesn't match this but it will in a second this is the difference of squares so you need another factoring pattern here this is 1^ squar basically minus sin s x so I'm going to rewrite this and then it's going to be completely obvious what's going on okay so first let's go ahead and write this as 1+ Sin of X Time 1 + sin of X okay so I'm just expanding this nothing illegal about that I have an exponent of two there so I'm allowed to do that okay then down here I'm going to factor this as 1 plus s of x * 1 - sin of X and now it is completely clear what I'm doing okay so this cancels with this and I have 1 + sin of X over 1 - sin of X again if we go back what did we have here for our left hand side when we worked it 1 plus sin of X over 1 - sin of X so what I'm going to do here I'll just put = 1 + sin x over 1 - sin of X and to complete this guy you might just want to write your basically three Expressions here so I'm going to say that the secant of X plus the tangent of X so plus the tangent of X over and I don't know if I can fit this all on the screen but I'm going to try so the secant of x minus the tangent of X let me scroll this down and try to write smaller is equal to we have this 1 plus the sin of X over 1us the sin of X so this is what we found from working the left side and the right side okay so this is your basically your common third expression okay and this is going to be your 1 plus your 2 * sin of X Plus sin SAR of X let me see if I can scooch this down just a little bit more and then over you had your cosine 2ar of x okay so this is going to basically do it for this guy again you're not going to probably see too many of these if you're on a basic test you're probably going to get maybe one of them towards the end they realize how much time these guys take okay so if you get something where you're just working and working working on one side you can't get it anywhere go ahead and get it in terms of sign or cosine stop and then start working the other side see if you can get it to match in this lesson we want to talk about the sum and difference identities for cosine all right so let's start out by looking at the sum identity for cosine so let's say you had the cosine of some angle a plus some other angle B this is equal to the cosine of a Time the cosine of B minus the S of a * the S of B so let me give you a quick example we're going to work on examples like this in a little while but just so you see what's going on let's say you had the cosine of let's say 45° plus 30° so we know this is the same as if you added those two angles and you said you wanted the cosine of 75° but you can break this up if you wanted to get an exact value and I'll talk more about that in a little while but basically if you wanted to do this it's the cosine of the first guy so the cosine of 45° then times the cosine of the second guy so times the cosine of 30° then minus the S of the first guy so the S of 45° and then times the S of the last guy or the second guy which is 30° so this is how you would have to set this up a lot lot of people incorrectly say this is the cosine of 45° plus the cosine of 30° that does not work okay so you can't do that you have to use this formula now for the other guy let's say you have the cosine of a minus B what's going to change is the sign here so I always remember this is the opposite sign cosine of a * cosine of B now plus s of a * s of B so for example if you did the cosine of 45° - 30° well now I'm looking for the cosine of 15° and it's still cosine of 45° * cosine of 30° this is now going to change to a plus the S of 45° * the S of 30° so it's just a change in the sign so this sign and this sign will always be different now let me get rid of this and let's talk about where this guy comes from what you'd want to do is draw yourself a little unit circle on a piece of paper and so again this point right here we know the x coordinate would be one and the y coordinate would be zero again we're on the unit circle if you went around and label these points this would be 0a 1 then this point right here would be -1 comma 0 and then lastly this point down here this would be 0 comma 1 so again we're on the unit circle so we know if we're on the unit circle and let's say starting here and going to here I have this Arc with length T well then I know the coordinates for this point Q sub one the x value is the cosine of T and the Y value is the S of T now if I keep going and let's say I end up at this point right here P sub one well now I have an arc that has a length of t plus s or you could say S Plus T so the x value here for this point is the cosine of S Plus T and the Y value here is the S of S Plus T and then if I come this way notice that I'm rotating clockwise so here I would say this is negative s so whatever this is right here it's the negative of that and so if I look at this point Q Sub 0 the x value here is the cosine of s and the Y value here is the sign of negative s Now using the negative angle identities you can really say that this is the cosine of s because the cosine of negative Theta is the same thing as the cosine of theta and then you can say that this is the negative of the S of s because again if you have the S of negative Theta it's equal to the negative of the S of theta so we're going to be using this right here instead of this right here all right so we have that settled and now what we want to move into is we we want to identify that we have these two equal arcs so hopefully you can see this The Arc Length if I think about here to here so this has s and it has t would be the same as if I consider let me change my color here from here to here right forget about the fact that this is negative s again this guy and this guy in terms of the arc length is going to be the same so you have S Plus T and here you have S Plus T so let me draw this out so that nobody's lost and probably I should change the color again let me use this like red here so let me trace this as good as I can so this Arc here would be equal to this Arc here I'm going to go underneath a little bit like that okay so we can see that those two arcs are equal and when you have equal arcs they are what we call subtended by equal chords so this chord here and this chord here these two are going to be equal right so you see up here I've written that the distance from P Sub 0 to P sub 1 so that's this right here is equal equal to the distance from Q Sub 0 to Q sub 1 okay so that's this distance right here so I went ahead and already set everything up for us so what you would want to do from this point is write out your distance formula remember we're on a coordinate plane and if we want to find the distance between two points we say the distance is equal to the square root of you have x sub 2 - x sub 1^ 2ar plus y sub 2 - y sub 1an squ now what I did was I went ahead and squared both sides of this so I'm working with the distance squared now so I'm saying that distance between the point P Sub 0 and P sub 1 that guy squared is equal to the distance between the point Q Sub 0 and Q sub 1 that guy squared so what we're going to do is just work on one side at a time and then I'm going to set these two guys equal to each other and you'll see you have your formula so first off I have again the distance squared here and I have my two points so let me write this out I'm not going to write the distance squared equals we don't even need that I'm just going to plug in here so I'm going to consider P sub 1 as my second point and P Sub 0 as my first point so let me get the x coordinate here so this would be the cosine of this s + T and then this guy right here would be minus you would want your one here okay so I took the x coordinate here and the x coordinate here let me wrap that and square it then we have plus so now I'm working with the Y's so the Y from this second guy here so that's going to be your sign of this s + T and then we would have minus your y here is so let's just write it in for right now but it's going to go away in a second so that's just setting this up again exactly matching this I didn't write the d^2 equals but we really don't need that so I'm just going to say this equals here so I'm going to use my formula remember if you have let me just get rid of this for a second so I can write this out so x minus y^ 2 this is x^2 - 2x y + y^ 2 so we're just going to use that formula pretty easy so I'm just going to say I have this first guy squared so cosine squared and then the angle would be as s + T there and then minus 2 * this * this now 2 * 1 is 2 so it's just 2 * the cosine of this s + T here and then plus this last guy 1 squared which is just one now I'm going to go plus now this one right here because you have minus 0 you could really just get rid of this and just say you have S of S Plus T being squared so I'm just going to write that as s^ squar of this s + t angle so now we have cine s of S Plus T and we have sin s of S Plus T remember if the angle is the same which we have S Plus T and S Plus T well then cosine squ theta plus sin squ Theta is equal to 1 so this plus this would give me 1 so really I can say that we would have 1 and then we have minus 2 * the cosine of this s + T and then + one so this is going to give me a final result for this side 1 + 1 is 2 2 so we'll say 2 - 2 * the cosine of s + T okay let me grab this and I'm just going to paste this in here so this is one side let me put equals and now we're going to work on this side so we have q Sub 0 and we have q sub1 so just plug in here forget about the d^2 equals I'm just going to find the x sub 2 which again I'm labeling this as the second point so I want cosine of T So then minus you want your x sub one here that would be Co s of s let me close that down and square it so then plus you have your y sub 2 so here that's going to be the S of T so the S of T and then you have minus your y sub 1 in this case that's going to be the negative of the S of s now minus a negative is plus a positive so you want to put a plus there and then the sign of s so be very careful about that step there a lot of people get confused remember this is right here A minus and in the formula you have a minus so minus a negative is plus a positive that's why we have a plus there okay so I'm just going to use my formula here so we know the formula for this one and this one if you have x + y^ 2 it's x^2 + 2x y + y^2 okay if this changes to a minus then only this changes to a minus everything else is the same okay so let's get rid of this and let's write equals here so this would be the first gu^ squ which is cosine 2 T and then minus 2 * this guy times this guy so the cosine of t time the cosine of s and then plus the last guy squar so this is cine s s okay then plus let's work on this guy now notice this is a plus guy here right so we have to think about a different formula so the first guy squared so s^ squar T then plus 2 * again the first guy times the second guy so we'll put the S of T * the S of s and then plus the last guy squared so this is sin s s okay so once again we have a situation where we can use that Pythagorean identity sin Square theta plus cosine sare Theta equal 1 so cosine Square t and then sin Square t so that's going to be one and then let me highlight this in a different color you also have cosine squ s and sin squ s okay let's put equals so we have cosine squ t plus sin Square t as I just mentioned that would be one and then let's go ahead and say cosine 2 s+ sin 2 s let's just put plus one for right now and then we'll go - 2 * the cosine of T time the cosine of s and then plus 2 * the S of T * the sign of s so we know that 1+ 1 would be two so instead of actually making a new line I'll just get rid of this and put a two here that way we can speed this up so I'm going to grab this and I'm going to paste this in here it doesn't quite fit on one line so we know that this is equal to this what I would do let me actually drag this a little bit further down I am going to subtract two away from each side of this equation so this right here is going to cancel and this is going to cancel so let me see if I can write it in one line now so -2 * the cosine of s + T is equal to -2 I'm going to flip the order here for the purposes of the formula you know that you can multiply in any order so I'm going to say this is times the cosine of s time the cosine of T then plus let me slide this down and see if I can fit this we're going to say 2 * your sign of s again I'm flipping the order for the formula time the S of T let's come down here a little bit so what I'm going to do now is just divide everything by -2 and that's going to get us where we need to go so this is going to cancel and you'll have cosine of S Plus T is equal to this is going to cancel so you have the cosine of s time the cosine of T remember it's the cosine of the first guy times the cosine of the second guy then positive over negative is going to be negative right right you can think about this canceling with this and giving you a ne1 so you're basically dividing by Nega 1 or multiply by Nega 1 whatever you want to think about there but essentially you have a minus here and then it's going to be the S of s times the S of T so again your sign is different you have a plus here and a minus here and then it's the sign of the first guy times the sign of the second guy if we grab this and we go back to our original formula you see that it matches the only thing that's changed is the angle here we used a and b here we used s and t so let me actually get rid of this and I'm going to show you where the second one comes from it's very easy if you understand the negative angle identities so let's say you said you had the cosine of a minus B but you wrote it as a + b like this so just following this formula here you would have the cosine of a then times the cosine of this negative B then you would have minus the S of this a Time the S of this this B with negative angle identities again the cosine of Nega Theta is the same thing as the cosine of theta and then the S of negative Theta is the same as the negative of the S of theta so what I can do here is just say this is the cosine of a times the cosine of B so that matches this part right here now this is going to come out right because again this is the negative of the sign of B so the negative times the negative would give me a positive so I can say this is plus now you would have the s of a Time the S of B so that's where this formula is coming from now let's look at a little example and here we're given the cosine of 75° so this type of example you're meant to give an exact value okay so the way you're going to do this is you're going to break 75 degrees up into two angles that's going to be on the unit circle okay where you're going to know what this cosine value is going to be so I can break this up one way to do it would be 45° and 30° so I'll just say this is equal to the cosine of and I'm just going to go 45° + 30° okay and then I'm just going to use my little formula so we'll have what you want the cosine of this guy so the cosine of 45° times the cosine of this guy the 30° remember your sign is going to alternate so because in this particular case we have a plus we want a minus there okay so you want to change your sign and then now I want go with the sign of these two angles so the S of 45° and then times the S of 30° okay so you're multiplying here you're multiplying here you're subtracting between the two okay so now what I want to do is just come through here and replace these with the actual values so if you go to your unit circle and you go to 45° or pi over 4 in terms of radians you see that it's 2 over2 for the cosine and then it's > 2 over2 for the sign remember this is the X that's the cosine this is the Y that's your sign so < TK of 2 over2 in each case so this is < TK 2 / 2 and this is < TK of 2 over 2 okay so what's going to be for 30° let's go back for 30° or pi/ 6 in terms of radians you've gotun of 3 over2 for your X or your cosine and then 1/2 for your y or your s so this is going to be the < TK of 3 /2 and this is going to be 1/2 okay so now all we have to do is just some basic arithmetic so let's come down here a little bit okay and then let's switch the color here so this would be theun of two * theun of 3 that would just be the < TK of 6 over 2 * 2 which is 4 then minus theun of 2 * 1 is just < TK of 2 and then over 2 * 2 which is four now the only thing you can really do here is write this with a common denominator so I'm just going to say that this is the squ < TK of 6us the < TK of 2 you can't combine there because the radicans are different okay you have the same index but different radicans so you can't really combine those so then this is over the common denominator four and this would be your exact value for the cosine of 75° all right let's take a look at another example so we're just trying to find the exact value again so we have the cosine of pi over 12 so obviously this is given in terms of radians now so what I'm going to do and this is just a personal preference I'm going to first convert this to degrees it's a little bit easier for me to find two angles to add or subtract depending on what you want to do so I'm going to multiply piun / 12 in terms of radians time 180° over pi and then basically I'm going to cancel this with this 180° / 12 would be 15° okay so basically if I had 15° what I'd want to do is add and subtract let's say I did 45° minus 30° that would give me 15° okay so I could say this is the cosine of if this is in terms of radians 45° is going to be pi over 4 okay so pi over 4 and then minus 30° is going to be < over 6 okay in terms of radians so I'm just going to use my formula from here and just say that I have the cosine of < over4 times the cosine of pi/ 6 because this is minus I now want to plus so I'm going to do the S of Pi over4 and then times the S of pi over 6 okay so let me put my equal sign there so we can keep going and then I'll put equals over here so if I think about pi over 4 or again 45° I know the cosine of 45° and the S of 45° that's going to be theun of 2 over2 in each case if I go back to my unit circle you can see for 45° or pi over 4 in terms of radians we got < of two over two and < of two over two so let's go back and just replace this with the < TK of two over two and the < TK 2 / 2 now for < / 6 or 30° we have again > 3 over2 for the cosine sign and 1/2 for the sign so this is going to be the < TK of 3 / 2 and then we're going to have 1/2 and then I'm going to put a plus here between them okay so let's go ahead and scroll down a little bit and do some arithmetic so this would again be the square otk of 6 over four now plus now I'm going to have the sare < TK of 2 * 1 which is the sare < TK of two over 2 * 2 which is four okay so again pretty much all we're going to do here is just write this with a common denominator I can't add the sare of 6 and the square of two I can't combine those radicals there because this and this is different right so you have different radicans same index with different radicans so I'm just going to write this as the square Ro of 6 plus the < TK of 2 over the common denominator of 4 so this will be my cosine of pi /2 radians okay let's take a look at one more of these so you might see the problem that we just worked on those two problems rearranged so now you see the cosine of 83° * the cosine of 38° plus this is the key here the S of 8 3° * the S of 38° so you see that you have the cosine of some angle times the cosine of some other angle okay and then you have the sign of that first angle so the angles here match and then the sign of that other angle okay so the angles there match you have a plus between them you're meant to realize that you could write this as the cosine of again this is a plus so I want a minus so I'm going to do the 83° minus the 38° okay so what is this going to be if I do 83° minus 38° I'm going to get 45° so this is the cosine of 45° and we already know because we've done this in the last two examples that the cosine of 45° is aare < TK of 2 over 2 so this is exactly equal to the < TK of 2 over two all right so another thing that comes up in this section we're going to revisit the co-function identities so before when we looked at something like the S of theta is equal to the cosine of 90° minus thet we said that Theta needed to be acute angle but now we're going to update that and say that Theta can really be anything as long as the function is defined so to prove this what we're going to do is use the cosine difference identity and over here let me just put equals here and I'm going to say that this is the cosine of 90° and then times the cosine of theta and then again if this is a minus you want to use a plus here this is the S of 90° and then times your s of theta now we know that the cosine of 90° is 0 and 0 time anything is 0 so you could write this out and say this is 0 * the cosine of theta and then plus we know the S of 90° is 1 so this is 1 * the S of theta and so again this is going to go away so you're just left with 1 * the S of theta which is just the S of theta so you can see that this is going to work for any value of theta again as long as the function is defined you'll be good to go so we can actually extend this to these other guys that we looked at earlier so we have the cosine of 90° minus thet is equal to the S of theta the S of 90° minus Theta is equal to the cosine of theta again s and cosine are co- functions same thing for secant and cosecant and for tangent and coent now let's look at a little example and we haven't gotten to solving trigonometric equations yet we should know that the solution I'm going to give you here is one of many many possible solutions or you could say an infinite number of solutions but I'm just going to give you one so here we have the S of theta is equal to the cosine of 55° so what you'd want to do is write this left side in terms of cosine so that you have cosine of some angle equals cosine of this angle here so I know that the S of theta is equal to the cosine of 90° minus Theta so I'm just going to replace this guy right here with this guy right here right since they're equal so I'm going to say that this is the cosine of 90° minus Theta and this is equal to this guy right here which is the cosine of 55° me get rid of this scratch work here and get rid of that and let me just slide this up here a little bit and so what you would do here you would set this guy equal to this guy right you set the angle measures equal so we would say that 90° minus Theta is equal to - 55° and so let me subtract 90° away from from each side of the equation this would cancel slide down here just a little bit we'll say the negative of theta is equal to 55° - 90° that's going to be- 145° and then to finish this up I'll divide both sides of the equation by1 this would cancel so you get Theta is equal to this will be 145° now again this is not the only solution okay again when we get to solving trigonometric equations we'll see that there's an infinite number of solutions if you want want you can go back and plug this in and you'll see that the S of 145° is exactly equal to the cosine of 55° okay now let's look at two examples where we're going to verify some identities these are a little bit easier than what we looked at in the last section where we were strictly verifying identities so let's start with we have cosine of we have theta plus pi okay is equal to the negative of the cosine of theta okay I'm going to flip this because the easier side is here on the right so I'm going to write the negative of the cosine of theta is equal to the cosine of you have your Theta okay plus your Pi now what I'm going to do is just put my equals over here I'm just going to slide this over for formatting okay normally they put it in line but I just want a lot of Rong so I'm going to expand this out using the sum identity for cosine so this would be the cosine of theta okay times your cosine of pi and then remember this is plus so you want a minus so I'm going to go the S of theta times the S of Pi okay now realize when you work with the cosine of pi or the S of Pi let's go back to the unit circle when we look at pi over here the cosine is NE 1 okay and the sign is zero so let's minimize this so the cosine of pi is Nega 1 so let's put times 1 like this and the S of Pi is zero okay so we know if you multiply anything by zero it's going to be zero right so basically you can get rid of this part right here right You' have - 0 which is just gone so basically I can say this is equal to the negative of the cosine of theta and so that matches exactly what we have here right negative cosine of thet negative cosine of theta so we're done okay let's look at another example of verifying an identity this one's just as easy you have the cosine 2 xus the sin^2 x = cosine of 2x now if it's not immediately obvious what you need to do remember if you have 2X this is the same as X Plus X okay so if you're in this section you're looking to think think about the sum and difference identities for cosine so think about things you can add or subtract so 2x is X Plus X okay so what I'm going to do here let's just go ahead and put equals here I'm going to write cosine of x + x like this okay and then I'm going to use my sum identity for cosine so I'm going to say this is the cosine of x times the cosine of x again then minus remember this is plus this is minus the S of x times the S of X again okay so once you write it like this it's clear that this is going to match right because cosine of x * cosine of x I can say this is the cosine 2ar of x and then minus I can say s of x * sin of X is sin^2 of X and now I can see that this side matches this side right cosine s x - sin s x cosine 2 x - sin 2 x so I can put my check mark there and I'm done okay let's look at one more problem type that you're going to see so we're asked to find the cosine of S Plus T you might also get the cosine of s minus t the procedure the way you do this typee of problem is the same okay so you have S in quadrant one and T in quadrant 3 so they're giving you information about sign so the S of s is 3 fths the S of T is -2 13 so the first thing we want to do is find the cosine of s so that equals what and the cosine of T so that equals what so one way you could do this is you could use your Pythagorean identity remember cosine in this case let's say squ of S Plus sin of s is equal to 1 okay so if I plug in for S of s right here I can figure out what cosine of s is okay so let's scroll down we'll come back up so this is going to be three fifths so remember this guy is squared so I'm going to plug in a three fifths here and I'm going to square that so plus you have your cosine squar of s and this equals 1 okay so 35ths squar is going to be 9 25ths so this is 9 25ths so plus you have your cosine 2 s here this equals 1 Okay so so I'm going to subtract that away from each side so I'm going to have the cine SAR of s is going to be equal to I'll go ahead and write 1 as 25 over 25 so I have a common denominator and I'm going to subtract away 9 over 25 right I'm just subtracting this away from each side so 25 - 9 is going to give me 16 and so I'll say I have cine 2 s is equal to again this is 16 over 25 if I want cosine by itself I'm going to take the squ of this side again you do plus or minus but because we're in quadrant one for S it tells us that I just want the principal square root of this guy so I can erase this and just say I have cosine of s is equal to this and so cosine of s will be the square of 16 is 4 the square Ro of 25 is 5 so you get four fifths there so this guy is going to be four fths okay what about the cosine of T well let's figure that out using the same process let's just change up this formula and I'm going to put T here and T here it's the same thing I'm just going to plug in for this so -12 13 but T is in quadrant 3 now so keep that in mind so let's come down here and let's plug in so we're going to have the cosine squar of T then plus and I hate using T because it gets confusing with this plus symbol so let me use the plus symbol in like black or something so it's clear that this is a plus I'll even Circle it that's a plus okay so then we have the s^ 2 T is equal to 1 I'm plugging in a -12 13 there so 1213 and this guy is being squared so if I square a negative I get a positive if I Square 12 I get 144 if I Square 13 I get 169 so I'm going to say this is cine s t and then let me be careful I'm going to put the plus sign here okay again that's a plus and then this is going to be 144 over 169 and then this is equal to 1 okay I'm going to subtract this away from each side and before I do that let me just get a common denominator going so let's go cosine s t okay this is going to go over here so I'm going to put equals and I'm going to do 169 over 169 so that's one it just wrote it with a with a common denominator with this guy then minus again I'm subtracting this away from each side 144 over 169 okay so if I do 169 minus 144 I get 25 so let's scroll down just a little bit and we'll say that cosine squar of T is equal to again this is going to be 25 over 169 now because we're going to be in quadrant three in terms of this T guy here cosine is going to be negative okay so I want the negative square root when I do this side so I'm going to say cosine of T is equal to the negative of the square < TK of 25 over 169 so let's go ahead and simplify that so we'll say the cosine of T is equal to we'll have the sare < TK of 25 is 5 the of 169 is 13 so the negative of 5 over 13 okay so that's going to be my cosine of t Okay so let's put -5 over 13 like this so now that you have all the information you can find cosine of S Plus T so sometimes they give you s of one of them and cosine of the other sometimes they give you s of of one and then s of the other just depending on the situation you're given you want to just plug into the identities that you know about come up with s cosine of each of the angles and then you can go through and use your sum or difference identities depending on the problem here we're going to use the sum identity for cosine and I'm just going to say that we have cosine of our S Plus T is equal to again if I take my cosine of s I know what that is that's 4 fths then times my cosine of T that's -5 over 13 this is a plus here okay so I want to make sure this is a minus you got to be really careful with the signs here because sometimes you'll make a mistake with that so now I want the S of s which is three fifths okay and then I want times the S of t which is going to be -12 13 okay so let's come down here a minute and just process this arithmetic so I'm going to say this is equal to If I multiply here you see that this five would cancel with this five so basically have a negative one there okay so this is basically -4 over 13 so -4 over 13 and then minus you're subtracting away this is 3 * -2 which is -36 I'm going to write this as over 5 * 13 which is going to be 65 now I canceled here because we were doing multiplication but it actually would have been smarter if I wouldn't have canceled because then I'd have a common denominator so let's undo this and say here we have 4 * ne5 which is -20 okay over 5 * 13 which is 65 in this particular case it's better to not cancel because now I have a common denominator and I can basically say what I have -20 minus a - 36 again you got to be really careful with that sign -20 minus a 36 which is going to be -20 + 36 which is 16 okay let me write this in a different color so this is going to be 16 and then over you're going to have 65 so this is going to be your answer for cosine of s + T in this lesson we want to talk about the sum and difference identities for S and also tangent all right so we talked about the sum and difference identities for cosine now we're just moving on and talking about the sum and difference identities for S and tangent just something you have to memorize so if we have the sign of let's say something like a plus b again don't distribute that and think s of a plus s of B okay it doesn't work that way you have to use this little formula here so we're going to say this is the S of a * the cosine of B plus the cosine of a times the S of B so notice how the sign here and the sign here is the same okay and then when you get to this one nothing's going to change for the sign so this is a minus and this is a minus this part right here is the same so the sign of let's say a minus B now we're going to have the S of a times the cosine of B minus the cosine of a * the S of B so with these you just have to write them down and just go over to a few times in your mind use it while you're doing the practice and you're going to memorize it okay so let's look at a little sample problem here you saw this in the last section where we would basically rewrite something to get an exact value you're going to do the same thing here so we have S of 165° so you're going to find two angles basically that are on the unit circle where you already know things about it so I'm going to do the sign of let's do 135° plus we're going to go 30° Okay so so basically if I use my little sum formula that I just talked about this is going to end up being what it's going to be the sign of the first guy so 135° times the cosine of the second guy which is the 30° then plus it's the cosine of this first guy the 135° and then lastly times the sign of this last guy the 30 Dees okay so some of you at this point are familiar with the unit circle others are not so let's just go really quickly so we can look at this guy just in case you need to pause the video and get the values when you're looking at 30° your cosine is going to be square of 3 over2 your sign is going to be 1/2 okay and then when you look at 135° your cosine is going to be the negative of theun of 2 over two and your sign is going to be the < TK of two over two so let's come back up and just punch these values in and we'll basically be done so I'm going to put equals here and then basically the S of 135° is the square 2 2 so I'm going to say this is the < TK of 2/ 2 times for the cosine of 30° I've got theare < TK of 3 over 2 then Plus for the cosine of 135° I've got the negative of theare < TK of 2 over 2 and then lastly for the S of 30° I've got 1/2 okay so now all we have left to do here is just basically go through and do some basic operations so we're going to multiply here so basically 2 * 3 really all I can can do is say that's the < TK of 6 over 2 * 2 which is 4 then plus if I do the negative of the > 2 * 1 it's just the negative of the square root of two so really I could just change this to minus okay and I'll put the square root of two over here over 2 * 2 is four okay so basically in this type of situation generally speaking you're going to write this with a common denominator so I'm just going to come down here get some room going and I'll just say that this is theare < TK of 6 minus the < TK of 2 all over the common denominator of four okay so this is the exact value of the sign of 165° all right so another type of problem you'll see is something where you're asked to simplify and basically you just need to realize here that you can use the difference identity for sign it's kind of going in reverse here but basically to solve this problem into simplify so if you notice here you have S of 3 Theta * cosine of -2 thetus the cosine of 3 thet * the S of -2 thet okay so if you notice here this angle right here and this guy right here it's the same you have the sign of this guy and the cosine of this guy and you have a minus Okay then if you look at this guy it's -2 thet and -2 Theta cosine of this guy s of this guy so this matches our little formula here the S of a * the cosine of B minus the cosine of a Time the S of B so in other words you see s of 3 * thet and then you see the cosine of 3 * Theta over here and then here you've got this cosine of -2 * thet okay and then here you've got this s of -2 * thet okay so basically what you're going to do is realize that it's this angle right here minus this angle right here so it's 3 * Theta minus a Nega 2 * Theta which is going to give us 5 * Theta okay so that's what that's going to work out to so I'll just put equals here I'm just going to put the sign of I'm going to take this first angle here this 3 * Theta minus a negative okay you have your 2 times Theta okay that guy right there and so this would give me again if I just combine this 3 * Theta basically minus a negative is plus a positive so you would have S of 5 * theeta all right Additionally you have the sum and difference identities for tangent so let's start with the sum guy so we have the tangent of let's say a plus b this is going to give us we have the tangent of a plus the tangent of B over 1us the tangent of a * the tangent of B and what's going to happen when we change over to this difference guy the signs here are just changed so before with when this was plus this was plus and this was minus okay now that this is minus this is minus and this is plus so again just something you have to write down and just keep practicing with and you will commit it to memory all right let's take a look at an example with this guy so suppose you're given the tangent of Pi / 12 so if you want to find the exact value again one way you can do this is by using these sum and difference identities so I like to work with degree when I do this it's just easier versus me trying to figure out what I need to add and subtract using fractions so if you can do this with fractions you know just skip this step that I'm about to do but I'm going to convert this first into degrees and then you can go back to radians okay so I'm going to go pi over 12 times I'm going to go 180° over pi and then basically this would cancel with this and give me 15° okay this is going to cancel out so pi over 12 in terms of radians is the same as 15 degrees okay so now that I got this in terms of degrees what two angles could I add or subtract to get 15° well one thing you could do is 45° minus 30° okay so I could write this as the tangent of 45° in terms of radians is going to be pi over 4 okay and then minus if I think about 30° in terms of radians it's going to be pi over 6 okay so pi over 6 all right so the next thing I want to do before I even use my formula I know I'm going to need the tangent of Pi 4 so I know I need the tangent of pi/ 4 and I know I need the tangent of piun over 6 let's go back to the unit circle I know some of you have this memorized some don't so let's just go back to the unit circle so if we're looking at our pi over 4 or 45° we have < TK of two over two and of two over two so that means the cosine and the sign are the same and basically the tangent is the sign divided by the cosine and since it's the same number right same number divided by itself is going to give you one okay so we know that I'll just write this down the tangent of pi over 4 is just going to be one okay because you have of 2 over2 id2 over two so that gives you one then the tangent if I think about 4 pi over 6 this is going to be what well you want the sign which is 12 divided by the cosine which is aun of 3 over2 so basically I would flip this guy okay because I'm dividing fractions so let's put the two up here and the square root of three over over here okay and then basically the twos cancel and I have 1 over thek 3 so 1 over thek 3 of course we're going to rationalize this so I'm going to multiply this by the square TK of 3 over theare < TK of 3 and so this is going to give me the square < TK of 3 over 3 okay so let me write this in a more compact way so Square < TK 3 over 3 Let's copy this okay and let's go back and let's just paste this in here real quick and so this is one this is one and this is of 3 over 3 okay so squ < TK of 3 Square < TK 3 over 3 okay so now that we figured that out we can use our little difference identity and for this guy right here let me put equals it's going to be and I'm just going to slide this off to the side for one second so this is going to be the tangent of the first guy so the tangent of pi over 4 is going to be one so it's going to be one then minus the tangent of the second guy so minus the tangent of pi/ 6 which is the square < TK of 3 over3 then this is over remember you have your one the sign is different here so it's going to be plus you're going to multiply them together right so the tangent of pi over 4 * the tangent of pi over 6 so basically 1 time this guy so it would be plus thek 3 over 3 okay so let's get rid of this we don't need it anymore now basically you have a lot of different options here when it comes to simplifying okay I am going to choose to basically multiply the numerator and denominator of this guy by three okay so this is going to get distributed here and here this is going to get distributed here and here so what is this going to give me let me scroll down and get some room going so 3 * 1 is obviously 3 and then minus this would cancel with this and I'd have the square root of three there okay so we'll put this over here and I'll kind of undo this real quick just showing that over here 3 * 1 is three then plus again the threes would cancel there and you'd have the square root of three okay so that is a little bit simpler but really in order to go anywhere what I'm going to do remember if you have a radical down here here you have this plus here if this is binomial like this then basically what you can do is you can multiply by the conjugate okay so I can wrap this and I can wrap this so I'm going to multiply this by I'm going to have three minus okay I'm changing the sign from plus to minus then the square root of three okay then over you would have 3 minus the > of 3 again this is just a complex form of one when you multiply these guys together you're just going to do the first guy squared minus the last guy squared remember the middle two guys are going to drop out okay so let me make sure I got my equal sign going so put that in there let's keep going all right so in the top realize this is 3us 3 quanti squared so I can use a formula on that take the first guy and square it so 3 squar and then this is minus so minus it's 2 * the first guy which is three times the last guy which is square Ro TK of three and then plus the last guy square of 3 squ is three okay you can simplify this right now 3^ s is n then basically 2 * 3 you know that's six so you'd have minus 6 * > 3 and then plus 3 now 9 + 3 is 12 so let me just go ahead and put this as 12 - 6 * the square root of three okay so we have that and I can even shorten this a little bit and put this like this now down here all I need to do is square the first guy so 3 squ is n and then subtract away if I Square this guy square of 3 squar is just three now I can just perform this operation 9 minus three is just six okay now we're not done because notice that you have 12 here which is divisible by six six here which is divisible by six and six here okay so what I'm going to do is I'm just going to factor I'm going to factor a six out from the numerator so this is going to be 2 minus < TK of 3 okay over six down here and now we're done right we cancel this and I end up with 2 - < 3 as my answer for the tangent of pi over 12 all right let's look at an example where we're going to verify an identity so we have the cosine of beta is equal to the S of we have this pi/ 6 Plus beta and then plus we have cosine of we have pi over 3 plus beta okay so basically we always want to work on the more complex side obviously the right side is way more complex so I'm just going to put equals over here okay and normally we line this up but I want to give myself lots of room so basically what I'm going to do is I'm going to use my sum identity here and my sum identity here okay so I'm going to first say that I have the sign of this pi over 6times the cosine of this beta okay then plus okay so now I'm going to do the cosine of this guy right here so let me change color so I'm going to go the cosine of this in this case pi over 6 and then times you have your s of beta so your s of beta now additionally I want to do the same thing with this cosine guy so I'm going to put plus here remember how to do this I'm going to go cosine of this guy so cosine of pi over 3 and then time time the cosine of the next guy so cosine of this beta here and then if this is plus it's going to go minus right so the signs are going to be different there and then now I'm going to do the sign of this guy s of pi over 3 times the S of beta okay so now what we want to do is think about what can we replace well think about the fact that you have the S of pi/ 6 and the cosine of pi over 6 so coming back to my unit circle again when we think about Pi / 6 the sign is a half and cosine is of 32 so let's just write this down so we have the S of pi/ 6 is 12 and we had the cosine of pi/ 6 that's going to be the < TK of 3 over 2 now additionally we were also working with this PI over3 right so we need the sign and cosine for that guy as well okay so the cosine for pi over 3 is going to be2 and the s for pi over 3 is going to be the square < TK of 3 over2 okay so let's go ahead and grab this guy and we're going to cut it away and I'm just going to come down here real quick and paste this in so we have it so let's just replace the S of pi/ 6 with 12 and the cosine of pi 6 with 3 over2 okay so let's come up here so let's go ahead and put equals instead of the S of pi/ 6 I'm going to put 12 and then times my cosine of beta okay plus for this guy right here I'm going to put theun of 3 over2 so the square of 3 / 2 time my S of beta okay and so let's get rid of these two right here we don't need them anymore and let me scroll down a little bit and I'll drag these up into our visual range okay so we can work with them let's come back up here real quick so now we want to replace this cosine of pi over 3 so I'm going to put plus this guy is going to be2 then times your cosine of beta okay so let's get rid of this and then lastly I'm going to put this minus here we have the S of Pi 3 which is theare < TK of 3 /2 so let's get rid of this and this is times our sign of beta okay and sometimes I put multiplication signs and sometimes I don't you don't need it sometimes I just put it there okay so let's scroll down a little bit get a little bit more room going let's think about what we can do so we'll notice here that this guy right here okay and this guy right here those can be canceled right you have the of 3 over2 * the S of beta and then you have the negative of that so I can just cross that out those are gone so basically I can say this is 12 * cosine of beta +2 time cosine of beta so let's stop for a second remember what we're trying to get we're trying to just get the cosine of beta so basically if you have a half of something plus another half of something you have one of that thing you basically have a common denominator here right you could say this is the cosine of beta plus the cosine of beta over two right you have two of this so let's go ahead and come down down here and just say that we have basically 2 * the cosine of beta/ 2 and we can cancel that and lastly we will say that we have the cosine of beta okay and so we verified our identity all right let's look at one other type of problem here we're going to be asked to find the S of theta plus beta the tangent of theta plus beta and the quadrant of theta plus beta so a very common type of problem unfortunately they do take a while to work out so let's go ahead and come down here and for this one I'd like for you to try it on your own before we even go we did a similar problem with cosine in the last lesson so we have S of theta is equal to 4 fths we have the cosine of beta which is equal to the negative of 513 and then we're told that Theta is in quadrant 2 and beta is in quadrant 3 okay so how do we figure this out well what we need is the cosine of theta so that equals what okay that equals what and we need the sign of beta so that equals what okay if we have these two pieces of information then we can use our little formulas that we've been talking about in the lesson okay so basically how do we get this if we know the S of theta is 4 FS well I can start off by saying well the sin squar theta plus the cine 2 theta equals 1 okay remember your Pythagorean identity so I can plug in here a 4 fths and that's going to be squared right so four fifths and this is going to be squared so this ends up being 16 over 25 plus the cosine 2 thet is equal to 1 again I'm just going to move this over here so let me scroll down a little bit I'm just going to say that I have cine squ Theta is equal to If I subtract this away from each side I have 1 - 16 over 25 if I make this as 25 over 25 then basically 25 - 16 is 9 so you would have 9 25ths okay but now the issue is I want cosine of theta not cosine squ Theta so basically I need to square root each side remember you get plus or minus over here and then based on what quadrant you're in that's how you decide what sign you're going to use now we're told that Theta is in quadrant 2 okay and in quadrant 2 cosine is negative so I'm going to go cosine of theta is equal to I'm going to use the negative of the square < TK of 9 over 25 and of course this is going to end up giving me the cosine of theta is equal to of 9 is 3 and then the 25 5 so 35ths okay so let's take this up so we're just going to cut this away and actually I'll just write it in here so I'll put -3 fths so that's the first piece of the puzzle so now we need to find the S of beta I'm going to use the same thought process so the cosine of beta is the negative of 5113 so again we have the sin squar in this case beta plus cosine SAR beta is equal to 1 and for cosine squ beta remember for cosine itself I'm plugging in a 513 and this guy is being squared don't forget to square okay so if I Square this guy I'm going to get 25 over 169 so I'm going to have sin SAR beta plus this guy here equals 1 same thing I'm just going to move it to the other side so I'm going to say s^ squar beta is equal to let me go ahead and put this as a common denominator so 169 over 169 same thing as one then minus this guy over here just moving to the other side so minus your 20 5 over 169 and so the s^ squar of beta is equal to 169 - 25 is 144 so you have 144 over 169 now we said that beta was in quadrant 3 and in quadrant 3 sign is going to be negative so I'm going to say the S of beta is equal to the negative of the square root of this guy so the 144 169 so the squ of 144 is 12 the squ of 169 is 13 so I'll go ahead and and say the S of beta is equal to the negative of 1213 okay so the negative of 12 13 so let's put this up here the negative of 12 13 okay so we are officially ready to answer this part up here the S of theta plus beta okay so basically you would have the S of theta plus beta is equal to the S of theta which is four fths multiplied by the cosine of beta which is - 513 okay then plus you have your cosine of theta which is - 35s and then times your s of beta which is -121 13 okay I'm going to wrap this in parentheses so I don't make a silly sign mistake now in a lot of cases you'll notice when you're multiplying here you can cross cancel here notice that you have 5 * 13 and 5 * 13 so don't cancel this because you're going to want to have a common denominator I make that mistake all the time so I'm just going to stop you from doing it okay so basically 4 * 5 is -20 and this is over 5 * 13 is 65 then plus negative * negative is positive so 3 * 12 is 36 over 5 * 13 is again 65 okay so to finish this up I'd have -20 + 36 which is 16 and then over the common denominator of 65 so s of theta plus beta is 16 over 65 so let's just erase all this work so sin of theta plus beta is equal to 16 / 65 okay so how do we get the tangent of theta plus beta well first I need the tangent of theta and I also need the tangent of beta so how do I get that remember for the tangent you divide the sign by the cosine so if I think about let me go down here because we're going to forget stuff so if I want the tangent of theta it's going to be the S of theta which is 4 fths so 4 fths divided by the cosine of theta which is - 35ths again we're dividing fractions so flip this guy I'm going to multiply by the reciprocal so times 5/3 this cancels and I get -4/3 okay so let me just scooch this down so this is going to be negative 4/3 okay and then we want the tangent of beta so how do we get that same way so I'm going to do ne213 let me change my ink here and then times over here I'm going to flip this so I'm going to do 135s and again stuff's going to cancel so this cancels with this negative cancels with negative that's positive so you have 12 fths right so this is positive 12 fths okay so let me drag this back up so let's cut this away and let's paste this in here okay and let's see if we can figure out what the tangent of theta plus beta is again I take the tangent of the first guy let me write this off to the side so the tangent of theta plus beta okay so the tangent of the first guy which is -4/3 then plus the tangent of the second guy which is 12 FS this is over we want one again this is plus because the same as this this is going to be minus right so this is always different this is always the same so one minus multiply the two together so it's -4/3 times your 125s okay let me get rid of this now we don't need it just so we have some room so let's move this over let's move this over okay put this over here here and I'm going to drag this up here a little bit and I'm just going to scroll down a little bit we'll come back up in a second so the best way to do this let me just put equals over here I'm not going to get a common denominator yet so I'm going to go -4/3 + 125s over for this one I'm just going to multiply -4 * 12 is going to be -48 this is over 3 * 5 is 15 and you've got one minus this okay so what is the LCD if I think about all the fractions involved if I think about three and that's it's a terrible three so let me make that better so if I think about this three here let me get off the marker mode so this guy the three the five and the 15 well it's 15 right so the quickest way to do this in my opinion is to multiply the top and bottom by 15 okay so I'll save you a lot of time we'll go ahead and distribute this here and here and then here and here okay so first off minus a negative is the same as plus a positive okay you can just change that right away so you have to deal with it and you'd have 15 ID 3 is 5 5 * -4 is going to be -20 then 15 ID 5 is 3 3 * 12 is going to be 36 so plus 36 you can go aad and do that if you want you know that's going to be 16 so this is 16 this is over if you multiply 15 by 1 that's 15 and then 15 ID 15 is 1 so you just have plus 48 there so what's 15 + 48 it's going to be 63 okay so this guy's going to be 16 over 63 and I could put equals there but really I'm just going to erase all of this and I'll just put a little line here put the tangent of theta plus beta is equal to and let me just paste that in my 16 over 63 okay so s of theta plus beta is 16 over 65 tangent of theta plus beta is 16 over 63 now we want to think about and I can just get rid of this we want to think about the quadrant of theta plus beta well I just want you to think about the sign the S of theta plus beta is positive right so s is positive okay the tangent of theta plus beta is also positive so the tangent is also positive okay so where is s positive and also tangent positive remember you can always draw your little coordinate plane and I know this is a terrible freehand draw but it's all students take calculus right all you have your s your tangent and your cosine okay so if we think about s it's positive in quadrants 1 and two and tangent is positive in quadrants 1 and three so the only guy that meets both criteria here is going to be quadrant one okay so the quadrant of theta plus beta okay is going to be one so this going to be we'll say is in quadrant quadrant I just probably put a capital Q so quadrant one in this lesson we want to talk about the double angle the product the sum and also the sum to product identities all right so at this point we've learned quite a few of the trigonometric identities we just keep adding on to the ones that we already know I'm just going to pop open this little worksheet here real quick so this is the whole worksheet if you go back up to the top this is all the stuff we've already learned so all of these guys we've already talked about and now we're getting into the double angle identities and then again we'll talk about the product to sum and the sum to product at the end of the video right now we're going to focus on these so you can either pause the video and write these down really quickly or you can use the worksheet that's been provided okay so the problem we're going to look at here is just going to be asking to find some trigonometric function values we have the cosine of theta is equal to 2 * 2 over 3 and we have that s of theta is greater than zero or you could just say positive right so basically if it's cosine of theta is positive and S of theta is positive that tells me that we're in quadrant one right so we can say here that Theta is in quadrant one so we're going to need that piece of information now what we're going to be asked to find is the cosine of 2 * Theta so we want to find the cosine of 2 * Theta we also want to find the S of 2 * Theta okay and then also we want to find the tangent of 2 * Theta and of course once you have these you can use your reciprocal identities to find the other ones I'm just going to do these three right here all right so the first thing I'm going to do what I'm going to focus on is finding the S of theta and the tangent of theta okay and then we're going to use our identities to find the cosine of 2 * th the S of 2 * thet and the tangent of 2 * thet okay so if I'm given cosine OFA how can I find S of theta again we've done this a million times one way we can do this we can say that cine 2ar theta plus our sin squ Theta is equal to one okay so our Pythagorean identity something we're really used to working with at this point so for this guy I'm going to plug in this 2 * < TK 2 over 3 and I'm just going to square this guy okay let me just copy this and go to a fresh sheet cuz we're going to run out room here so let's come down here and just paste this in so we have lots of room and then I'm just going to say plus my sin^2 Theta is equal to 1 okay so from here if I squared two I would get four if I squared the square root of two that's two okay and if I squared three I would get nine so basically 4 * 2 is 8 so this is 8 9ths then plus your sin^2 Theta is equal to 1 I'm just kind of drag this up here and from here I'm just going to say that I have sin^2 thet is equal to I'm going to subtract 8 9th away from each side of the equation so I'm going to put 1us 8 9ths okay and again all I did was subtract it from both sides of the equation and so now what I have is my s^ 2ar Theta is equal to if I write this as 9 over 9 so I have a common denominator and then I'm subtracting away this 8 9ths here this on the right hand side is going to be 1 nth so I would say sin^2 thet okay is equal to 1 9th okay now from here in order to get rid of this kind of squaring operation I'm going to take the square root of each side normally again you do the plus or minus on the right hand side but because we're in quadrant one I know that sign is positive so I want the principal square root so I'm just going to say that the S of theta here is going to be the Principal square root let me show this fully so the principal square root of 1 nth okay so the S of theta is equal to S of 1 is 1 squ of 9 is three okay so let me just write this up here that the S of theta so the S of theta is equal to 1/3 okay and we'll come back up and we'll use this let's come back down here for a second let me erase all of this so I'm also going to need my tangent of theta so the tangent of theta we all know that this is the S of theta divided by the cosine of theta so the S of theta we just found was 1/3 okay and the cosine OFA if I go back up here it's 2 * 2 3 okay but because I'm going to be dividing fractions I'm going to go ahead and flip that guy and multiply so this is 3 over 2 * < TK of two okay so what's going to happen is the threes are going to cancel and I have 1 over 2 * the < TK of two I want to rationalize the denominator here so I'm going to multiply by theare < TK of two over the square root of two okay so basically if I think about what I'm going to have here in the numerator it's just the < TK of two in the denominator it's 2 * the of 2 * of two of 2 * 2 is 2 so 2 * 2 would be four okay so the tangent of theta is the < TK of 2 over 4 so let's just write this here that the tangent of theta is the < TK of 2 over 4 okay so now let's go ahead and find the cosine of 2 * Theta the S of 2 * Theta and the tangent of 2 * Theta so for the cosine of 2 * Theta let's go back to our little sheet here so we have cosine of 2 a you could plug in Theta for a or X or whatever you want to do it's cosine squ of a minus sin s of a so that's pretty easy to do so let's make a little border here and actually I'll probably better off just going to the other sheet so we'll say cine of 2 * thet is going to be equal to cine 2 thet minus s^ 2 Theta okay so I'm just going to plug in so cosine of 2 * Theta is equal to remember our cosine of theta was 2 * the < TK 2 over 3 okay this guy is going to be squared then minus remember our sin Square Theta you might remember this was 1 nth but basically s of theta was 1/3 right so I'm going to put 1/3 here and square it and that's going to give me 1 nth okay so if you remember from earlier when we squared this we got 8 9ths right because 2^ 2 is 4 sare < TK of 2 squar is 2 4 * 2 is 8 okay and basically 3 S is 9 so this is 8 9us let me put equals over here I'll just write the whole thing out so cosine of 2 Theta is equal to 8 9th minus again if you square this you get 1 nth okay so this gives me a final result here of seven n so this is my cosine of 2 Theta okay so let's come up here and put that our cosine of 2 Theta is equal to 7 9ths now if you're asked for the secant of theta you would just flip that and get 97 okay but we're not asked for that here so then now I want to find the S of 2 thet okay so what's this going to be equal to let's come back down here so what is my S of 2 Theta well let's go back to our worksheet here and we see that the S of 2 a is equal to 2 * the S of a * the cosine of a so again you could just plug in you could say the S of 2 * Theta is equal to 2 * the S of theta * the cosine of thet so this is equal to 2 * the S of thet time the cosine of theta well again we know the S of theta is 1/3 so this is 1 3 and we know the cosine of is 2 * < TK 2 over 3 so let me put my 2 in here and let me put s of 2 Theta like this so let me write this again over here let me put s of 2 * Theta is equal to basically I'm just going to multiply across 2 * 1 * 2 is going to be 4 then times the < TK of 2 over 3 * 3 is 9 okay so sine of 2 th is equal to 4 * 29 so let me just move this down a little bit or actually I can put it over here so the s of 2 * thet is = to 4 * < TK 2 over our 9 and again you could flip this to find the cosecant of 2 Theta okay but again you'd have to rationalize the denominator there we're not asked for that here so we're not going to do that okay so lastly we want the tangent the tangent of 2 Theta so how do we find this well essentially all I need to do is take this guy right here and divide it by this guy right here or I can use my formula so let's see this both ways so for the tangent of 2 * th the first way we're going to do it probably the easier way you basically take your s of 2 * thet which again is 4 * < 2 over 9 so 4 * < tk2 over 9 we're going to multiply it by the reciprocal of this cosine of 2 thet which is 7 9th so if you flip that you get 97th again I'm flipping that because if you're dividing fractions you take the first fraction multiply by the reciprocal of the second so this is going to cancel we're going to be left with four 4 * > 2/ 7 so that's the quickest way that's the way you probably want to do this so let's come back up here so 4 * of 2 over 7 now the other way and I just want to show you this with the identities it's a little bit longer but I think you should see it let's pop this open we have the tangent of 2 a is equal to 2 * the tangent of a over 1 minus tangent s of a okay so let's come down here and say that the tangent of 2 * Theta is equal to 2 * the tangent of thet over 1 - tangent tangent 2 thet okay so what is the tangent of theta if we go back up the tangent of theta was < TK of 2 over 4 so let's come back over here and I'm just going to put this over here so the tangent of theta was a < TK of 2 over 4 so we would start with and let me just put equals over here 2 times this guy right here which is < T of two over 4 okay over 1 minus tangent squar Theta so I'm just gonna Square this guy so if I Square let me just show this so so if I square the square root of two I'm going to get two if I Square Four I get 16 okay so I'm going to go ahead and say this is multiply across I'll say this is 2 * < TK of two over four over you would have 1 minus again if I Square squ t of two I get two if I Square Four I get 16 okay let me scroll down and get a little bit of room going here so we can see what's going on it's probably quicker for me to just multiply this by 16 over 16 that will cancel all my denominators so 16 time this guy the 16 would cancel with the four and give me a four so you'd have 4 * 2 which is 8 so let me put 8 * the square < TK of two here and then down here 16 * 1 is 16 and then minus 16 times this guy right here the 16s would cancel so I just have two okay so that's probably a quicker way versus what I was going to do which is basically to work with these guys individually okay so now 16 minus 2 is going to be 14 so this is 8 * < tk2 over 14 okay and basically I can cancel this 8 and this 14 they're each divisible by two so this is going to be four and this is going to be seven right so we get 4 * < TK 2 over 7 and you see that's what we got before right so it's much quicker if you have the S of 2 * Theta and the cosine of 2 * Theta to just work with that versus using the formula but sometimes you're told to show it on a test or something like that so if you do it the alternative way using the double angle identity it works the same all right so now we're going to look at a similar problem but it's going to go in reverse so now we're given the cosine of 2 * Theta in this case it's equal to the Nega of 7 over 25 we're told that Theta is in Quadrant 4 now you might be thinking that the cosine of theta should be positive in Quadrant 4 but remember this is the cosine of 2 * Theta okay so not the cosine of theta this is cosine of 2 * Theta so what we're want going to find here is the six trigonometric functions those values for Theta okay so basically basically I would start off by finding my S of theta I can then use that to find the cosine of theta and from those two guys I can basically find all the rest right really easy to do so how am I going to find the S of theta given this information let's go back to our little worksheet or you could say handout and basically if you look here you have this guy right here which would work pretty nicely you have the cosine of 2 a is equal to 1 - 2 * sin of a so let's use that so we have have the cosine let me do this on another sheet so we have the cosine of 2 * in this case I think it was Theta so it was Theta is going to be equal to 1 - 2 * sin^2 thet okay now I know what cosine of 2 * Theta is right I'm given that as -7 over 25 so just plug in for that so we're going to say this is -7 over 25 okay is equal to 1 - 2 * sin^2 th all right so from here let's scroll down a little bit and get some room going I'm going to basically flip this around I want what I'm solving for on the left so I'm going to add 2 * sin 2 thet to both sides so I would have 2 * sin 2ar Theta okay is equal to I'm going to leave this guy over here and I'm going to add 7 25ths to both sides so basically I'm going to have 7 over 25 plus one over here okay so all I did was just kind of swap things around it's just adding and subtracting the same thing to both sides of an equation which is legal okay so basically what I want to do at this point is go ahead and add 725s + 1 so I'm just going to write this as 25 over 25 and this would give me 32 over 25 okay so I would have over here 2 * sin^2 thet is equal to 32 over 25 I have this two that I want to get rid of so I can multiply both sides by a half so let's multiply this side by a half and this side div by half and basically this is going to cancel out and over here this would cancel with this and give me 16 okay so now let me keep scrolling down here we have that sin^2 Theta is equal to you have 16 over 25 okay so you have to take the square root of each side again if we go back up we are told that Theta is in Quadrant 4 we know that the S of theta if it's in quadrant four is going to be negative right because sign is positive in quadrants 1 and two so if we come back here what I want to do is take the negative square root of this right hand side so I'm going to say the S of theta is equal to I'll go the negative of the square < TK of 16 over 25 so the S of theta is equal to the negative of square of 16 is 4 < of 25 is 5 so s of theta here is equal to the negative of 4 fths okay so let's go back up and let's put right here the S of theta is equal to -4 fths all right all right so how do we find the cosine of theta and all the rest of them well basically at this point we can just use our regular identities that we've been working with I can use the Pythagorean identity to get this guy so we know that sin sare theta plus cosine squ Theta equal 1 so I'm going to have my -4 fths for my S of theta I'm going to square that plus my cosine 2qu Theta this equals 1 okay so if we Square this we get our 16 over 25 then plus our cosine s Theta this equals 1 Let's subtract this away from both sides so I'm going to have cosine 2 thet is equal to I'm just going to write this as 25 over 25 to speed this up and over here I'm going to put minus again I'm subtracting this away from each side so 16 over 25 okay so from here it's pretty easy we have cosine 2 Theta is equal to 25 - 16 is 9 so this is 9 over 25 okay so we're in Quadrant 4 co sign is going to be positive in Quadrant 4 okay so what I want to do is take the principal square root of this right hand side so I'm going to say the cosine OFA is equal to the principal < TK of 9 25 so now let's finish this up and say the cosine of theta is equal to of 9 is 3 of 25 is 5 so this is 3 fths so let's come back up and this will be 3 fths okay so then the tangent of theta is found by what you basically take this guy right here this Nega 4 fths and you multiply by the reciprocal of this this 5/3 okay and this cancels you get 4/3 so this is -4/3 and then the other guys are found with the reciprocal identities so let's just go ahead and do the cotangent really quickly so the cotangent of theta that's found by flipping this so it's - 34s then if I want let me scroll down just a little bit if I want the cosecant of theta that's found by flipping this this s of theta so this is neg 54s and then lastly if I want the secant of theta that's found by flipping the cosine of Theta so that's going to be 5/3 okay let's look at two examples of verifying some trigonometric identities you always want to get practice doing this because it keeps coming up again and again and again and the more you work on it the better you're going to be when you need it for calculus okay so we have the cotangent of x * the quantity 1us the cosine of 2x is equal to S of 2x so the first thing I'm going to do is flip this I'm going to put s of 2x here basically on the left hand side for formatting and going put is equal to let's just go ahead and putang of x * the quantity 1 minus the cosine of 2x okay so how can we approach this problem the first thing I always like to do if I see something like the cotangent of X or the tangent of X the first thing I'll try is I'll try to put things in terms of s and cosine it doesn't always work that way but in this case it will work out this way so we have the S of 2x is equal to I'm going to go ahead and say that the cotangent of X is the cosine of x over the S of X and this is times the quantity 1 minus the cine of 2x now you have 1 minus something and I want you to look at the identities here if you go back here notice you have this 1 - 2 * sin^2 of a here for the cosine of 2 a okay so if I plug this in here I can get that one to drop out so I'm going to go ahead and write this as the quantity make sure you use parentheses here 1 - 2 * sin 2ar in this case of x X okay so you need parentheses there because you're subtracting the whole thing away what I'm going to have here let me put my equals over here and I know normally you line this up but I want a lot of room so I got to do it over here so cosine of x over sine of x times in this case you could go ahead and write this as 1 - 1 + 2 * sin^2 X and see how the ones are going to cancel out there so let's scroll down a little bit more get some room I'm going to say is equal to I'm going to to go cosine of x over s of X Time 1 - 1 is 0er right so that's gone you have 2 * sin^2 X now you can do some cancelling right this is going to go away I can get rid of one of these right so basically what I can do here is say that I have 2 * sin of x * the cosine of x now let's stop for a minute if we go back up we're trying to get the S of 2x okay so that's what we want so is there an identity that matches that if we go back we have that the S of 2 a is equal to 2 * sin of a * cosine of a so we're basically done right we can basically replace this right here with the S of 2x so we can go ahead and put a big fat check mark there to show that we're done okay let's look at another example again I just want to do a lot a lot of examples of these so we have 1 plus the cosine of 2x is equal to theang of x * the S of 2x okay so which side is more simp Le I would say that because this has coent involved and this just has cosine right this is cotangent sign over here and you have cosine so I would not mess with the left hand side I would mess with the right hand side so I would put equals over here and I would change this into again cosine of x over s of X just to get started again it doesn't always work out that way and then you have the S of 2x so be careful here don't go through and cancel this with this immediately because this is not like having sin squar of X right this is s of 2x so you basically want to use your little formula where the S of 2 * X is basically 2 * the S of X time the cosine of x so let's put cosine of x over s of X and let's say basically over here we have instead of s of 2x I'm going to put time 2 * the S of X time the cosine of x okay so now we can cancel this with this again don't cancel it here that's not correct you have to cancel it here when it's s of X and S of x or if you had sin squ of X or something like that then you can cancel but s of 2x you're not going to be able to cancel with s of X doesn't work that way so let's come down here a little bit okay and let's go ahead and put our equal sign and now basically what I'm going to have is 2 * cosine 2 of X so you might think you have a little stopping point here because I'm trying to get 1 plus cosine of 2x but again if you look at your identities you're going to find something that can match so if I come back here right now we have 2 * cosine ^ 2 of x okay if I add one to each side it's exactly what I'm looking for right cosine of 2 a + 1 is equal to 2 * cosine of a okay so basically I'm just going to replace this guy right here with its alternative form I'm going to say this is equal to 1 plus the cine of 2x okay so straight from your identities handout we'll go ahead and put a check mark there because we're done okay let's look at some problems that involve the product to sum identities and also the sum to product identities these come in a variety of forms sometimes you're asked to find the exact value sometimes you're asked to simplify these guys are really really easy to work with as long as your teacher allows you to have the sheet in front of you otherwise you do have to memorize it okay so what we have here is two times s of 105° time cosine of 75° so basically this is a product okay and we're going to want to change this into a sum to get an exact value so let's go to our little worksheet here and there's quite a few of these so let's go to basically where we have our s of a times our cosine of B like this this is going to be2 times inside these brackets here you have the sign of a plus b plus the sign of a minus B okay so let's just put this down here and I'll just write it out really quickly for you so basically if we have the S of a times the cosine of B okay which is what we have right here then basically this is going to be2 times the sign of let me put this in Brackets a plus b okay and then you have your plus your sign of let me slide this down this is not going to fit so your sign of your a minus B okay so let's close that down so the first thing is you can cancel this immediately by saying you have two times this 1/2 here or you can cancel it later it doesn't really matter you can put this as two times and then you can put brackets here and then put one half and then put more brackets and then put your sign of what's a in this case that's going to be 105° what's B that's 75° so 105° plus 75° is 80° okay let me just write this like this I'll just put s of 80° plus the S of what's a minus B what's 105° minus 75° that's going to be 30° okay so you can put these brackets here again it's not going to matter because when two gets multiplied by half it's basically going to be one right so it's basically gone so we can just say say that we have the S and I put 80° this should be 180° so the S of 180° plus the S of 30° so at this point you should be very familiar with your unit circle the sign of 30° you should know is 1/2 and the sign of 180° you should know is zero so this basically turns into 0 plus 12 which gives me 1/2 okay so just a little formula where you start off with something that looks kind of hard to work with you plug some things into the formula and you get an exact value of a half okay let's look at an example of what we call a sum to product here we're just going to be asked to simplify so you have the cosine of 13 Theta minus the cosine of 5 Theta okay so basically we want to change this into a product so let's go ahead and go back and look at now if we come down here we have the sum to product identities we have the half angle identities but we'll get to that in the next lesson so what we have here is cosine of some a minus cosine of some B this equals -2 * the S of this a + b/ 2 * the sign of this a minus B over 2 Okay so let's go back and let's just write this down here so basically the cosine of a okay minus the cosine of B is equal to -2 * the S of you have this a + b/ two okay times the S of you have this a minus b/ 2 okay so I'm just going to plug in a is going to be 13 Theta and B is going to be 5 Theta okay so let's think about this I want to put equals and I'm going to carry this over here and let's just go ahead and say that we have -2 times the S of what's a plus b well it's 13 Theta + 5 Theta so that's going to be 18 Theta over 2 okay so I can go ahead and reduce that and say that's basically 9 * thet right so s of 9 * Theta and then basically you can say times the sign of so the sign of we have a minus B now so 13 Theta minus 5 Theta is going to be 8 Theta okay over two again 8 divided two is just four so you can put that like that that's basically all we can do if we're asked to write this as a product so we have -2 * the S of 9 Theta * the S of 4 Theta in this lesson we want to talk about the half angle identities all right so we're going to wrap up our section now on the trigonometric identities and get ourselves prepared for solving trigonometric equations but before we do that we need to talk about these half angle identities so let me quickly go back to our handout so we have all of our trigonometric identities listed here so these are the ones we've already talked about let me just keep scrolling down here we're going to come down to all the way down where we have these half angle identities okay so you see you have ones for cosine s you have three for tangent again something you need to basically memorize if you want to go quickly through these problems but in a lot of cases they're going to let you have these handouts so it's something you can just refer to back and forth okay so I'm going to come back to these when I need them the first one I'm going to need is this guy so you can pause the video and write it down or you can have your hand out whatever you want to do but the cosine of some angle a over 2 is equal to plus or minus the square root of 1 plus the cosine of a over two now this plus or minus here is there to tell you that you want to take the appropriate sign based on the quadrant that you're in remember cosine is going to be positive in quadrants 1 and four okay so let's go ahead and minimize this let's go to the first problem here where we're told to find the cosine of 67.5 de now before I start let me just put this off to the side let me put my identity here so the cosine of let's say a over two is equal to we have plus or minus you have the square root of you're going to do 1 plus the cosine of this a over you're going to have two okay so let me just slide this down just a little bit so this is what we're going to be using here so basically if I want to write this angle here as some angle over two I would first multiply by two to figure that out right so 67.5 * 2 is going to give me 135 right so I would say this is equal to the cosine of 135° over 2 okay so now this guy right here matches my format right here okay so I can use my little identity here so let's just put equals here and I'm going to put since we're trying to find the cosine of 67.5 Dees remember it has nothing to do with this guy that's being divided by two this is what we want here okay so this angle right here is going to be in quadrant one so we know cosine is positive in quadrant one so we want to use the principal square root so I'm going to say this is equal to the principal square root of one plus the cosine of here's where it gets tricky this is a this is a okay so you want the cosine of 135° and then over two okay so this is what we're going to be figuring out now hopefully at this point you've memorized the unit circle but if you haven't I have it here so basically if you go to 135° and you look at your x coordinate here this is going to be the cosine of 13 35° so you have negative of the < TK of 2 over 2 and all we need to do is replace that here okay and once you have that it's basically just a lot of simplifying okay so let's just put the square Ro T of 1 plus again this is the negative of the square < TK of 2 over two and then all over two and I should probably make this a little bit bigger so it encompasses the whole thing let's continue this over here so the quickest way to do this in my opinion you could get a common denominator there but I'm just going to multiply by 2 over two okay so 2 * 1 over here would be two let me make my square root symbol and then two times you'd have netive < TK of 2 over two the twos would cancel so you're basically subtracting away theun of two so I'm going to put minus the < TK of Two And this is over 2 * 2 here is 4 okay so we can quickly simplify this by realizing again you could write this as the square Ro T of 2 minus theare < TK of two over the square OT of four we know the square Ro T of four is two okay so I can come down here let's finish this up and just put this is equal to the square < TK of 2 minus the square < TK of two you can't really do anything else there over the square Ro of four is two okay so that's going to be your exact value if you're looking for the cosine of 67.5 de all right let's take a look at a similar problem now we're going to have the tangent so we have the tangent of 7 pi/ 12 and for this one again if you get radians if you're really good at working with fractions you can do it in this way I always just convert these into degrees because it's just easier for me okay so I'm going to first say we have 7 pi/ 12 times again to convert this you want to put Pi in the denominator and you want to put 180° in the numerator always remember that 180° equals Pi radians okay so this is your unit fraction it's basically equal to one and we can use this to cancel so this cancels with this and0 ID 12 is 15 okay so you can just cancel this out and put 15 15 * 7 is 105 okay so this will end up giving me 105 degrees so let's erase this real quick and I'm just going to put that this is equal to the tangent of 105° and then I'm just going to double this number Okay because I'm going to use this half angle formula so I'm going to say this is equal to the tangent down here of 105° * 2 is 210 so 210° over two okay so which formula do we want to use let's go back to our worksheet so you have several here to choose from you can pick this one right here I think it's probably the easiest the tangent of a over two there's no square root symbol to work with it's the S of a so the sign of whatever this is up here over 1 plus the cosine of a so let's go ahead and write this maybe off to the side we have that the tangent of a over 2 is equal to again this is the S of a over you have 1 plus the cosine of a so what do I want to do here again this is going to be what I'm plugging in for a here and here and here okay so basically I would say this is the S of 210° over you have 1 plus the cosine of 210° okay so we don't need this anymore I'm just going to erase it for room okay and then I'll just put equals here what is the sign of 210° let's go back to our little unit circle so you have the S of 210° is - one2 and you have the cosine of 210° is < TK 3 /2 so let's now plug in so the S of 210° is2 and then this is over let me scroll down and get a little bit of room here so again this is over 1 plus what's the cosine of 210° it's the negative of the < TK of 3 over 2 so let me put minus here instead of plus so the negative of the sare < TK 3 over two okay so now all we really have to do is simplify this guy pretty easy overall I'm just going to multiply the top and bottom by two so this gives me what this is going to cancel right the twos would cancel cancel and I just have1 up here and then this is over down here 2 * 1 is 2 and then minus 2 times this guy right here the twos would cancel and you would just have the square root of three okay now you don't want to leave this this way because you don't ever want a radical in the denominator so you need to rationalize the denominator so let's put equals over here and I'll put -1 over 2 minus the < TK of 3 okay remember how to rationalize this when this guy is basically two terms you want to use the conjugate right so the terms would be the same the two would be the same and the square root of three would be the same but you're going to choose the alternate sign okay and the reason for this is when you go through and do the foil down here you're going to have those middle two terms drop out so you're just going to have the first guy squared minus the second guy squared okay so let's go ahead and put equals here in the numerator we know what this is going to be it's just going to make everything negative so it would be let's say -2 then minus the > of 3 down here you would have the first guy squared so 2 squ is four and then minus the last guy squared which would be three well 4 Min - 3 is 1 so basically anything over one is just itself so I can just say the answer here is -2 minus the sare < TK of 3 all right let's take a look at another common problem for this section so your book will probably say something like finding function values of s over2 given information about s okay so in this particular case we are given that the S of theta is 4 fths and Theta is basically in quadrant too right because it's greater than 90° and less than 180° so you can basically say that Theta is in quadrant 2 okay so we have that information now what if I asked you to find let's say the S of theta over 2 and then let's say we want the cosine of theta over 2 and also the tangent of theta over 2 and of course you can get the other ones with the reciprocal identities we're just going to do these three because basically once you have those you can get the other ones easily okay so first and foremost where is Theta over two so Theta over 2 is in what quadrant so this is very confusing for a lot of people but basically to determine this you would just divide everything here by two okay because that would give you Theta over two now 90° divid 2 is 45° okay so that would be less than Theta over 2 which is also less than 180° / 2 would be 90° okay okay so this tells me that Theta / 2 is greater than 45° and less than 90° so it's got to be in quadrant 1 so it's in quadrant one so keep this in mind when you're working with these guys because you've got to choose the appropriate sign okay for your trigonometric functions now if we go back to our identities you'll see to find cosine of a over2 or S of a over2 you need to have the cosign of a okay in each case so that means we first need to find cosine so let me go to a fresh sheet we know that the S of theta is 4 fths and we know that from the Pythagorean identities we can say that 4 fths which is basically sin^2 theta plus cine 2 Theta is equal to 1 okay so what I can do here is just go through and find cosine of theta so basically we've done this a million times we know if we Square Four we get 16 we Square Five we get 25 okay plus cosine squar Theta is equal to 1 I'm going to subtract this away from each side of the equation so this is going to be my cosine 2 thet is equal to this will be -16 over 25 and then plus one okay so let's come over here I know that I need a common denominator now right so you can just erase this right now just put 25 over 25 so 25 - 16 is 9 so I'm going to say cine s Theta is equal to 9 over 25 so this is where a lot of students struggle I'm going to put the cosine OFA is equal to plus or minus theare < TK of 9 25 now which sign should I choose let's go back up and let's remember that we're working with Theta for right now not Theta over two okay so this is where people get confused they make a sign mistake so it's Theta Theta is in quadrant 2 okay if I'm in quadrant 2 cosine is negative okay so that means that in this particular case I want want to choose the negative square root okay so I want to say the cosine of theta is equal to the negative of the square < TK of 9 over 25 so the cosine of theta here is equal to -3 over 5 or you could say - 35ths so let's go back up so let me write this in here the cine of theta is going to be equal to- 35s okay so now we have that piece of information and I want to find the S of theta over 2 the cosine of Theta 2 and the tangent of theta 2 again we know that Theta over 2 is in quadrant 1 so all of these guys if there's a square root involved you want the principal square root okay so if I go back to my little sheet we have the S of a over two is equal to again we just want the principal one the square root of 1us the cosine of a over 2 so let me first write this out so we would have the principal square root of 1 minus the cosine of in this case this is your a right you got to substitute this this is Theta this will be Theta here and and then this will be over two okay so cosine of theta we know is- 35ths so I'm just going to plug that in so minus a negative is plus a positive so I'm going to put plus 3 FS now how you go about doing this let me make this better so three fifths like this how you go about simplifying this is up to you of course you could write one as 5 over five and then basically add your fractions or you could multiply this here by 10 over 10 it's whatever you want to do whatever's quicker let's just do it this way so I'm going to say 10 * 1 is 10 okay and then 10 / 5 is 2 2 * 3 is 6 so 10 basically plus 6 would be 16 so let's say this is 16 over 10 * 2 is 20 okay so I'd have the square root of this guy let me kind of slide this down a little bit kind of working with some narrow space here so let's come down here and let's put this is equal to of course 16 and 20 are both divisible by four so divide 16 by 4 you get four divide 20 by 4 you get five right so what is the square root of 4 FS well the square root of four is two and then over the square OT of 5 now I need to rationalize the denominator so I'm going to multiply this by theare > of 5 over theun of 5 and so this gives me 2 * theun of 5 over five okay so all that just to get this answer here of 2 * theare < TK of 5 over five okay now for the cosine of theta over two we have a similar formula we come back here again we're using the principal square root it's 1 plus the cosine of in this case it's Theta here it's a but in in our case it's Theta over two so let's come back and say this would be what the square root of you have the one plus what's the cosine of theta well it's negative 35ths so in this case you don't have the minus and negative it's straight up minus 3 fths and then this is over two okay let me make this a little bit better and make this longer like this let's come down here so let's say 1 - 35ths let's do this an alternative way let's just say this is 5 fths okay so 5 - 3 would be two so basically you would have the square root of you would have two fths if you're dividing by two you just multiply by half so what you'd have here is the twos cancel and so you'd have the square root of 1/5 so let's erase all of this and put the square root and I made that terribly so the square root of 1/ and of course theun of 1 is 1 so 1 over the < TK of 5 you need to rationalize the denominator so this is times the > of 5 over the square otk of 5 and so this becomes the square > of 5 over five okay let's erase all this and let's slide this all the way down Okay so this going to be right there so Square < t of 5 over five and that's probably too tight so Square < TK of 5 over five okay so for the tangent of theta over 2 again you can come back here and use one of these guys guys but again remember you can just use your definition of the tangent okay so I can just divide this guy by this guy so I can basically say 2 * the < TK 5 over 5 if I'm dividing with fractions I just multiply by the reciprocal so I'm going to flip this and I'm basically going to have five over theare < TK of 5 and so we see that this would cancel okay this would cancel and just get two okay so the tangent here is going to be two so now we found all of the guys that we were looking for so if we're looking 4 the S of theta / 2 is 2 * 5 5 the cosine of theta / 2 is 5 5 and the tangent of theta over 2 is just going to be two okay so again if you wanted to find the other ones you could flip this guy right here in order to find your cosecant of theta over 2 flip this guy right here in order to find your secant of theta over 2 and flip this guy right here to find your cent of theta / 2 all right let's look at one more problem here and this just involves very verifying an identity it's just important to get a lot a lot of practice with this not only for calculus but because in this section in trigonometry it just keeps coming up over and over and over again and there's no General set of rules to follow for this it's all just practice right so manipulating things trying to make sure it works out and then just keep going as long as you're making proper algebraic substitutions you will eventually get where you want to go okay so if I look at this I have 1 plus the S of beta is equal to this quantity of beta/ 2 plus cosine of beta 2^ 2 okay so obviously the left side here is much simpler so I'm just going to work on the right side normally you put your equal sign in line here but I'm going to have to move this all the way over here just because I'm going to run out of R okay the first thing I'm going to do you remember if you have something like x + y quantity squared this is the first guy squared plus 2 times the first guy times the second guy plus the last guy squared okay so I'm just going to do that here just to expand this guy out so I'm just going to say that we have basically the S squar of this beta/ 2 plus 2 * this guy times this guy so this s of beta/ 2 and then times cosine of beta/ 2 and then plus the last guy squar so cosine squar of this beta/ 2 okay so from here you want to think about if I look at this guy what do I have here that I can maybe do something with so nothing really with the half angle identities you actually have to go all the way back up to the Pythagorean identities remember we used this earlier the sin Square theta plus the cosine Square theta equals 1 okay so if we think about this we have sin square of beta over 2 so this is just some angle plus cosine square of that same angle beta over 2 so this plus this would be one okay so I can just basically say that that's 1 plus your 2 * s of beta/ 2 time the cosine of beta/ 2 let me get rid of this marking here because kind of in the way and from here you're really thinking about now your double angle identities from the last section okay remember what you're trying to get to if you look at this you have 1 plus the S of beta so for this guy right here we have the S of 2 a is equal to 2 * the S of a * the cosine of a okay so this is what I have here right I have 2 * the S of beta/ 2 time the cosine of beta 2 so I'm going to go and write it like this s of 2 * beta/ 2 okay so I want to put this as 1 plus we'll say the S of 2 * beta over 2 the twos are going to cancel and look at what I have I have 1+ the S of beta okay if I go back up I'm looking for 1 plus the sign of beta so bam we got it and I know these things are kind of hard I've obviously already worked the problem so I know what to do but when you're working with these don't get frustrated again you can always try to put things in terms of s or cosine perform any factoring perform any squaring and just get as much practice with these as you can because when you get to the test you have to get them done in a reasonable amount of time in this lesson we want to talk about inverse trigonometric functions all right so before we get into the inverse trigonometric functions I want to recap some things that we talked about earlier in the course it's going to be really important that you understand this information in order to fully understand this lesson so first and foremost we discussed the idea of a one:1 function so in a one:1 function each x value corresponds to exactly one yval and additionally you could say in a one: one function each yvalue corresponds to exactly one x value so often you'll hear for each X there's one y and then for each y there's one X now when you have a one: one function you can verify that using something called the horizontal line test so for the horizontal line test if you have the graph of a one:1 function no horizontal line will impact that graph in more than one location so a common source of confusion when you work with inverses would be the notation that's involved so I have here if a function f is a one: one function then the inverse of f is denoted as here this is f inverse so again this is the inverse of my function f now this right here is what you have to get straight this is not an exponent of NE -1 that is a very common source of confusion later on in this lesson when we get to inverse trigonometric functions you'll see something like Y is equal to the inverse sign of X and that's one way to write it we have a different way to write and I'll get to that later on but you need to know that this is not an exponent of ne1 this is telling me I have the inverse s of X now provided that your function let's just call it f of x is a one:1 function you're going to be able to find the inverse using this little procedure so to find find the F inverse of x from F ofx again f ofx is a one:1 function here we're going to replace f ofx with Y we're going to swap X and Y then you're going to solve for y and then once you've done that you're going to replace y with f inverse of X so this right here this swapping X and Y this leads to the following so we have that the domain of f so the domain remember is the set of allowable X values will become the range of f inverse so the range is the set of allowable y values so again the x's and the Y's were switching places so that's what's causing this the domain of f or the X values will become the range of f inverse or the Y values so the x's and the Y's are switching places similarly the range of f will become the domain of f inverse again the x's and the Y's are switching places so this tells us if a comma B is on the graph of f then B comma a is on the graph of f inverse so notice that these guys switched roles so here this is your X and then here it becomes your Y and then here this is your Y and then here it becomes your X so again the domain of f becomes the range for f inverse and the range of f becomes the domain for f inverse so your X's become the Y's and the Y's become the X's very important and then I have here the graphs of F and F inverse are reflections across the line Y = X so there are many cases is where we're going to have a function that's not a one: one function but we still want to find the inverse so in that situation what can we do well let's start off with something that we did earlier in the course we looked at the squaring function so this graph right here is for f ofx isal to x^2 so clearly this is not a one: one function let's just use the horizontal line test real quick to prove that so this would be the line Y is equal to 4 now if you look at this horizontal line it is impacting this graph of f ofx = x^2 in two locations so right here and then right here so again if you have the graph of a one:1 function no horizontal line will impact that graph in more than one location so this right here will not be a one: one function if you look at this point right here from the origin we're going two units to the left and four units up so this is -2 comma 4 if I look at this point right here from the origin I'm going two units to the right and four units up so this is going to be 2 comma 4 so the issue is that this yvalue of four is associated with two different X values here with -2 and here with positive two and that's coming from that squaring operation if I plug in a 2 for X 2^ 2qu is four if I plug in a negative -2 forx -22 is also 4 so when this happens what can we do well like we saw earlier in the course we can oppose a little domain restriction so for this graph right here we start with f ofx = x² squared so that's the squaring function but you would notice that it seems like I've cut off anything to the left of zero so basically we would say that X here is greater than or equal to zero so this is going to create a one: one function and this is the key the range is going to be the same so we're going to see that's going to be important as we get into these trigonometric functions and the inverse trigonometric functions that we create from restricting the domain so the range here is the same before it was from 0 to positive Infinity and now it's still from 0 to positive Infinity now let's just prove that real quick I'll just put in this same horizontal line here so again this is y is equal to 4 and notice that now when I draw this line it's going to impact the graph only in one location and that's true for any horizontal line that you would draw here so this is now the point 2 comma 4 my point -2 comma 4 that was on the other one where I didn't restrict the domain that's gone because X can't be -2 here x can only be zero or anything larger larger so I only get 2A 4 so basically in the inverse I'm going to have 4 comma 2 right so 2A 4 and then 4 comma 2 so for each X there's one y and now for each y there's going to be 1 x so I'll be able to find an inverse okay so let's just go through that real quick hopefully you know how to do this from earlier in the course if not let's just go through it once so that later on it'll be easy so basically you start off by replacing f ofx with Y so let me write here that we're going to find the inverse and would start by replacing again f ofx with Y so y = x^2 then the next step is to swap X and Y so we're going to say that we have X is equal to y^2 so this y became X this x became y that's all you're doing then your next step you're going to basically solve for y so first I would write this as y^2 = X as my personal preference goes I like what I'm solving for to be on the left and then I would use the square root property so you would say that Y is equal to plus or minus the square < TK of X so here's where we have to pause and think in this function right here that's domain restricted F ofx = x^2 we said that X is greater than or equal to zero well we know that the domain from this function right here becomes the range for the inverse so specifically here I would say that Y is greater than or equal to zero so I don't need this plus or minus here because I can't have the negative of the of X because I'm specifically saying that Y is greater than or equal to zero so I'm going to get get rid of this plus or minus and say that Y is going to be the Principal root of x so let me replace this with Y is equal to the principal square root of x and I'll get rid of this and then step four is just to replace your y with your F inverse of X and so this is equal to the square root of x all right so before we move on it might be helpful to look at this graphically this is especially useful when you're trying to figure out if you got the right inverse remember that a function and its inverse are going to be reflect across the line Y = X so you can always look at this graphically and see that so this graph right here for this guy in Orange this is our F ofx = x^2 where we said that X is greater than or equal to zero if you're going to do this on Desmos you want to do this as F ofx = x^2 and then you're going to open up some brackets and you're going to put in your restriction so X is greater than or equal to zero and then just close those and it'll give you this graph here in Orange okay then the graph in red this is just the line Y is equal to X or you can put F ofx equal x whatever you want to do and the graph in blue this is our inverse so F inverse of X is the square root of x I'm just going to put this as G of X so f ofx and g of X those are going to be inverses and this is equal to just the square root of x now what you can do is look at a few points and just prove a few things so first off let's just start with this point right here so this point right here I'm going three units to the right and nine units up from the origin so this is going to be the point let me Arrow to this so this would be 3A 9 now again if I reflect this across the line yal X I'm going to get this point right here so this point right here would be 9 comma 3 so notice that the x value which was three is going to become the Y value here and the yv value here which is nine is going to become the x value here so the x's and the Y's are switched then let's just do one more so this point right here if you look at at it is going to be 2 comma 4 so this point right here is 2 comma 4 and then if you look at again reflecting across the line yal X you're going to get this point right here and now instead of 2 comma 4 you're going to get 4 comma 2 so this is always something you can use if you're trying to figure out if you got the right inverse it's very helpful to graph your function graph your inverse and graph the line yal X and just look at if your function and the inverse are reflect flections across the line yal X then you know you got the right one all right now let's move on and talk about how to find the inverse trigonometric functions we're going to start off with how to find the inverse sign function and so to do that we're going to begin with our graph of f ofx equals the S of X or we could say Y is equal to the S of X so we studied this graph earlier in the course at this point you should know that the domain is the set of real numbers and the range is from NE 1 to POS 1 so let me write this out so the domain is going to be the set of real numbers so from negative Infinity to positive Infinity so that means I can plug in any real number I want for X here so a negative 0 positive as long as it's a real number I'm good to go then in terms of the range so in terms of the range we can see that's from -1 to positive 1 and you should know that from the unit circle so let's go ahead and put from -1 to positive1 clearly this guy is not the graph of a one:1 function if I draw a little horizontal line like this well you can see that I'm going to hit here and here and here and here and here and here and here and remember this continues forever to the right and forever to the left so I'm going to keep hitting over and over and over again so obviously this is not the graph of a one to one function it fails the horizontal line test so what can we do when we encounter this situation again just like with the squaring function we restricted the domain such that we had the same range so let me actually erase this and show you what we're going to do we're going to take a piece of this graph so we're going to go from right here this is pk/ 2 to right here this is pi over 2 let me write that a little bit more neatly so again this is pi/ 2 here so you're going to take right here to right here so just a piece of that graph right there and now it's going to pass the horizontal line test so it'll be a 1:1 function it'll have the same range from -1 to pos1 so let me come to this graph here it's a little bit easier to see things so this is called the restricted sign function so basically it's your sign function with a restricted domain so we have y is equal to your s of X and then we would say that X is greater than or equal to piun / 2 and then less than or equal to pk/ 2 so obviously the domain is from piun / 2 to pi/ 2 with those guys included let me just write this out so the domain is going to be from piun / 2 to piun / 2 with both of those guys being included and this is the key here the range is the same so if you go back to the yal S of X again the range is from -1 to positive 1 well here even though we've restricted the domain the range remains from negative - 1 to positive 1 so let me write that the range is going to still be from -1 to positive1 okay so now let's cover the the inverse sign function so we're going to start with the restricted sign function so that's going to be this graph in Orange let me just Trace over what the computer Drew here so this is my graph in Orange a lot of stuff is overlapping here so it's pretty hard to see it clearly but this graph again is y is equal to my sine of x where X is greater than or equal Tok / 2 and less than or equal to < / 2 so that's the restricted sign function if you reflect it across the line Y X you're going to generate this graph here in green so let me go over that so from here to here so a lot of overlap here you can do this on Desmos if you want a cleaner picture but in terms of the video I think it's going to come through at least good enough to where you can see what's going on so this right here is going to be the graph of the inverse sign function typically you'd write that as Y is equal to the inverse sign of X like that you could also so write this as Y is equal to the AR sign of X although if you're in a trigonometry course this is typically more common but some books use this it just depends so you need to know both of those in case you run into one and not the other okay so basically like we saw earlier you're just taking this point right here which is from the restricted sign function and you're reflecting it across the line yal X to get this point right here on your inverse sign function so notice that this x value of Pi / 2 becomes the yvalue of pi/ 2 here in the inverse and the yvalue of one becomes the x value of one in the inverse we can also take this point right here and again reflect it across the line yal X to get this point right here so you see this x value of piun / 2 becomes the yvalue of piun / 2 and then the yalue of 1 becomes the x value of-1 okay so let's look at a cleaner graph here so this is the graph of Y is equal to your inverse sign of X and you can see that the domain of this guy came from the range of your restricted sign function let me actually put a little border here and let me write Y is equal to the S of X and let me put that X is greater than or equal to piun / 2 and less than or equal to piun / 2 and let's consider that the domain of this guy so the domain of this guy is the same as the range from this guy so the domain here is from1 to POS 1 so this is going to be from 1 to POS 1 well here if you look at the range if you look at the range it's from -1 to positive 1 then if we think about the range for this guy so the range this is coming from the domain of the restricted sign function so it's going to be from piun over 2 to piun over 2 so from piun / 2 to piun / 2 and then the domain here of course we know this at this point let's just write it one more time so again this is from negative pi over 2 to Pi / 2 so just like we talked about the domain from the restricted sign function becomes the range in the inverse and then the range in the restricted sign function becomes the domain and the inverse so the X's become the Y's and the Y's become the X's all right now let's look at a few sample problems so we're going to start with y equals the inverse sign of 1/2 so we just want to figure out what is the value for y here so if you were to key this into your calculator so if you were to type in the inverse sign of/ 12 you would get pi/ 6 if you're working with radians or you would get 30° if you had it set to degree mode so why does it give you that answer so why does it give you Pi / 6 and not the 5 pi/ 6 that you would see on the unit circle so we can see that for Pi / 6 I have a sign value of 1/2 but then also for 5 pi/ 6 I have a sign value of 1/2 as well well what's going on well to see this let's think about deriving the inverse algebraically we saw a graphical representation of it but now let's do it algebraically and we'll get a little bit more insight into what's going on so let's say you start with again y equal the sin of X and of course we need to make this 1: one so we can have an inverse so X is greater than or equal to piun / 2 and less than or equal to piun / 2 so to find the inverse if you had F ofx you change it to Y well that's already done for us here and then basically you'd swap X and Y so this y becomes X and is equal to the sign of this x becomes y okay well now this domain is going to become the range so this is y is greater than or equal Tok / 2 and less than or equal to < / 2 now this is a form that you're going to see in your textbook what they'll say is from here in order to solve it for y we're just going to say that it's Y is equal to the inverse sign of X so this right here is just solved for y it means the same thing as this now I'm going to concentrate on it in this form for right now it's a little bit easier to think about it so basically let's say that we start with this right here so the domain restricted sign function we know that a point on that graph is going to be pi/ 6 comma 12 so where did I get that from well first off Pi / 6 is within the domain it's greater than or equal Tok 2 and less than or equal to Pi / 2 so if I plug that in for X I get a true statement so it's within the domain and the sign of pi/ 6 is2 so this would be a point on the graph well when I get to the inverse these are switched so now this is 12 comma pi over 6 in other words when I look at this guy I am asking the question let's say I plugged in a 1/2 here for X I'm saying what angle in the interval from piun / 2 to piun / 2 is going to have a s value of 1/2 if we go to the unit circle I'm going to look at this right here so this is what I'm looking for right here so this is Pi / 2 and then this down here if you're rotating clockwise this is going to be netive piun /2 so this is what you're looking for here if your argument is positive like we have here we're going to be looking at things in quadrant one if your argument is negative you're going to be looking at things in Quadrant 4 now this is an exception to what I just said if you have the inverse sign so if you have the inverse sign of one that's going to give you Pi / 2 and of course that's a quadrantal angle if you had something like the inverse sign of 0 that's going to give you zero so Z is not a positive number but if you plug that in you're going to get something that's a quadrantal angle and then also the inverse sign of let's say -1 that's going to be piun / 2 and that's coterminal with 3 piun 2 so again these are quadrantal angles but if I say something like the inverse sign function is going to give you something in quadrant one if the argument is positive and quadrant four if the argument is negative you need to just know these exceptions let me get rid of this real quick and let's go back to what we were talking about so here we have the inverse sign so y equals the inverse sign of 12 or we wrote this as X is equal to the S of Y I where specifically Y is greater than or equal Tok / 2 and less than or equal to > 2 so again where in this interval am I going to have a sign value of 1/2 well again this argument is positive so I'm only going to look in the first quadrant and I see that I have a sign value here of 1/2 and that is going to be right here at pi/ 6 so that's why I'm getting this pi/ 6 and I'm not getting this 5 pi over 6 here as an answer if you notice you have a sign value of2 for 5 pi over 6 and a lot of students will always complain why am I not getting the 5 pi/ 6 you're not getting it because it's not within the range of the inverse sign function so to answer this if you have y equals the inverse sign of2 let's just write the inverse s of 12 is going to be equal to < over 6 all right let's look at y equals the inverse sign of > 2 over2 so this one's a little bit more tricky let's come down to the unit circle let me write Y is equal to the inverse sign of > 2 / 2 again you can stick with it in this form if you want you're just looking in Quadrant 4 for an angle whose s value is going to be2 over2 that's what you're looking for you can again write it as the S of Y is equal to X where Y is greater than or equal to PK / 2 and less than or equal to < / 2 and of course you could just plug in aun2 over2 so again Y is the angle in this interval where we have a s value of2 over2 now because this is negative I want to work in Quadrant 4 let me just highlight the whole thing first let me highlight the whole thing first so we get comfortable with this and basically if I go into Quadrant 4 and I'm looking for a s value of2 over2 it's going to be right there so this is going to be at 7 pi over 4 now is 7 Pi over4 the answer no you cannot say that the inverse sign of the negative of the2 / 2 is equal to 7un 4 that is incorrect because it's not in this interval what you have to do is actually rotate clockwise here let me pick a good color here that's going to show up so this right here you need a clockwise rotation so the reference angle there is pi over 4 so this rotation clockwise this angle is going to beunk over 4 so this would beunk over 4 as the answer now another way you can do this is a longer way but essentially what you could do is you could figure out okay I'm going to be in Quadrant 4 so I want this 7 pi over 4 answer but I need to put it into this interval since I started here and rotated all the way around by 7 pi over 4 what I have to do is now rotate clockwise by 2 pi so that's going to be me taking 7 pi over 4 and subtracting away 8 pi over 4 to get my pi over 4 so in other words you're going to go this way clockwise by by 2 pi so that's just subtracting 2 pi away so coming back here and finding a co-terminal angle so ending up right there so now I'd get the angle that I want which is going to be netive piun over 4 I think this is a little bit more complicated but sometimes people need to do that to figure out what's going on what I always do is figure out what the angle is so in this case okay it's 7 pi over 4 the reference angle is pi over 4 and I know it's negative so it's just negative pi over 4 I think that's pretty easy to understand all right now let's talk about the cancellation properties for sign so these are some simple rules you can use to quickly evaluate things so the first one's very straightforward so you have something like the S of the inverse sign of X that's going to be equal to X for X is greater than or equal to1 and less than or equal to postive 1 so why do we need this restriction here so first off if you consider the outer part here so the sign of so this right here is the argument for the sign function does it matter what I plug in for the sign function in terms of the domain no because it's all real numbers so I can plug in whatever I want a negative zero positive whatever I want the problem actually comes from the inverse sign function so here if I look at this x here what can I plug in for that well remember the domain of the inverse sign function is going to be from -1 to postive 1 so that's where you have to be careful if you did something like the sign of the inverse sign of two well you're not going to be able to get an answer there because the inverse sign of two you can't type that into your calculator and get an answer it's just going to give you an error so what you would want to do is make sure that it's within this interval here and then if you had a problem like the sign of let's say the inverse sign and then for this let's say we go with what we just did so let's say we plug in2 well this is going to just give me 1/2 in other words these guys are just going to cancel away and I'll be left with this 1/2 if you want to do this the long way the inverse sign of 1/2 we just saw that was pi/ 6 so this would be the sign of pi/ 6 which we know is2 so I can just skip this step right here and just go straight to saying that the answer is 1/2 all right so the next one is a little bit more challenging to understand so what you want to concentrate on is the first part so the S of X so think about that for a moment we know that again the domain for the sign function this is not a restricted sign function or anything like that I can plug in whatever I want for X it does not matter any real number so plug in whatever you want the result of the S of X is going to be from Nega 1 to positive 1 now with the inverse sign function remember the domain or what you can log in is going to be from Nega 1 to postive 1 so I'm good to go in terms of plugging things in for the inverse sign function I'm not going to run into any issues but the problem is that the range or the output for the inverse sign function is only going to be from negative piun / 2 to Pi / 2 so if you're in that interval you can just cancel this with this and say it's equal to X if you're not you need to evaluate the S of X first and then take the inverse sign so let's look at two examples so the first one is we have the inverse sign of the S of piun 3 so to answer this the first thing you would do is say is piun 3 greater than or equal Tok / 2 and less than or equal to piun / 2 well this part I know I know that negative piun over 3 is certainly less than pi/ 2 because a negative is always less than a positive now what if you compared this one well the numerators are the same forget about the fact that they're negative for a moment let's say you had pi/ 2 and pi over 3 and I just put a question mark here so if I look at this the numerators are the same the denominators are different so same numerator that means that the larger denominator is going to belong to the smaller fraction so I can say that this guy right here is greater than this guy right here but now if I multiply both sides by negative 1 that's going to change so this right here this right here and this gets flipped so now this guy which was the larger fraction because now that that it's negative it's going to actually be the smaller fraction because it's further to the left of the number line so that means that netive piun / 2 is less than negative pi over 3 so this inequality is going to be true so that means that I can just say that the answer here let me slide this out of the way is just going to beunk over 3 if you go to the unit circle and you want to define the S of piun over 3 again I'm just looking for an angle in Quadrant 4 with a pi over 3 reference angle so that's going to be right here at 5 piun over 3 but again I'm not thinking about it as 5 piun over 3 because I'm rotating clockwise so that's my piun over 3 angle so the S ofk over 3 is the same as the S of 5 piun over 3 the co-terminal angle so that's going to be the negative of the 3 over 2 so if you went through and said well what is the inverse sign so what is the inverse sign of the negative of the 3 over 2 you're going to go right back to that piun over 3 so again if you're in this interval you can just shortcut all that and just give this as the answer right away so the answer here is just piun over 3 all right let's look at another one so this one will give us a little bit of trouble so we have the inverse sign of the S of 5 pi/ 6 so can we just say the answer is 5 pi over 6 no and the reason for that is 5 pi over 6 is not in the range of the inverse sign function we cannot get that as an answer so because of that again what you'd have to do is first find the S of 5 pi/ 6 so coming to the Circle here's your 5i 6 and all of you know that that's 12 so basically you would have to transform this into the inverse sign of s of 5 pi over 6 is 12 and then we know that this is pi/ 6 but again if we just come back down you're working here and this guy is going to be positive so you're looking in quadrant one for an angle with a sign value of2 so you've got that right here and that's going to be at pi/ 6 so that's the angle I'm looking for so coming back we just answer pi over 6 so when this guy right here is not in the range so from NE piun 2 to Pi / 2 of your inverse sign function you've got to do additional work first you're going to do the inside so evaluate the sign of 5 pi/ 6 that gives me 1/2 then take your inverse sign of 1/2 and that's going to give me pi/ 6 all right now let's move on and talk about finding the inverse cosine function so the process is exactly the same as what we did for finding the inverse sign function there's just a few differences in terms of what part of the graph we're going to take in order to create a one:1 function so we would start with the graph of y is equal to the cosine of x so we've studied this graph before we know that the domain is all real numbers and the range is from negative - 1 to positive 1 so let me write here that the domain is going to be the set of real numbers so from negative Infinity to positive Infinity so I can plug in whatever I want for x there so a negative zero a positive whatever I want as long as it's a real number and then for the range so for the range we can see that this is from -1 to positive 1 with both of those being included now again this function is not a one:1 function you can see that from the graph if I make myself a little horizontal line here you're going to see that it hits here and then here and then here and then here and again this graph continues forever to the right and forever to the left so it's going to keep hitting over and over and over again so obviously this is not the graph of a one: one function so again what do we do well we're going to restrict the domain such that we have the same range so for this one what we want to actually do is start at zero so an x value of zero and then go to an x value of pi so this is the piece of the graph we want so this is going to be from 0 to Pi and notice that your range is not going to change it's still going to be from Nega - 1 to positive 1 so this would create a one:1 function for us and to see that more clearly this is the graph of our our restricted cosine function so this is y is equal to the cosine of x where X is greater than or equal to Z and less than or equal to Pi so you can see this would be a one:1 function now and let me just write here for clarity that the domain obviously it's right here but let's just list it so this is going to be from 0 to Pi where Z and Pi are both included and then the range this is the key here it's going to be the same as your regular y equal the cosine of x so it's still going to be from - 1 to posit 1 with both of those guys being included now what we can do if we want to find our inverse is reflect this graph across the line yal X so if we start with this graph again in purple this is the graph from the previous page this is my Y is equal to the cosine of x where X is greater than or equal to Z and less than or equal to Pi so again the restricted cosine function let me just go over this for clarity so this graph right here again you're going from 0er to Pi and the range and this is the key is still from -1 to positive 1 so when you reflect this across the line yal X that's this guy in Orange here you're going to get this graph that's in teal so this guy right here is the graph for y equals the inverse cosine of x so if you started with something like let's say this point right here which is 0a 1 and you reflect across the line Y = X you're going to get to this point right here which is 1 comma 0 and that's going to be on your inverse so the x value of 0 becomes a yvalue of 0o and then a yvalue of one becomes an x value of one so those x's and y's are switched then if we look at this other point right here which is pi comma 1 again if I reflect it across the line Y = X I'm going to get this point right here which is- 1 comma Pi so here the x value of pi becomes the yv value of pi in the inverse and then the yalue of1 becomes the x value of1 in the inverse to get a better look at this we can see this graph on its own so this is for y is equal to your inverse cosine of x and let me put a little border just like we did last time let me put the Y is equal to the cosine of x where X is greater than or equal to zero and less than or equal to Pi and again for this one we know that the domain so we know that the domain because it's listed specifically right here is going to be from 0 to Pi we'll notice that the range here for the inverse cosine function is going to be from 0 to Pi so let me write that the range for this one is going to be from 0 to Pi so the domain becomes the range and the range is going to become the domain so the range here so the range in this case is going to be from Nega 1 to positive 1 and you're going to see in the inverse again the domain now is from negative 1 to positive 1 so let me put that the domain is from -1 to pos1 all right let's look at a few problems now so here here we have y equal the inverse cosine of the negative of the 32 so again a lot of you can immediately say that the answer is going to be 5 pi/ 6 but just to go through this the long way so we understand where this stuff is coming from what we want to do is think about this y equals the cosine of x so just start with this like we did graphically and we're going to restrict the domain so X is greater than or equal to zero and less than or equal to Pi if you go through this is going to get cut off so let me just slide down here we'll come back up so if you go through and find the inverse of this domain restricted function you swap the X and the Y so you end up with X is equal to the cosine of Y and of course this right here which is the domain becomes the range so now you have y is greater than or equal to zero and less than or equal to Pi so if I have something like let's say the negative of thek of 3 / 2 is equal to the cosine of Y where Y is greater than or equal to Z and less than or equal to Pi well when it's written this way you can say that Y is going to be the angle between 0 and Pi where 0 and Pi are included where your cosine value is 3 over2 well again if you just think about the unit circle and we come down here well I'm working from Zer to Pi so now I'm looking here and if your argument is positive you're going to look in quadrant one and if it's negative you're going to look in quadrant 2 again there are exceptions to what I just said just like previously if you have something like the inverse cosine of let's say one well this is going to be zero so that's going to be a quad Al angle if you have the inverse cosine of 0 that's going to be pi/ 2 so that's a quadrantal angle and then lastly if you have the inverse cosine of something like -1 well that's going to be Pi so basically other than these guys you can say that if your argument is positive you're looking in quadrant one if your argument is negative you're looking in quadrant 2 the argument we have let me get rid of this is negative so we're looking for the inverse cosine of the Nega of the 3/ 2 so that tells me I want to look in quadrant 2 so looking in quadrant 2 where coine values are negative I see that's going to occur right here so there's my cosine value of3 over2 and the angle that's associated with that is 5 piun / 6 so the result here would be 5 pi over 6 in other words on the graph of this guy right here the restricted cosine function you have the point 5 piun / 6 comma 3 / 2 so when you get to the inverse cosine function these guys are going to be swapped so you're going to have 3 / 2 comma 5 pi over 6 so when we answer this we basically just say that and I think this is cut off too far so let me just come down here and I'll just answer it formally I will say that if we have y is equal to the inverse cosine of the negative of the 3 over 2 then we're going to say that this is equal to 5 piun / 6 all right let's let look at another one so we have y equal the inverse cosine of2 over2 so again you can work off the unit circle a lot of you know this off the top of your head but we're working from Zer to Pi now because that guy was a negative argument so in other words we had Y is equal to the inverse cosine of the negative of the > 2 / 2 again when this guy is negative I want to work in quadrant two that's where cosine values are negative so just find where you have A2 over2 for the cosine value and the Y there is going to be the associated angle so basically I'm going to find that right here let me just highlight that so you have a cosine value of2 over2 with your 3 Pi so with your 3 pi over 4 so the answer here for y is just going to be 3 pi over 4 again if you look at the restricted cosine function so y equal the cosine of x where X is greater than or equal to 0 and less than or equal to Pi well a point on that graph is going to be 3 pi 4 comma > 2 / 2 so that's a point on the graph so when we think about the graph of the inverse cosine function so y equals the inverse cosine of x well these X and Y values are going to be swapped so now you have this neg2 / 2 and then comma 3 piun over 4 so the inverse cosine of2 over2 is going to give me 3 pi over 4 all right just like with s you have the cancellation properties for cosine and the inverse cosine function when you put those guys together so basically if you have something like the cosine of the inverse cosine of x it's equal to X but only for X being greater than or equal to 1 and less than or equal to postive 1 so the reason for that is again the domain for the inverse cosine function is going to be from negative 1 to positive 1 you can't take something like the inverse cosine of two again if you plug that into your calculator you're going to get some kind of error so what you would do let's say we had something like let me just slide down here real quick quick the cosine of let's say the inverse cosine of let's say something like2 well again this right here would be in quadrant 1 so I'm thinking about where is the cosine value 1/ 12 in quadrant 1 well that's pi over 3 so this going the long way would end up being the cosine of pi over 3 and I'm right back to 1/2 so again if this value satisfies this inequality so it's greater than or equal to 1 and less than or equal to postive 1 you can just shortcut this process and immediately give the answer of 1/2 all right so the other property involved is going to give you a lot of trouble if you're not paying attention so the inverse cosine of the cosine of x equals x but this is the key this is only for X being greater than or equal to zero and less than or equal to Pi now what's causing this this is your regular cosine function on the inside so we know that the domain is all real numbers so really it doesn't matter what I plug in for X could be zero could be negative could be positive it doesn't matter we know that the range range or the outputs here are going to be from -1 to positive 1 and those guys are going to be included so let me actually just mark this out real quick so we can think about this so I know that what I'm going to be plugging in here for the inverse cosine function is going to be from -1 to positive one and again that is the domain of the inverse cosine function so I'm okay there I'm never going to violate the domain so let me erase this and now let's think about this when I take the inverse cosine of something from ne1 to positive 1 with those guys being included I know that my range or my set of output values is going to be from 0 to Pi so that's why we have this for X is greater than or equal to Z and less than or equal to Pi now if that was confusing let me just give you an example here and hopefully this will make sense so here we have the inverse cosine of the cosine of 7 pi over 4 now 7 pi over 4 is not in the interval from 0 to Pi where 0 is included and Pi is also included if you go to the unit circle again this is where we're working right here so this is from 0 to Pi now if we think about 7 Pi 4 that's down here okay so that's not in that interval so what you'd have to do is say that you have the inverse cosine of the cosine of 7 pi over 4 you need to evaluate this first so the cosine of 7 piun over 4 that's going to give Youk 2 over2 so we're going to say this is the inverse cosine of the < TK of 2 over 2 and again looking at the correct interval here you see where the cosine value is2 over2 that's going to be right here and that's going to be with P pi over 4 so you'd want to answer piun over 4 it is incorrect to say that the answer is 7 piun over 4 because that is not within the range of the inverse cosine function so you have to be very very careful with that type of problem if you had gotten the problem the inverse cosine of let's say the cosine of pi over 4 well in this particular case now you're within that interval so these would just cancel away and you would say this is pi over 4 again if you go through it the long way which is fine you can still do that the cosine of P piun 422 and then the inverse cosine of2 over2 will get you back to P Pi / 4 all right now let's talk about the inverse tangent function so we're going to use the exact same process that we used with the inverse sign function and the inverse cosine function so we're going to start off with a graph that we already know about so that is y is equal to the tangent of X now when you graph this guy hopefully you've seen this already you know that you have a series of vertical asmp tootes and that's going to be at multiples of pi/ 2 so something like this is at Pi / 2 and then this guy right here is going to be 3 pi/ 2 and this guy right here is going to be 5 Pi / 2 and if you were going this way so this right here would be netive piun over 2 and then this right here would be -3 Pi / 2 and then this guy right here would be -5 pi over 2 so again what causes that well if you go back to the unit circle if you look at something like P piun / 2 or 3 piun / 2 the cosine value is zero and and this is going to be true for pi/ 2 3 piun / 2 5 piun / 2 7 piun / 2 so on and so forth so a lot of books will just say odd multiples of pi/ 2 so what happens is when you look for the tangent of something like that if the cosine value is zero you have the S value divided by the cosine value and again if you're trying to divide by zero that's undefined so coming back up that's why you see these vertical asmp tootes here so for the domain so for the domain and I'm going to write this in set build or notation so I'm going to say that it's the set of all X such that X is not equal to we're going to start with Pi / 2 and then just say plus pi n and here specifically n is just going to be any integer so if you added pi to pi over 2 you would get 3 pi over 2 then if you added let's say 2 pi to pi over 2 you would have 5 Pi / 2 so on and so forth now in terms of the range there's no restriction right so this is from negative Infinity to positive Infinity so for the range we're going to go I'll just use interval notation it's a little bit easier so so from negative Infinity to positive Infinity so basically all real numbers now when we look at the graph of yal the tangent of X obviously this is not a one toone function it fails the horizontal line test so let me make a little horizontal line here and it would hit here and it would hit here and here and here and here and of course this graph continues forever to the right and forever to the left so it's going to hit over and over and over again so obviously this is not the graph of a one:1 function so what do we do well just like before we're just going to take a piece of this graph such that we have a restricted domain but we keep the same range so what we've chosen to do in this situation is go from netive pi/ 2 to Pi / 2 with those values being excluded because again tangent of negative piun / 2 or tangent of Pi / 2 that's undefined because you'd have division by zero so this is the piece that we're going to take right here so that's what we're looking for and let me just show you that using a better graph so this right here would be your Y is equal to the tangent of X where X is greater than and this is strictly greater than piun / 2 and less than and that's strictly less than pi over 2 so this is your restricted tangent function and just like before if you reflect that across the line Y = X you'll get your inverse tangent function so this right here in Orange this is your y equals again the tangent of X where X is greater thank / 2 and less than < / 2 and then the graph in green here this is going to be your Y is equal to the inverse tangent so the inverse tangent of X so let's go down to this little graph here where things are easier to see so let's write that this is y equals the inverse tangent of X and for the domain you'll see that it's the set of real numbers so I can plug in whatever I want for X there there's no problems so I'm going to say that my domain is just going to be from negative Infinity to positive infinity and if we think about where that comes from well think about the range of the restricted tangent function it's from negative Infinity to positive Infinity so when we get the inverse the domain becomes the range and the range becomes the domain so this range here from negative Infinity to positive Infinity became the domain here of negative Infinity to positive Infinity now similarly when we look at the r here this is coming from the domain of the restricted tangent function which was from 2 to piun / 2 with those guys being excluded so this is really important you want a parenthesis here for Pi / 2 it's not included and then a parenthesis next to pi/ 2 because it's not included so this value up here and this value down here they're not included so you would have a horizontal ASM toote here and also here so these graphs will approach those values but never actually touch them okay let's take a look at some problems so we have y equals the inverse tangent of square < TK of 3 so again when I look at something like this I always go Y is equal to the tangent of X and I'm going to say x is greater thunk / 2 and less than piun / 2 and again I'm just going to swap X and Y to find the inverse so I'm going to say x is equal to the tangent of Y where specifically again this was the domain so now it becomes the range so Y is greater thank / 2 and less than Pi / 2 so if I plug in aunk of 3 for X there so we'd say the > of 3 is equal to the tangent of Y and again where Y is greater than < / 2 and less than pi/ 2 well let's think about this for a moment I'm asking for the angle in the interval from piun / 2 to piun / 2 such that the tangent value is going to be the sare < TK of 3 well you can use special triangles we'll go to theit Circle in a moment if you look at a tangent value of square Ro of three so that's going to be right there I don't think my highlighter really shows up there so maybe I can use a different color like black I think that might show up a little bit better so basically you can see that's going to happen at 60° or if you want to convert this into radians that's going to be pi over 3 Additionally you can always go to the unit circle and use that the values aren't directly there but if you look at pi over 3 again that's in your interval you're working from piun / 2 to Pi 2 The key thing is here that Pi / 2 and negative Pi / 2 are not included okay so it's not like sign where they are included here they're excluded so if I look at that that's within the interval and again if you look at the tangent of pi over 3 well really quickly you can just look at well it's the S of piun over 3 which is the 3 / 2 / the cosine of piun over 3 which is basically going to be 12 so times the reciprocal which is 2 over one those twos cancel and you get theun of three so that tells me that the inverse tangent so the inverse tangent of the of 3 is going to be < over 3 because again it's in this interval so let's come back and I'm just going to answer that this guy right here is going to be pi over 3 okay let's look at another one so this will be real easy we have y equals the inverse tangent of the negative of the of 3 so first off what you could do to solve this problem the inverse tangent function and the inverse sign function are odd functions so you could say that this is equal to the negative of the inverse tangent of the Square t of 3 and we saw earlier that the inverse tangent of theun of 3 was pi over 3 so this would be the negative of pi over 3 or just negative pi over 3 so that's the quickest way to solve that given the fact that you know that the inverse tangent of 3 is pi over 3 now the other way let's just do this the longer way just so we can see it the first thing is if I know that Y is equal to the inverse tangent of the square root of 3 is going to be equal to pi over 3 we already know that well then what I can do is go to my unit circle and look for an angle in Quadrant 4 where tangent is negative and I'm going to want a pi over 3 reference angle so here I'm going to rotate clockwise to get to right there so this is not going to be 5 pi over 3 it's going to be piun over 3 again you can start with 5 pi over 3 if you want and just subtract away 2 pi so that's 6 piun over 3 and you can get the ne piun over three like that if you want again it's up to you how you want to do things but essentially you want to make sure that you're in that interval again from piun 2 to pi/ 2 let me highlight that real quick so I'm going to make sure to not touch pi/ 2 or negative pi over 2 those are not included so as long as you're in this interval you're going to be good to go so basically because the tangent of piun over 3 is equal to the negative of thek of 3 well because again this is in that interval you can say that the in inverse tangent so the inverse tangent of the negative of the of 3 that's going to beunk over 3 so coming back up the answer here once again is piun over 3 all right let's look at some cancellation properties with Tangent and inverse tangent so these are very similar to what we saw earlier so the tangent of the inverse tangent of X is equal to X for any X that's a real number so that's basically what this is saying this says X is an element of the real numbers whenever you see this r that looks like this this means the set of real numbers so let me write that out so the set of real numbers and this right here if you've never taken a course on sets this just means is an element of now when we think about the inverse tangent function remember the domain is the set of real numbers so I can plug in whatever I want there for x and the output will be between piun / 2 and pi/ 2 with those values being excluded so that means again you could just mark this out and say okay well the tangent of between piun / 2 and pi 2 that's always going to be defined so I won't really have any issues there so I can just basically cancel things away and say that I just have X so if you wanted to see an example of that you could just quickly do something like the tangent of let's say the inverse tangent of let's just go ahead and do something like maybe the square < TK of 3 over3 something like that I think most of you know that the inverse tangent of the sare of 3 over 3 is going to be pi/ 6 so this would be the tangent of pi over 6 and again the tangent of piun 6 is going to get me back to the < TK of 3 over 3 so you could have just shortcutted the process and just said the answer is > 3 over 3 all right let's look at the other one and this one is where people get into trouble so the inverse tangent of tangent of X is equal to X for X is greater than NE piun over 2 and less than pi/ 2 so let me walk you through this one so basically let's start off with the tangent of X so the inside part here we know that for this guy right here x cannot be P piun / 2 3 piun over 2 5 piun over 2 so you could say Pi 2 plus pi n where n is an integer or you could say something like odd multiples of Pi / 2 whatever you want to do but basically assuming that this guy right here you don't plug in something that gives you undefined as a result well then we know that the range for the tangent of X is going to be all real numbers so assuming that you didn't plug in something that violated the domain to where you got undefined the result of this guy right here would be all real numbers now when you take the inverse tangent again the domain is all real numbers and when you plug in all real numbers for this guy you're going to get basically between piun / 2 and P Pi / 2 as a result so that's why we say this is for X is greater than piun / 2 and less than Pi / 2 because again that's from the range of the inverse tangent function so let's look at an example so here we have the inverse tangent of the tangent of 4 Pi 3 so the tangent of 4 piun over 3 that's defined but it's not within that range from piun / 2 to piun / 2 so coming back to the unit circle again thinking about from pi/ 2 to negative pi/ 2 again those guys are excluded so I'm trying to stop short there and maybe I touched there so let me try this one more time I want to make that crystal clear that pi/ 2 and negative pi over 2 are not included so you're looking at this angle right here which is 4 pi over 3 so that's not in that interval so what you would have to do is first say that I want the tangent of 4 pi over 3 now this has a pi over 3 reference angle and you're in quadrant so we know that this would be equal to the tangent of pi/ 3 because again the tangent values are positive in quadrants 1 and three so essentially if I look at this guy right here the tangent of pi over 3 it's going to be the same as the tangent of 4 Pi 3 again the period for tangent is going to be Pi so the tangent value is going to be the same there so this is going to be equal to as we saw earlier the square root of three so essentially when we solve this problem if we say something like the inverse tangent of the tangent of 4 pi over 3 what you need to do is the tangent of 4 pi over 3 first so this is the inverse tangent of again that would be squ of three so squ of 3 and then this equals what well where is the tangent value square of three in this interval here well it's going to be at pi over 3 so that puts me right back to pi over 3 so you got to be really careful with these types of problems now again if you had gotten the problem as something like let's say the inverse tangent of and let's go tent ofk 3 we saw that earlier well this guy right here is in your interval so we know that that's going to be right here again rotating let me do this in a different color so rotating clockwise like that so that's my Nega pi over 3 so I can just take that in this case that'll be the answer so this is just PK over 3 all right now let's look at some problems that deal with the inverse cotangent function the inverse secant function and the inverse cosecant function so these problems are pretty easy if you already understand the inverse inverse sign function the inverse cosine function and the inverse tangent function you can actually convert them over and use those guys to get your solution so let me start with Y is equal to the inverse cotangent let me make that a little bit cleaner there of x if you're looking this up online or if you're using a textbook that's different from mine you might see a different graph and the reason for that is very simple when you derive this guy you start off with yal the cotangent of X so we know that that that guy right there is not a onetoone function we also know it's not defined for pi n or n is any integer because if you have something like Zer or Pi or 2 pi or 3 Pi your s value is going to be zero and you'd have division by zero which is undefined so what you could do is take a piece of that guy let's say from 0 to Pi where Zer and Pi are not included and basically if you restrict that domain there you're going to get a one:1 function and if you reflect that across the line yal X you're going to get get this graph here so in this case the domain is now going to come from the range of that guy so you're going to have all real numbers so from negative Infinity to positive infinity and then the range again is coming from the domain of that guy so from 0 to Pi where zero and Pi are not included so the way you can evaluate this is to say it's X is equal to the coent of Y where Y is greater than zero and less than Pi so again Z and Pi are not included okay let's look at a quick little problem and I'm going to show you an issue that's going to come up if your argument is negative for the inverse cotangent function so we have y equal the inverse cotangent ofk 3 so first off I know that basically I can say that this is X is equal to the coent of Y where Y is greater than zero and less than Pi well the X here is the negative of the square root of three so basically I'm asking for what angle between Z and Pi is going to have a cotangent value that's NE 3 again if you come back to this little table some people have this for cotangent already but I'm just going to use this one if you look at this value right here for 30° or pi/ 6 radians well the tangent so the tangent of pi/ 6 is going to be theare < TK of 3 over3 so that tells me the coent of pi over 6 is going to be again take the reciprocal of this so 3 over theare < TK of 3 now you could do this in different ways you could rationalize the denominator or a quicker way would be to say that 3 is the < TK of 3 * the > 3 so you can do it that way if you want so this would cancel and you get the square < TK of three so pi/ 6 would be my reference angle in quadrant 2 so in quadrant 2 what angle has a reference angle of pi/ 6 well it's going to be 5 pi/ 6 so this right here would just be equal to 5 pi over 6 again it's got to be within this interval for you to give that answer another way you could have done this is by by using the inverse tangent function so let's say you started with the inverse cotangent of your negative square root and let me make that a little bit better of three can you say this is equal to the inverse tangent so the inverse tangent of the reciprocal of this so the negative of 1 over the < TK of 3 if you rationalize the denominator that is the square < TK of 3 over three well the answer to that question is no you need to issue a little bit of a correction because this right here so this right here is going to give you something in Quadrant 4 rotating clockwise so this guy right here would give you this guy right here so this would beunk over 6 so a pi over 6 reference angle in Quadrant 4 and you've got to rotate clockwise but the answer is actually going to be right here at 5 pi/ 6 because with the inverse cotangent function I'm working from 0 to Pi so I'm working right here so what you would want to do is make yourself a little correction and add Pi so I'm going to add Pi so I can swing around right there and that's going to give me the correct answer so I'm going to add pi to issue a correction here and so this would be piun over 6 plus pi let's write that as 6 pi over 6 and so that would be equal to the 5 pi/ 6 that is the correct answer so if you're using your calculator you need to be aware of this if your argument is positive you're going to be in quadrant one so you're okay there you don't have to issue a correction because the inverse tangent of something that's positive is going to give you something in quadrant one as well but if this guy is negative you're getting something in Quadrant 4 for the inverse tangent function and you're getting something in quadrant 2 for the inverse cotangent function so you have to be very very careful and issue a correction there now if you do the inverse cotangent of zero you're just going to end up getting pi/ 2 as an answer so that's one you just need to memorize all right so the next graph is going to be for the inverse secant function so y equals your inverse secant of X so if you look at the Domain so the domain so this one let's just do it in interval notation so from negative Infinity to negative 1 so let's go from negative Infinity up to an including ne1 so again this is included here you have a point there this would be -1 comma pi and then basically the union with positive one is included and out to positive Infinity so basically the values between and not including Nega 1 and postive 1 those are excluded from the domain so then for the range let me write that down here so for the range you can see that this guy is going to be from zero to Pi where basically this Pi / 2 let me write this in so this is not confusing so this is pi over 2 here that's not going to be included let me write up here that this is going to be Pi so the range you're going to have let me move this up a little bit so it fits basically from zero and including zero going up to and excluding Pi / 2 and then the union with you're going to have again Pi over2 is excluded going up to and including Pi okay let's take a look at a little sample problem we're going to convert this over and solve it very quickly so here we have y equals the inverse secant of < TK of 2 so essentially what I'm going to do is actually say that this is equal to the inverse cosine of I'm just going to take the reciprocal of this so 1 over theare < TK of 2 so if you rationalize the denominator there this becomes the < TK of 2 / 2 and so this we know is going to be pi over 4 so for these problems you don't have to issue a correction basically you're just going to key it in like this so if you have the inverse secant of something you basically are going to take the reciprocal of that value and have the inverse cosine of that and that's how you're going to get your answer all right let's take a look at the inverse cosecant function so again with the inverse cotangent function the inverse secant function and the inverse cosecant function it might look different in your book or your online resource so just telling you that in advance so this guy right here for the domain pretty easy to see basically it's going to be from negative Infinity up to an including -1 so this right here this is a point that's included so this would be -1 comma piun / 2 and then basically the Union with this point right here this is POS 1 comma piun / 2 so basically you're going to take from positive 1 out to positive Infinity so that's your domain and then in terms of your range so in terms of your range well that's going to be from Pi / 2 to Pi / 2 where you're going to exclude zero so basically let's say we have from pk/ 2 and then we're going to go up to an exclude zero and then the union with again we're excluding zero and we're going up up to piun / 2 so piun over 2 and piun /2 are included but zero here that's going to be excluded okay let's look at a problem so here we have y equal the inverse cosecant of a 2 * 3 over 3 again all you have to do for this problem is say this is equal to I'm going to use my inverse sign function so my inverse sign function and I'm just going to take the reciprocal of the argument here so it's going to be the negative of 3 over 2times the Square t of three so let me go ahead and write this as the < TK of 3 * the 3 and I'm going to cancel this with this and so this becomes the negative ofare of 3 over2 so the negative of the sare < TK of 3 over2 and this one you might benefit a little bit from the unit circle so again with the inverse sign function you're working in quadrants 1 and four if you're in Quadrant 4 you've got to rotate clockwise so where is my S value 3 over2 it's going to be right there at 5 piun over 3 but again you've got to rotate clockwise so the reference angle there is p piun over 3 so you'd say this is PK over 3 so that's what you want for your answer let me just move this down here this would be equal to piun over 3 so these types of problems are really easy if you already understand the inverse sign function all right now let's move on and look at some typical examples for this section so what if you see something like the inverse sign of the tangent of piun over 4 well the idea for this type of problem is to evaluate this part first so what's inside or what's going to be the argument for the inverse sign function and so you would say what is the tangent of piun over 4 a lot of you know that's -1 but again just in case you don't you're going to look at your unit circle and you're going to say well where is netive pi over 4 well again I'm looking in Quadrant 4 and I want a pi over 4 reference angle so that's going to be right here at 7 pi over 4 but what I'm doing is I'm rotating clockwise so I want this guy right here again rotating clockwise let me do this in a different color here so this shows up really well so this right here is the angle that I'm looking Fork over 4 now of course this is co-terminal with 7 piun over 4 so if I want the tangent ofi over 4 I can just say that's equal to the tangent of this 7 pi over 4 and I can just grab that from the unit circle again here you want the S value which is2 over2 / the cosine value which is < TK of 2/2 so I'm dividing opposites so I'm going to get -1 as a result so coming back up this would be the inverse sign of this right here is -1 and again a lot of you already know this is going to be Pi / 2 but again because we just started with this section let me get rid of this you're thinking about where your s value is -1 so that's going to be right here but again that's not in the interval of the inverse sign function so I can't give this 3 pi/ 2 two as a result so what I want to do is think about rotating clockwise so clockwise like this and basically that's going to give me an angle that is piun / 2 and of course that's co-terminal with 3 piun over 2 so the S value is going to be the same it's going to be-1 so coming back up I want to give an answer here that's going to beunk / 2 all right let's look at another common type of problem for this section so you might see something like the tangent of the inverse sign of 34 4 so what can you do for this type of problem well of course you could pull out your calculator and you can key in the tangent of the inverse sign of 3/4s and it will give you an approximation but in most cases you're asked to find the exact value so for that what I recommend is to make a little substitution for this part right here so the inside right here which is the argument for the tangent function you're going to let something like let's say Theta be equal to this guy right here so the inverse sign so the inverse sign of 34s and so since Theta is equal to this I can say that now I'm trying to figure out what is the tangent of theta so I'm just going to say for right now this equals what so let me slide down and let's think about do we have enough information to figure out what the tangent of theta is well of course we do first off we know that if Theta is equal to the inverse sign of 34s then we can say so I'm going to put then we can say that the S of this angle Theta is equal to my S value is 34s again if the S of theta is 3/4s well then the inverse sign of 34s is going to get me back to Theta now we also know that if sign is positive I'm going to be in quadrants one or two but specifically here because I'm working with this inverse sign function and the argument is positive here I know I'm going to be in quadrant one this is always going to be true unless you have the inverse sign of positive 1 which is Pi / 2 that's a quadrantal angle but that's the only exception so for this particular case I know that Theta is in quadrant 1 and I went over my border a little bit so let me put the Border like this and so now I can go back to what I talked about earlier on in the course when we said that okay if the S of theta is equal to 34s well that three is going to be your opposite or your Y and that four is going to be your hypotenuse or your R so if I want to figure out the tangent of theta remember this is going to be your y or your opposite over your X which is your adjacent so I already know that Y is three so I can just plug that in right now and say this is three and I just need to figure out what is X so you're going to use your Pythagorean theorem you don't have to do this but I went ahead and made a little sketch so remember for the Pythagorean theorem you have x^2 + y^2 is equal to R 2 so in this case we know that the y or the opposite side is three so I'm just going to plug that in there we know that the r or the hypotenuse is four so I'm going to plug that in there we're going to find that the x or the adjacent side is going to be square of 7 right now I don't know that I'm going to solve for that so here I would have x^ S Plus 3^2 is 9 and this equals 4 S that's 16 so what I'm going to do is subtract 9 away from each side of the equation and so this right here is going to cancel you'll have x^2 is equal to 16 - 9 is 7 now by the square root property you would say that X is equal to plus or minus theare < TK of 7 now I'm in quadrant 1 where X values are positive so I don't need the plus or minus here x is going to be the Principal < TK of 7 so now I know that the adjacent side here or the unknown X is going to be sare < TK of 7 and just coming back up I'm going to put that this right here is the squ < TK of 7 and let me move this over here I can just answer that this right here is 3 over the > of 7 so 3 over theare < TK of 7 and of course you would want to rationalize the denominator so times the < TK of 7 over the > of 7 so my final answer here for the tangent of the inverse sign of 34s is going to be 3 * the < TK of 7 over 7 all right now let's look at another type of problem so this one is just a little bit more tedious than the previous problem so we have the sign of the inverse sign of 12 plus the inverse tangent of -3 so for sure you're going to get a like this in your textbook and what you want to do is start off by asking can I replace anything in my problem so in other words I know that the inverse sign of 1/2 is pi/ 6 let me just come down to the unit circle again this is new for us I know a lot of you already know that but again I'm working from piun / 2 to piun / 2 and I am asking the question what is the inverse sign so what is the inverse sign of 12 so I'm looking for a sign value of 1/2 so that's going to occur right here and so what I want for the associated angle is Pi / 6 so the S of Pi 6 is 12 so the inverse sign of 1/2 is going to be Pi / 6 again it's got to be in this interval from piun / 2 to piun / 2 so this right here is going to be piun / 6 so coming back up here I'm just going to replace this with pi/ 6 so this is the S of your pi over 6 Plus now this one I'm going to make a substitution for it let me just write it in for right now so the inverse tangent of NE 3 okay so just like in the previous problem we have something here where we're going to make a little substitution so you have the inverse tangent of -3 let me slide down here real quick and I'm just going to put that we are going to let something like beta be equal to the inverse tangent of-3 so what I'm going to do is replace this right here with beta so I'm going to say that now I'm looking for the S of < / 6 X Plus beta remember when we think about this beta here if we let beta be equal to the inverse tangent of -3 that tells me that the tangent of beta is equal to -3 now where is beta remember the inverse tangent function is going to give you something in Quadrant 4 rotating clockwise if the argument is negative so that tells me let me put a little border here that beta is in Quadrant 4 Okay so we're going to use that in a little while so the first thing is I'm going to grab this and copy it and I'm going to go to another page and let me paste this in here when you have the sign of a plus b in this case it's Pi 6 Plus beta but the sign of a plus b is equal to the S of a * the cosine of B plus the cosine of a * the S of B this is your sum identity for sign that we talked about earlier in the course so we're going to say that this s of you have this pi/ 6 Plus beta this is equal to the sign of the first guy which is I / 6 and then times the cosine of the second guy which is beta and then plus the cosine of the first guy which is pi/ 6 and then times the S of the last guy which is beta so the S of beta so this equals what in terms of the S of Pi / 6 we know that's a half and the cosine of pi 6 that's 32 most of you don't need this at this point but again if you don't have a unit circle here's Pi / 6 so the S value is 12 and the cosine value isun 3/ 2 so let me come back up here so the S ofun / 6 is 12 and then times your cosine of beta and then plus the cosine of pi/ 6 is the < TK of 3 / 2 and then times your s of beta so what I need to do is I need to figure out what is the cosine of beta so the cosine of beta is again that's your adjacent over your hypotenuse or we say the X over the r and then what is the S of beta so the S of beta is equal to your your opposite over your hypotenuse or your y over your R well let's go back up and let's remember that we know that the tangent of beta is equal to ne3 you can go ahead and say that this is -3 over 1 so this right here is the Y value it's going to be -3 because again we're in Quadrant 4 so the Y values are going to be negative and the X values are going to be positive and this right here will be your x value so that's positive one so let's come down here and I'm just going to write in that the tangent of AA is equal to -3 / 1 again this is your Y and this is your X so we can come through here and say that X is 1 so I know this is a one and then Y is -3 so I know this is -3 now how do we get R again go back to the x^2 + y^2 = r 2 and again you can make a sketch it's not necessary but you can do it so let's put x^2 + y^2 = r 2 we're going to find out that r or this hyp huse here is of 10 but to get that X or your adjacent is going to be POS 1 so this is 1 2 plus your y or your opposite is going to be -3 remember you're in quadrant four so the Y values are negative so this is -3 make sure to wrap that and this is going to be squared and this equals your R being squared so that's going to be for your hypotenuse that guy being squared okay so this is pretty straightforward 1^ 2 is 1 and then plus -3 again this is wrapped being squ that's POS 9 and this equals R 2 so you get R 2 is equal to 10 Again by the square root property you get R is equal to plus or minus the < TK of 10 now your R is always positive so this is just the principal sare OT of 10 so your X is 1 your Y is -3 and then your R is going to be square < of 10 so coming back up if I know that R isquare of 10 I'm just going to replace that here so this isquare of 10 and then this right here is square < TK of 10 okay I'm going to get rid of this we don't need this anymore and I'm going to R rationalize the denominator in each case so this is time thek 10 over the > of 10 so this equals theun 10/ 10 and then here time the of 10 over thek of 10 so this equals -3 * the of 10 over 10 okay so I just need to replace here and here so what I'm going to have let me put equals you have 1 12 times the cosine of beta is theare < TK of 10/ 10 and then plus you have the RO 3 / 2 * the S of beta that's 3 * thek 10/ 10 so this equals I would just multiply and say this is the < TK of 10 over 20 just multiplying 1 * the of 10 that's of 10 and 2 * 10 is 20 so then you have plus you have the 3 * -3 * of 10 so let's go ahead and write that as a negative here and I'll put 3 * the < TK of 30 of 3 * of 10 is 30 and you really can't simplify that any further because 30 is 6 * 5 6 is 2 * 3 so you really can't do anything more with that so then this is over 2 * 10 that's going to be 20 so what I'm going to do is write this with a common denominator so I'm going to say this is the < TK of 10 - 3 * the < TK of 30 over the common denominator of 20 all right so let's come back up here to the top and let me just paste this in so I'm going to say that the sign of the inverse s of 12 plus the inverse tangent of -3 is going to be equal to this < of 10 - 3 * of 30 over 20 in this lesson we want to talk about solving trigonometric equations using linear methods all right so let's start off with the easiest possible scenario we're going to solve some trigonometric equations again with these linear methods let's say you had something like 2 + 4 * the S of theta is equal to 4 so what we're trying to do here we're trying to find values of theta okay that I could plug in here and make the equation true now before I go through this let's just look at something from basic algebra so we'll remember this if I had something like 2 + 4x is equal to four how would I solve this well I'm trying to find values for X that make the equation true right so I would first try to isolate X so I would subtract two away from each side of the equation and so this is going to cancel I would say 4 X is equal to 2 divide both sides by four and I get that X is equal to 1/2 okay so I plug a 1/2 in there and basically you have what you have two plus basically four * a half is 2 so 2 plus 2 gives me four so that is the correct solution with this guy it's not as straightforward it's not as easy because remember you're plugging in an angle measure there sign of that guy would give you a number okay so you're looking for sign of some value I can already tell you that this needs to be a half for this to be true so where is s OFA equal to a half okay that's what you're going to be looking for but to solve this fully let me subtract two away from each side of the equation this cancels we get four * the S of theta is equal to 2 we divide both sides by four and basically we're going to get that the S of theta again is equal to a half okay so from this guy right here this is where you need to basically do a little bit of detective work okay there are so many ways to solve this it's not even funny you might want to use a calculator and use your inverse sign function you might want to use the unit circle there's a lot of different things you could do you could use the special triangles that you remember all kinds of ways so let me talk about two ways that you can solve this first let's think about the unit circle approach so if you are lucky and you get an easier problem where basically you can find this from the unit circle you can go to the unit circle if I'm looking for S of theta is equal to a half where you can look for the y-coordinate and find out where it's a half okay so for 30° or Pi 6 in terms of radians the y-coordinate is a half right so that tells me that s of 30° is equal to a half then also I know that sign is positive in quadrant 2 so remember when we think about reference angles if the reference angle is 30° then sign of that guy is also going to be a half right so where would I have a reference angle of 30° well with 150 degrees right because 180 degrees minus 150 degrees the reference angle will be 30° so s of 150° s of 150° and I'm messing up my right in there so s of 150° would also be a half right you can see that here this is a half this is 150° or 5 pi over 6 if you're talking about radians now when you talk about Solutions over the interval from if you think about 0 to 360° your solutions would be 30° and 150° but when you think about a general solution remember when you you think about s and cosine the period is 360° or 2 pi okay so that means if I started here and I rotated around 360° well instead of 30° now I'd have 390° and s s of 390° would also be a half okay if I added 360° again then s of in that case 750 degrees would also be a half you know so on and so forth same thing with 150° if I had 360° I'm going to be at 510° sign of that guy would be a half as well okay so let's take that information and come back here and give a solution okay so first Let's Pretend We're restricting the domain so 0 to 360° or again you might see this in terms of radians so you might see 0 to 2 pi in terms of radians whatever you get you want to match that in terms of your answer okay so if you get degrees put degrees if you get radians put radians okay so in this case I'll just do both so in this particular case I'll say that Theta is equal to we had 30° and then also 150° and again you know the reference angle for both of those going to be 30° and then if we look at in terms of radians we could say Theta is equal to pi/ 6 this is 30° in terms of radians and then 5 pi over 6 and this is 150° in terms of radians now this is the solution for the restricted domain if you want a general solution so let me put the general solution okay so this is going to keep working as we either rotate clockwise or counterclockwise so remember if you go 360° doesn't matter which direction if you go negative Direction so you're going clockwise or you go the positive direction so you're going counterclockwise you're going to keep getting these guys okay so I'm going to start with let me make some set braces here okay this is typically how you'll see it I'm going to take this solution you could do it degrees or you can do it radians I'll do both so 30° plus the 360° okay when you go around again you're going to get that same value and then times n or n is just any integer okay and then I'm going to do 150 degrees and then plus this 360 Dees let me slide all of this down because it's not going to fit so let me move this down and let me put this comma back where it should be right there okay so this I'll close my set braces and I forgot my n so let me put the N there okay so that's your general solution in terms of degrees with radians you would say pi over 6 plus remember 360 degrees is just 2 pi okay so I'm going to put 2 piun N and then I'll do 5 piun / 6 + 2 piun N okay so there's your general solution there's your solution for having this domain restriction okay where basically you're between Z and 2 pi or if you think about this in terms of degrees 0 and 360 degrees now let's talk about another way that you could have gotten this answer in a lot of cases especially harder cases you're not going to be able to use the unit circle okay so what you want to do is use your inverse sign function so let me use a fresh sheet here let's say I looked at my calculator and I did inverse sign of in this particular case we were looking at a half okay so you punch this into your calculator and we already know you're only going to get one value right you're just going to get 30° okay but again if I think about s of theta is equal to 12 there's an infinite number of angle measures there that's going to be true for this and if I have one of them then I can find all of them again because where is sign positive it's positive in quadrants one and two okay so I have the answer from Quadrant One I have my answer from Quadrant 1 it's going to be 30° and again it repeats so 360° in either direction so if I go plus 360° or if I go minus 360° I'm going to again run into another solution but in quadrant 2 I just need an angle where the reference angle is 30° okay so again that's going to be found by taking 180° so I would take 180° okay minus 30° and that gives me 150° okay so this is the angle I'm looking for so in quadrant 2 I get 150 degre so that's another way that you could solve this you could also use special triangles if you wanted to okay so there's a lot of different ways to think about this guy okay let's look at another example so we have the cosine of theta now instead of the S of theta so we have -4 - 4 * cosine of thet is equal to -3 - 2 * the cosine of theta and again if you were working with this let's say you had something like -4 - 4x is equal to -3 - 2x well I'd want to get all the terms with X on one side all the numbers on the other same thing here I'm going to get all the terms with cosine of theta in involved on one side all the numbers to the other so let me add four to both sides okay it let me add 2 cosine Theta to both sides okay and basically we see that over here this is going to cancel over here this is going to cancel so I have -4 + 2 which is -2 times the cosine of theta is equal to-3 + 4 which is 1 okay to isolate this guy I'm going to divide both sides by -2 and I'm basically going to find that the cosine of theta is equal to -2 again another easy problem where you can get this from the unit circle but let's go ahead and use our inverse cosine function so let's go cosign inverse of -2 and let's do a little bit of detective work if you punch that in on a calculator you're going to get 120 in terms of degrees okay so let me erase this so I can drag this up we'll have a little bit of Rong okay let me bring this up and let's think about this so I have 120 degrees there okay so in terms of the reference angle for that remember if you're in quadrant 2 the reference angle is 180° minus your angle which is 120° so this is going to be 60° so this is my reference angle okay so what I want to do now now again because this is negative here cosine is negative in quadrants 2 and three so I have my quadrant 2 solution my quadrant 2 solution is 120° for my quadrant 3 solution I want to think about an angle in quadrant 3 okay that has a reference angle of 60° how do I find that well in quadrant 3 I would want to take 180° and I would want to add to that 60° which is my reference angle okay which would give me 240° okay this is 240° and the reason that works is again if I had a 240° angle I said hey what's the reference angle you would subtract 180° to get 60° okay so this is my answer in quadrant two this is my answer in quadrant 3 okay now again if you wanted a general solution so let's pretend that first you got a domain restriction from 0 to 360° like this and then also in terms of radians we have 0 to 2 pi okay like this so for the degrees you would just put that Theta is equal to 120° okay and then 240° and then for your general solution again all you have to do in this particular case is let's use some set braces here and we'll say 120° plus 360° * n where n is any integer and then you'd have 240° plus 360° * n or n is any okay if you want to do this in terms of radians we can do that as well no problem so we would say something likea is equal to within this domain 120° is going to be 2 pi over 3 okay and then we also have 240° which is going to be 4 pi over 3 okay for the general solution I just go 2 piun over 3 + 2 piun N and then I go 4 4 piun / 3 + 2 piun N okay so again the period for cosine and S is 360 degrees or 2 pi in terms of radians so that's what you want to add so you have 2 pi and then n is just any integer okay so that's how you want to do that now if you wanted to solve this more quickly especially if an easy problem use the unit circle so where's the x value one2 well again you come over here -2 is 120° or 2 pi over 3 radians okay and then also 140° or 4 pi over 3 radians you got -2 again is the x value okay so this is the quickest way to do it the other way is something you need to understand because in some situations you're not going to be able to use the unit circle they're going to at some point give you harder problems okay let's look at one more problem and this one has tangent involved remember the period for tangent is now going to be Pi okay or 180 degre so you have to keep that in mind here so if I have something like the negative of the of 3 + 4 minus tangent of th is equal to 4 + 8 * the tangent of theta well immediately I can just get rid of the four from each side right same thing on each side just get rid of it and what I want to do now let me go ahead and subtract 8 times the tangent of theta away from each side of this guy so I forgot my 8 so minus 8 times the tangent of theta okay so this will cancel out over here okay and I'm just going to add 3 * of 3 to both sides so what I'm going to end up with is you put a Nega 1 there you'll say you have -9 * the tangent of theta is equal to 3 * 3 divide both sides by9 okay let me scroll down just a little bit and I'll say that I have the tangent of theta is equal to the three is going to cancel with the nine this is a one this is a three so let's say the negative of theun of 3 over three okay so now with Tangent it is a little bit more challenging to think about these you could use your special triangles in this particular case and just think about the of 3 over 3 the tangent of 30° is 3 3 so let me just write this over here the tangent of 30° is equal to the < TK of 3 over 3 so again I would say that tangent is going to be negative where let me draw this out here so we remember this so we remember the all students take calculus so tangent is negative in quadrants four and two okay so let's think about this in quadrant 2 where would I have a 30° reference angle again if I take 180° and I subtract off 30° that gives me 150° okay my reference angle here for 150° angle would be 30° so my value here if I did tangent of 150° because it's negative I would get the negative < TK of 3 over3 okay so let me erase this for a second let me just put one answer here so Theta here would be 150° or again if you want to write this in terms of radians this is 5 pi over 6 okay whatever you want to do so let me erase this for a second let's think about in Quadrant 4 so remember I said the period when you work with Tangent is 180° so I can just add 180° to this okay so 150° plus 180° is going to be 330° okay also what you could have done and this is 11 pi/ 6 in terms of radi also what you could have done is again look for a 30° reference angle in Quadrant 4 so in Quadrant 4 I would just go 360° minus 30° is going to give me 330° so this is the angle I'm looking for all right so obviously these would be your Solutions if you had a domain restriction okay but if you didn't have a domain restriction let me just write this over here so this is 0 to 360° and this one is 0 to 2 pi okay in terms of radians if you didn't have a domain restriction you have to be careful because a lot of students will come through here they're used to working with s and cosine so they go 150° plus 360° * n okay this is actually wrong okay the period for tangent is 180° so you want to put 180° like this now do you need to list 330° plus 180° * n no you do not and the reason for this is if you go back to this one right here let's say n is one well you're basically adding 150° to 180° you get to 330° so you don't need to put this guy in here it's just Overkill okay just put it like this and be done with it okay then the other one over here for radians you're just going to put 5 pi over 6 plus you're going to do PI okay which is 180 degrees in terms of radians times n okay that's all you need to do because again if you added basically one you can get a common denominator say 6 over 6 so if you had 6 piun over 6 plus 5 5 pi/ 6 you would get 11 pi over 6 okay so again it's Overkill to list both of them in that case you just want to list it like this as simply as you possibly can in this lesson we want to talk about solving trigonometric equations by factoring all right so now we're just going to look at the next scenario we already talked about how to solve these basic trigonometric equations using linear methods now we're just going to be factoring so we have three * cosine theta plus 2 * cosine s Theta equal 1 so a very easy problem you want to get this if you see something with a squared term in it you want to think about factoring right away okay that's usually how you're going to solve that and you want to think about getting it in the form of ax^2 + bx+ c equal 0 okay remember this is a quadratic and you're able to solve this by factoring and then setting each factor equal to zero right that's the zero Factor property so I'm going to do the same thing here I'm going to move this to the left basically so I'm going to say I have 2 * cosine 2 thet and then plus this 3 3 * cosine of thet and I can add one to both sides of the equation so say plus one and this equals z so I if I can Factor the left side again I can use that zero Factor property there is a tremendous roadblock to factoring this when you first start I don't know why it is but when I work with students and I'm just going to do it on the first one I tell them just make a little substitution so in this case you're working with cosine of theta so just let something like U be equal to the cosine of theta okay so everywhere you see cosine of theta replace it with so here I have cosine squ Theta which is basically cosine of theta squ so I would say I have 2 u^ 2 plus in this case 3 U and then + 1 equals 0 so from here could I factor this and solve it using factoring yes of course I could so this guy is a prime number and this guy is one so this is really easy to factor this is going to be 2 U and this is going to be U we know that off the top and we know that because this is positive one and all the signs are positive the only possibility is that this this would be one and this would be one so basically you would check this and say okay well 2 U * U is 2 u^2 the outer would be 2 U the inner would be U so those combined together would give me the 3 U and the last would be 1 * 1 which is one so we're good to go so you could have factored this the same way using the same principles with the cosine of theta in there but again I use U there just to avoid the little roadblock of first factoring a trigonometric expression like this so what I want to do now is because U is cosine of theta I just want to plug back in so I'm going to say 2 * everywhere there's a u I'm plugging in a cosine of theta then plus 1 and then times you have your cosine of theta plus one and this equals z so now that we've got this guy factored let's set each factor equal to zero let's see if we can get a solution so I'm going to start with this one I'm going to say 2 * cosine of theta + 1 = 0 Let's subtract one away from each side of the equation we're going to get 2 * cosine of th is equal to1 divide both sides by two and I'm going to get that the cosine of theta is equal to -2 so you're going to see the same numbers pop up over and over again you probably at this point know that the values for this in terms of degrees would be 120° and 240° but I'll go to the unit circle in a moment and show you so let me just erase this kind of make this a little bit more neat and I'll just say that cosine of theta is equal to negative one2 let me just slide this over just a little bit and I'll delete this from here and I'm going to go to the unit circle all at once so let's do this side now so we have cosine of theta basically you're just subtracting one away from each side so let's let's do the full thing so plus 1 equals 0 subtract one away from each side you get cosine of theta is equal to 1 you probably know already this happens at 180° or Pi if you're thinking about radians so let's move this up and say we have cine of theta is equal to 1 so let's go to the unit circle real quick and get all all of our Solutions again you can also do this using your calculator you can do it with special triangles there's so many ways to do it but for the first two they're very easy I'm going to give you a final problem where you have to use your calculator there's no other choice okay so we'll see that so let's go down to the unit circle so again if we're looking for cosine of theta is equal to -2 and then cosine of theta is equal to1 where you're looking for an x value of - 1/2 so that's going to happen here again at 120 Dees or 2 pi over 3 radians and then also remember this guy would have a reference angle of 60° so you would want 240° because that also has a reference angle of 60° so this is your negative - 1/2 there again 240° or 4 pi over 3 radians so let's write 120° and then I'm going to write 240° out here I'll put a little space here because I'm going to put this 180° here for the negative 1 right so negative 1 is your x value there so let's put 180° there and let's just put the radians in there as well generally you're going to give your answer based on the Restriction you get right if they tell you the intervals between 0 and 360° you put degrees if they tell you 0 to 2 pi you put your radians so let's go ahead and put we have 2 pi over 3 we have pi and then we have 4 pi over 3 so let's go ahead and copy this we're going to need that and we're back here so let me just go ahead and paste this in bring this over here so this is going to be our solution again if you restrict the domain so for this one let's say that we restrict the domain from 0 to de to 360° okay so basically in that interval for this one this is going to be if you restrict it from 0 to 2 pi okay in terms of Radiance So based on what you get that's how you're going to get your answer if you're asked to get a general solution remember these guys are going to repeat right so with the S or cosine if you're working with one of those the period okay is going to be 360° or 2 pi in terms of radians so just take this if you wanted a general solution so let's put General solution here here you would take these guys so I'm going to say 120° plus 360° * n or n is just any integer you're just going to take all of them so then 180° plus 360° * n and then lastly 240° plus 360° * n okay let me close the set Braes then for this one remember 360° is like 2 pi in terms of radians so what I want to do here is say 2 pi over 3 and then plus we'll have 2 pi * n okay so n is any integer same thing so Pi + 2 pi n and then lastly 4 piun over 3 + 2 piun N so we have our general solution here and then we have our solution where basically we're in an interval you can say you're restricting the domain all right for the next one let's look at four equals s of beta + 2 * sin 2 Beta + 3 again if you have something squared and then something to the first power I'm always thinking factoring right I'm always thinking ax^2 + bx+ Cal 0 so I'm going to write it like that I'm going to first let's go ahead and flip the sides I want everything to be on the left side that I'm working with so let's go ahead and say we have 2 * s^ s beta plus let's say s of beta I'm going to subtract four away from each side of the equation so this would end up being basically negative one over here because if I subtract four away from each side this would be -1 and this equals 0 so now what I have here I should be able to factor so again if you're struggling with this if you just can't get past the mental road block go ahead and take s of beta and set that equal to a variable like you but I think we can get past it right so I'm just going to put sine of beta here and S of beta here to start now this two in front that's a prime number so I've got to stick that in one of these so I'm just going to stick it here so we have 2 * s of beta * s of beta which will give me 2 * sin beta okay so that works now we have alternating signs here let's think about this we have a -1 as a final term so it has to be 1 * 1 but we have to think about the signs there is it going to be plus here and minus here or vice versa well the outer here would be 2 * s of beta but it would be negative and then the inner would be plus the S of beta so that's not going to work because the middle term's positive okay so that means I need a plus here and a minus here so let's erasee this and now you see it works out right because the outer would be 2 * s of beta and then the inner would be minus s of beta and so if I do the subtraction I get my plus sign of beta and then- 1 * POS 1 is NE 1 so we have correctly factored this guy same thing so 2 * s of beta minus one setting that equal to zero let's go ahead and solve it add one to each side of the equation you get 2 * s of beta is equal to one and I'm going to divide each part by two here and basically I get that the S of beta is equal to a half so let's erase this and you see these values so often it's almost like you start dreaming about them we know that this is going to be 30° or 150° again if we're in that interval from Zer to 360° where you could say pi over 6 or 5 pi over 6 in terms of radians so let me write this over here so s of beta equals 12 so right now I know beta is equal to let me just do this in solution set notation we know we have 30° and we have 150° okay so so far that's what we got let me put this in terms of radians also so let's go pi over 6 and then 5 pi over 6 so what's going to be the result from over here let me draw a little line we have the S of beta + 1 equal 0 again we're just going to subtract one away from each side so the S of beta is equal to1 and again if you know your unit circle this is basically going to give us 270° or 3 pi over 2 so let's get rid of this and let's put 270° and then over here let's put 3 pi over two let me just erase this real quick and let me just drag this out of the way so we put this over here so let's put or and then let's come down here and again the more you reference the unit circle the more you work with it the more you memorize it again we were looking for sine of beta to be equal to2 so that's going to be here you're looking for your yvalue so 30° or pi over 6 and then if I go in quadrant 2 where sign is positive again what angle is going to have a reference angle of 30° 150° so it's going to be here okay it's 150° or 5 pi over 6 then if you're looking for a yvalue of ne1 it's going to be at 270° or here or again 3 pi over 2 in terms of radians okay so let's talk about the general solution this might not be obvious but when you look at these sometimes they're going to have this pattern I know when we work with s or cosine they repeat every 360 6° but if you notice this one if you started let's say you went through and you say okay well 30° plus 360° is 390° then 150° + 360° is 510° and then 270° plus 360° is 630° so I want you to notice that everything that I have here is just differing by 120° so if I start here and go here I added 120° I added 120° I added 120° so on and so forth for a general solution in this particular case I can say 30° okay plus I know the period of sign is 360° but because of this I can say 120° * n so 30° plus 120° * n this would be a simpler way to write this then for this guy if you want to put this in terms of radians let's go ahead and just say that you have pi/ 6 plus you're going to do 120 degrees in terms of radians so that's 2 pi over 3 so 2 piun over 3 and then time n okay so let's close that up so this would be our general solution and then this is if we had a domain restriction so let's say from 0 to 360° and then here let's do from 0 to 2 pi all right let's look at one that's much more challenging so here we're not going to be able to use our fancy unit circle I think we can on one part but on the other we can't so we have tangent Square beta plus tangent of beta - 2al 0 so same deal I'm going to be able to factor this and I'm just going to set up it's already it's already in line for me the way I want it so I'm just going to put the tangent of beta and the tangent of beta so I basically need a positive 1 and a -2 so how am I going to get that well basically I want to do positive2 and Nega 1 so this guy is factored pretty easy to do that let's go ahead and start with the easier one so let me start with the tangent of beta - 1al 0 so we'll say that tangent of beta if I add one each side would be equal to one so so where is the tangent of beta equal to 1 you might remember from your special triangles that one solution here would be 45° remember tangent is positive in quadrants one and three remember the all students the all all students take calculus so it's positive in quadrants one and three okay so I need to find the quadrant 3 solution so I want the angle in quadrant 3 with a 45 degree reference angle so I can go ahead and do 180 degrees plus 45 degrees which is going to give me 225 degrees so let me put 225 degrees to finish that up and then I can erase it so let's put beta is equal to 45° or 225° in terms of radians you could write it like this you could put pi over 4 and then you could do 5 pi over 4 so that takes care of this part now over here this is where we're going to struggle a little bit so we have the tangent of beta plus 2 we're gonna set that equal to zero so let's go down here get a lot of room we'll come back up so the tangent of beta plus two is equal to zero let me subtract two away from each side of the equation I get the tangent of beta is equal to -2 so there's nothing you can pull from special triangles or the unit circle where you're going to figure this out okay so what I would do here is go through and basically use your inverse tangent function on your calculator okay go ahead and use this guy okay and I'm going to use pos2 not -2 if you punch this up you're going to get 63 point and I'm going to round this let's say 43 so this is in terms of degrees so now what this is telling me is that the reference angle with wherever I need to go is going to be 6343 de now tangent of beta here is negative where is tangent negative it's negative in quadrants 2 and 4 so I'm looking for a solution in quadrants 2 and in quadrants 4 and what I want is the angle with a reference angle of 63.4 3° so in quadrant 2 I would do 180° minus 63.4 3° which I'll say is 116.5704 I would do 360° minus this 6343 de and that would give me 296 let me write this down so 296 . 57° so let's erase all of this we don't need it anymore and let's just bring this up here and I'm going to combine all of this for one solution so let me get this over here and let me come up here and I'm basically going to say and let me kind of drag this over here so I'm going to do one for degrees and one for radians so beta could be 45 deges 225 deges you could have 11657 degrees and this is an approximation so maybe you want to notate that this is approximately here and then this is an approximation also so 296.55 de and then in terms of radians if you wanted to convert this over again for each of these you would multiply by you have degrees here so you want degrees down here and then you want pi up here so you would basically say that the degrees are going to cancel so let me just show that real quick 116.5704 so let me slide this down and I'll just say it's about 2.34 so let's get rid of this one and let's do this one so let's move this up and basically we're going to multiply by again you want pi over 180° we know that the degrees here are going to cancel and basically I'm doing the same thing on my calculator 296.55 ID 180 then times pi so let's just go ahead and say this is going to be 5176 so 5176 six and again you just want to show some way like hey this is an approximation so I'll just put approximately like this just showing that this is not an exact value now remember if you're working with Tangent the period is 180° or it's Pi radians okay so this would be for in the case of from 0 to 360 in terms of degrees and I made that terribly okay so 360 degrees and then this would be basically from zero let me put my degree symbol there 0 to 2 pi and then for the general solution I'm going to take 45° so let's go ahead and say 45° plus I'm going to do 180° * n and then I have another guy here so I'm going to do 116.5704 way of notating this isn't standard this is my way of notating that hey this is an approximation it's not an exact value so then down here for this guy I would start with pi over 4 and then plus the pi * n and then for this one I would take my 2.34 and then plus my Pi n okay like this now again if you want to just notate that this is an approximation and this is an approximation just to let your teacher fully know that you're aware that this is an approximation and not an exact value all right let's take a look at one more problem so somewhere in your textbook they're going to tell you that you don't want to divide both sides of an equation by a trigonometric function such as the cosine of beta so when you do this you could potentially remove some of your Solutions these are going to be solutions that make the divisor zero so what I'm going to do is solve this in two different ways and I'll show you how this can cause problems so the first way is the correct way and that's with factoring so if you have the cosine of beta time the cotangent of beta and this equals the cosine of beta you could subtract the cosine of beta away from each side of the equation so over here this would cancel and you'll have zero so on the left you would have the cosine of beta then times the coent of beta and then minus this cosine of beta and what I'm going to do because we're going to factor this out let me put times one so nobody gets lost and I'm going to set this equal to zero let me actually get rid of this because we're going to come back up and use that in a moment so let me slide down here just a little bit and I'm just going to factor out the cosine of beta so this right here is going to get pulled out in front of some parentheses so you have the cosine of beta times the quantity inside you would have the coent of beta and then minus if you pull this out you would have your one and then close that down and this equals zero so we're now going to solve this using the zero product property or you can call it the zero Factor property so for that I'm going to take the cosine of beta and set that equal to zero then I'll say or I'm going to take this other Factor so the coent of beta minus one and set that equal to zero well for this one you can add one to both sides so the cotangent of beta would be equal to one and we know that the tangent of beta is equal to 1 over the cotangent of beta so the tangent of beta here would be one over one which is one so let me write this as the tangent of beta equals 1 so the first one is the cosine of beta equals 0 so I'm looking on the unit circle for an x coordinate of 0 so that's going to occur right there at pi/ 2 or 90° and then right there at 3 piun / 2 or 270° now for the tangent of beta equals 1 we've seen this several times already we know that in the first quadrant this is going to be pi/ 4 again if you skip the other problems you can just do the inverse tangent function with one as the argument you'll get pi over 4 or or 45° you could also think about this with the definition of tangent so you could say that the tangent of beta is equal to the S of beta over the cosine of beta and in this case if you replace beta with that angle which is pi over 4 you can go to the unit circle and actually see why we're getting a value of one and that's because we have the same non-zero number divided by itself the S value of square2 over two the cosine value of2 over two so of two over two divid of2 over two will give you one and then the period for tangent is pi so if I add Pi radians or 180° that's going to give me my solution that's in quadrant 3 so that's going to be at 5 pi over 4 and the tangent of 5 pi over 4 again you would have the same nonzero number divided by itself and so you would get one so let me get rid of this and let me highlight this over here as 5 Pi over4 or 225° and I'll get rid of this and this and this so if you wanted to write out your solution you could say that beta is equal to let's start with pi over 4 and then let's go to Pi / 2 and then let's go to 5 pi over 4 and let's go to 3 pi over 2 if you want to work with degrees we could say that this is 45° this is 90° this is 225° and then lastly 270° okay let me grab this real quick and let me come back up here and paste this in here and I'm just going to show you real quick that these Solutions all work just so that when we get to the next part you're not lost in terms of whether we did it correctly or not so the first solution was pi over 4 so let's plug that in there and there and there and let me make that a little bit better there let me slide this out of the way so we can fit this we know that the coent of Pi / 4 is 1 and the cosine of piun / 4 is2 over 2 so this would be the > 2 / 2 * 1 is equal to the2 over 2 so that is a valid solution now similarly if you do 5 piun over 4 the only thing that's going to change is the cosine of 5 piun over 4 that's going to be a negative > 2 over2 so if I did 5 piun over 4 5 piun over 4 and 5 piun over 4 this would be2 over2 * 1 = 2 over2 so that one's going to work as well because you get the negative of the 22 equals the negative of the 22 now the other ones are going to make the cosine of beta equal to 0 so whether you did Pi / 2 or 3 pi over 2 when you plug that in let me just do Pi / 2 you're going to get 0 equals 0 and let me make that a little bit better there so cosine of Pi / 2 is 0 cotangent of Pi / 2 is actually also 0o so you would get 0 is equal to over here cosine of piun / 2 is is zero so this one would work and so with this one it's the same thing so you can see these Solutions work in our interval so from 0er and including zero going out to 2 pi 2 pi is excluded and then for this one you would have from 0 degrees where that's included going out to 360° where that's excluded okay let me get rid of this and let me show you what can happen if you solve this in a different way so let's say instead of factoring I decided I was going to divide both sides of the equation by the cosine of beta so let's divide this side by cosine of beta and let's divide this side by cosine of beta and you're going to see a problem happen so this would cancel with this and you'd have the coent of beta is equal to well this would cancel with this and I have a one so it equals one okay well we know this turns into the tangent of beta equals 1 and we know we get this solution right here of pi over 4 and we get this solution right here which is 5 Pi over4 but what happened to these Solutions right here of Pi / 2 and 3 pi over 2 well those are the solutions that make this divisor zero cosine of piun / 2 is 0 cosine of 3 piun 2 is 0 so that's what's going on you're going to potentially remove Solutions if you use this strategy so what we want to do is stick to factoring when we try to solve this type of problem because you're not going to run into these types of issues so I just want to note real quick for those of you that are interested you can look at this problem on Desmos what you would want to do is go back to the fact that you have the cosine of beta I'm just going to call it the cosine of x time the coent of X is equal to the cosine of x now what you're going to end up doing is graphing this guy on the left and this guy on the right separately so the graph in red is yal the cosine of x * the cotangent of X and the graph in green is yal the cosine of x so if you want to solve it this way you're looking for the points of intersection between those two graphs so you're basically asking the question for which X values are the Y values going to be the same those are going to satisfy the equation so the ones we found we found pi/ 4 we found Pi / 2 we found 5 pi over 4 and then we found 3 pi over 2 so you can continue with that if you want this is something you can look at on Desmos on your own if you want a visual representation of what's going on in this lesson we want to talk about solving trigonometric equations using square roots squaring and identities all right so let's go ahead and start off with 4 * sin^2 beta - 1 = 0 so just like we've been doing before we're going to add one to both sides of the equation we're going to have 4 * sin^2 beta is equal to 1 okay so what I want to do now remember you're always trying to isolate your trigonometric expression so I want s of beta by itself so what I'm going to do here is just divide both sides by four okay and so that's going to give me the sin s beta is equal to to 1/4 okay let me scroll down just a little bit and get some room so now because this guy is squared I want to take the square root of each side remember how this works if I take the square root of the left on the right I want to go plus or minus right to account for the positive or the negative so I'm going to say that the S of beta is equal to plus or minus the square < TK of 1/4 okay if you simplify this you'll say the S of beta is equal to You'll Go plus or minus theare < TK of 1 is one and the Square < t of 4 is two so you can basically say this is 1/2 okay so let me actually erase all of this and let's come up to the top and just paste this back in here so we know this gives us two scenarios right so I want to split this up and say the S of beta is equal to2 or the S of beta is equal to the negative of 1/2 so let me erase this now okay so we have our two guys there so basically what I'd want to do is go to my unit Circle I know at this point a lot of you already have it memorized but some don't so let's go down and let's look for where the S of beta is equal to a half and then also where the S of beta is equal to- one2 so basically you're looking for your y-coordinate to be a half or negative one2 so that's going to happen here okay that's also going to happen as we go around it's going to happen here as we go around it's going to happen here and then as we go around it's also going to happen here okay so notice everywhere where I circled has a 30° reference angle right so basically it's going to be if you want to put this in terms of radians or degrees we'll just do both so 30° then you have 150 Dees you also have your 210° and then you have your 330° okay so that's in terms of degrees and then in terms of radians you would have your pi over 6 okay you would have your five pi over 6 you would have your 7 pi over 6 and then last you would have your 11 Pi okay over 6 so these would be your Solutions if you said between 0 and 360° or also if you said between zero and 2 pi in terms of radians so let's just cut this away I'm just going to cut this come back up here and paste this in okay so these are going to be again our Solutions let me just erase this nonsense here these are going to be our solutions for basically if you restricted this over an interval again you have from 0° to 360° that's this guy right here here and then this guy right here would be from 0 to 2 pi okay so sometimes they ask you this question to solve over a specific interval and then other times they'll say hey what's the general solution so if we want the general solution let's just put this right here really quickly remember when you work with sign the period is going to be 360° but what you notice here is that as you jump here from 30 de to 210° you're increasing by 180° okay so I would start start this by saying I have 30° plus 180° time some integer n and then you'd have 150° so 150° plus this 180° okay times n because again if you look at 150° and 330° those are different by 180° okay and what's causing this if we go back to the unit circle if I Circle this one again you'll notice that these guys are across from each other this is 180° right so basically these are across from each other and the these are across from each other okay so you can basically say that you start with 30° and you add 180° to get your next solution then you add to 180 degrees again gets the next solution so on and so forth if you started with 150° you add 180 degrees to gets your next solution then 180 degrees again so on and so forth okay if you wanted to put this again in terms of radians we're going to use the same concept here I'm going to go pi/ 6 so basically you're 30° then plus 180° in terms of Radian is going to be Pi okay so you're just going to do PI * n and then for 150° this is going to be 5 piun over 6 so 5 pi over 6 and then I'm just going to go plus my Pi * n okay so here's all the solutions that you could possibly want you have your degrees across that interval from 0 degrees to 360° and then you have your radians across that interval from 0 to 2 pi and then you have your general Solutions so there just basically if they didn't ask you to solve over an interval this is You' want to list this for degrees and then for radians all right let's take a look at another example so here's one where we're going to need to use some identities so we have the negative of cosine 2 Beta minus 2 * s of beta is equal to -2 so notice that you have S here and then also cosine here and cosine specifically is squared okay so what you're meant to think about if we go back to our worksheet or our little handout on the identities if you go to the Pythagorean identities remember cosine squ Theta is the same as 1 - sin 2 th okay so you want to look for possibilities there where you can substitute things in so remember the negative here this is a big deal if you have a negative there you're putting a negative put this in parentheses what I'm going to do because the negatives got to get distributed to everything so I'm going to go one minus my sin SAR beta okay again that's in parentheses then minus your 2 * your s of beta is equal to -2 okay distribute the negative to everything so I'll say -1 and then plus s s beta then - 2 * sin of beta is equal to -2 okay so from here we already know what to do we want to end up with s of beta equals some number and then we want to solve right so I'm going to tell you in advance this is factorable if you have something that is quadratic in form if it's not factorable you can use the quadratic formula okay but here we're going to be able to factor so what I'm going to do is I'm going to add two to both sides of the equation and I'm just going to rearrange things I want the square term all the way to the left so I'm just going to say that I have sin^2 beta and then this is gone I'm going to put zero here okay over here I'm going to put - 2 * the S of beta and then if I have negative 1 + 2 that's positive 1 okay so this equals zero and the idea here is that I can Factor the left side okay so I can Factor this again if it's a roadblock for you when you're trying to factor these guys go ahead and take something like you okay and say s of beta equals U and then it's just like if you were trying to factor U ^2 - 2 U + 1 okay if that was equal to zero you could solve that in a breeze right it's just a roadblock because we're now dealing with trigonometry so how could we Factor this guy right here and again if it's if it's troubling you just look at this one well first off if I had sin squ beta and there's nothing out here there's basically a one well I'm going to have the S of beta times the S of beta okay so that's my first term in each case now for this part right here again you're just looking at the -2 and the positive one okay so I need two numbers that are going to multiply together to give me pos1 but sum to -2 well that's going to be -1 And1 okay so let's go ahead and get rid of this and basically we could write this okay as the S of beta minus onean squar okay so this equals zero and the reason I write it like this is you can write it like this or this remember this is basically from special factoring when you think about this it's really only going to be one of these guys that we have to set equal to zero right because you have a duplicate solution so basically all I want to here is just again take one of these and say s of beta minus 1 is equal to 0 I'm going to go ahead and add one to both sides and just say the S of beta is equal to one okay now this is going to be really easy to solve for us where does the s of beta equal 1 well if we're looking for the S of beta equals 1 well again where's the y-coordinate 1 well right here right so at 90° or pi over 2 in terms of radians and that's basically the only place there so your your general solution would just be you rotating around another 360° before you get there okay so let me go back up so let's just put our solution here so again if we're in the interval if they give you degrees you want to answer in degrees so if they say over the interval you have 0 degrees and then to 360 Degrees again if you get this then your answer should be in degrees and you'd want to put 90 degrees here okay if you got an answer or if you got a test question where let's say it was with radians so let's say they said 0 to 2 pi like this okay then now you want to give your answer in radians so you could say beta is equal to Pi / 2 or you could do the solution set notation just ask your teacher what they want okay so I'm just going to do this so pi over 2 okay so that's my solution basically in these intervals now again a general solution is where we keep thinking about when we're just keep rotating around and around and around so basically your general solution if it's in terms of degrees well it would be 90° plus another four rotation is 360° times some integer n okay and then for the radians you just use this Pi / 2 and then plus again one full rotation in terms of radians is 2 pi so you would do 2 pi and then times some integer n okay so this is your general solution here sometimes they ask for that and then these are solutions if you have a specified interval sometimes they ask for that okay so let's look at a very I would say challenging but a very tedious type of problem now this one involves squaring and you'll recall from basic Al albra if you square both sides of an equation you lose information and so you have to basically check your Solutions in the original equation right you sometimes get these extraneous solutions and so they're not going to work as solutions to the original equation okay so we're going to see that here we're going to see that it just takes a long time to go through everything so we have the negative of cosine Theta is equal to the < TK of 3 * the S of theta okay so the first thing I'm going to do is square both sides okay because I want to get rid of this radical and also I have cosine and sign so so I'm thinking Pythagorean identity okay so I'm going to square both sides let me write this as the negative of cosine of theta okay this is going to be squared this equals the sare < TK of 3 * the S of theta this is going to be squared okay so the negative is going to go away negative squar is just going to be a positive right 1 * 1 is positive 1 then cosine of theta squar we could just say this is cosine s Theta like this over here the sare root of 3 being squared is 3 S of theta being squared is just sin s Theta okay so at this point we know that cosine s thet is 1us sin sare thet okay so let's just go ahead and make that substitution I'll say this is 1 - sin^2 thet is equal to 3 * sin^2 thet okay so let's scroll down and get some room going all right I'm just going to subtract 3 * sin^2 Theta from both sides of the equation and this is going to cancel and I'm just going to put zero here so I'm going to have one minus remember you can treat this as a NE 1 so1 - 3 or1 plus3 however you want to think about this this is going to be -4 okay so basically -4 * sin^2 Theta is equal to zero okay so how are we going to solve this I want you to think back again to basic algebra okay this is where your skills really come into play let's say it's something like a 2 minus b^ 2 how would you factor this remember this is the difference of squares so it's a plus b okay that quantity time a minus B that quantity well here you have the difference of squares right I could write this out and say that this is 1 2 minus you could do 2 * the S of theta okay quantity squared like this and this equals zero okay so now it's apparent or it's very clear that this is the difference of squares sometimes you have to rewrite things like that so this would be one + 2 * the S of theta * 1 - 2 * the S of theta okay and we'll set this equal to zero so basically I'm going to take each one of these guys and set it equal to zero so I would have 1 + 2 * the S of theta is equal to 0 or you'd have 1 - 2 * the S OFA is equal to0 let's go ahead and solve they're very easy to solve I subtract away one from each side over here and over here and over here and this is going to cancel and this is going to cancel okay so over here I have two times the S of theta is equal to 1 over here I have let me put the or I have -2 * the S of theta is equal to 1 okay so basically at this point I'm just going to divide this by two okay and this by two and then this by two or this is -2 over here and this by -2 okay so this cancels and I have S of theta is equal to the ne2 then or over here this is going to cancel right you have negative divid by negative so that's positive so you get the S of theta is equal to2 right so you can condense this down and say basically you have plus or minus 12 okay so let's get rid of this and let's just copy this real quick let's go to a fresh sheet so we have a lot of room to work because the problem here is you're going to get all these Solutions okay so if the S of theta is a half or it's negative a half remember we already went through this so let me write this again so s of theta is a half or S of theta is negative a half basically you would have a 30 Dee reference angle in each case right so you would have here and then basically you go around you'd have here and then you go around and you would have right here and then you go around and you would have right here okay so we already know this because we saw a problem with this earlier but basically 30° 150 de 210° and 330° okay so let me write this down so we have our 30° we have our 150° we have our 210° and then we have our 330 de okay and I'll write it in radians when we finish up so let me just keep this in degrees for now let me cut this away and let me just paste this in right here okay let me get rid of all this nonsense now when we Square both sides of an equation the solutions here that we get are proposed Solutions so we have to get our original equation going so we can check it so cine of theta okay is equal to theare < TK of 3 times the S of theta okay so we're just going to use degrees you can use radians it doesn't matter you're going to plug in for Theta okay so basically if I plugged in 30° here and 30° here what would we get well again I could go back and forth between the unit circle but a lot of you already know that the sign of 30° is a half okay so this is going to be half so this would give me the > of 3 over 2 so let's just write that as the > of 3 over two okay over here I'd have the negative of cosine of 30° well the cosine of 30° you know is a positive number right because 30° is in quadrant 1 all of these guys are going to be positive in quadrant one right so there's no way this can be true because this is the negative of a positive number okay so you can go ahead and Mark this out the actual answer here is of 3 over2 right but it's the negative of that so it's the negative of the 32 is equal to the > of 3 over2 so that's false right so that solution does not work okay for the next one me erase this and set this back up so we had the cosine or the negative of the cosine of theta is equal to we had the square otk of 3 times the S of theta so we know that the sign of if I put in 150° here we know the sign of 150° is basically a half right because the S of 30° is a half the S of 150° because s is positive in quadrant 2 150° has a 30° reference angle so this is a half as well right so this is basically going to be the > of 3 over 2 okay over here if you think about the cosine of 150° remember cosine is negative in quadrant 2 Okay so this will save you a little bit of work you'd have the negative of a negative which is positive so you know this is probably going to work right so the cosine of 150° we know that this guy is going to be the negative of the of 3 over2 okay but you have the negative of the negative so be really careful so the negative of the negative of the of 32 so if I apply the negative I basically would have the of 32 is equal to theare of 3 over2 okay so this one checks out so 150 degrees is a valid solution then for 210 deg let's go ahead and set this back up and I probably shouldn't erase this each time but it's the negative of the cosine of theta is equal to the < TK of 3 * the S of theta so 210° if you put this in here okay and you put this in here again we know know that the reference angle is 30° right so what is the sign of 30° well it's going to be 1/2 but we're in quadrant 3 now so we know this is negative so basically this is -2 so this would be the > of 3 * -2 so 3 over 2 and this is negative okay so over here 210 again this many degrees I know that I have a reference angle of 30° so the cosine of 30° is square < TK of 3 over 2 and because this guy's in quadrant 3 cosine is negative so this is the negative of the squ < TK of 3 over2 so this isn't going to work out right because again if I apply the negative this becomes positive so again we have one that doesn't work so this one doesn't work and then for 330 degrees let me just do this one last one so the negative of cosine of let's just go aad and say 330° is equal to you have theare < TK of 3 times your s of 330° again I've got a 30° reference angle and I'm in Quadrant 4 so I know this is netive 12 right so might as well just put 3 over2 okay like this and I'll just put this out here okay so over here again I have a 30° reference angle but cosine is positive okay in Quadrant 4 so I would say squ of 3 over two so squ of 3 over two but it's the negative of that because I've got this negative hanging out so these guys are equal so this one does work out okay so your two valid Solutions here are 150° or 330° okay in terms of degrees and then if you're working with radians it's 5 piun / 6 or then 11 piun over 6 so let's go ahead and write our solution here we'll say from the interval if you're in this interval 0 deges and then 360° we'll say the solution set is basically 150° okay and then 330 degre only okay so just those two guys and then if I'm in this interval from 0 to 2 pi okay if they give it to you like this always answer based on what they give you as the interval okay if they don't give you anything then you can just ask your teacher like hey what do you want this at so basically at this point I want to do my radians so I'll go ahead and say that this is 5 pi over 6 and then 11 pi over 6 let me close that down and then again if you want a general solution you want to realize here that you can just add 180° each time again I know the period for sign is 360° but you have to pay attention to this type of stuff if you're writing a general solution so the general solution here would be let's go ahead and say 150° plus 180° * n okay so you don't want to do 360° there because again if I add 180° I get to 330° then if I add 180° to that I'm going to be at 150° plus 360° which is 510° you know so on and so forth okay then in terms of radians I'm going to go ahead and say that it's 5 pi over 6 Plus 180 degrees is going to be Pi radians and then times n okay so these are your general Solutions and then these are your Solutions if you're given a specific interval to solve over just get with your teacher in terms of what you need to do in this lesson we want to talk about solving trigonometric equations with half angles all right so let's start off with a very easy example so we have the S of beta/ 2 is equal to the < TK of 2 minus the S of beta/ 2 we have a little domain restriction or you could say you're solving over an interval in this case it's from 0 to 2 pi with zero being included okay now I want to make something perfectly clear you are solving for for beta okay so you want to plug something in here for beta so basically here and here and you want the result of this to be true okay you are not in the end solving for beta over two but we need to use beta over two in order to get a solution okay so what we're going to have to do and let me write this down here I'll show you we need to basically change this up just a little bit and it explain why so let me rewrite this real fast I'm going to say that I have the S of beta/ 2 is equal to theare < TK of 2 minus the S of beta / 2 okay so the first thing I'd want to do is get that trigonometric expression by itself and in order to do that I'm just going to add the sign of beta over 2 to both sides of the equation okay so that's how I'm going to get things started so let me scroll down just a little bit here get some room going and I'm just going to show this cancels over here so s of beta over 2 plus s of beta/ 2 is 2 * the S of beta/ 2 and this equals now the square root of two okay if I want to isolate this completely I'm multiplying this by two okay so I can just divide both sides by two and then I'll take care of that so I'll divide this side by two and this side by two and basically this is going to cancel with this okay so that's gone and I've isolated the sign and let me make that a little bit better so the S of beta/ 2 so that's on my left side on the right side I have the square < TK of two over two okay so let's pause for a minute I'm actually going to erase all this and let's think about what we have we have this right here which is beta over two and we have the sign of that guy and we're saying that it's equal to the < of 2 over2 so before we start thinking about how we're going to solve this let's first modify our little domain to work with what we're working with here we're working with beta over two so I'm going to write this as an inequality I'm going to say that zero is less than or equal to I'm going to do this beta guy here which is less than 2 pi okay because that's what we're initially work with again this is for beta so now we have beta over two so we divide everything by two here okay to consider the interval here so this is just going to be zero this will be beta/ 2 and this is just going to be Pi okay so essentially what I'm looking for here is I'm looking for where beta over 2 sign of that guy would be square of two over2 now in this particular case if you think about < TK of two over two in terms of sign if you go to your unit circle you can see in quadrant one you have < of 2 over2 here with pi over 4 in terms of radians or 4 de with degrees and then you would want a 45° reference angle in quadrant 2 okay so in terms of degrees that's going to be 135 degrees or 3 pi over 4 in terms of radians right so right here okay and you can see this right here okay so that's pretty obvious and you can see that that's Falls in line with our interval for beta/ 2 because it's from zero to Pi okay so we're going to take pi over 4 and 3 pi over 4 so I'm going to say that the S of pi over 4 is equal to the < TK of 2 over2 and also that the S of 3 Pi over4 okay is equal to theare < TK of 2 over2 okay so what I want to think about here now that I have everything set up remember you can set these guys equal to each other because I know that the S of Pi over4 is equal to thek of 2 over two and I know that beta over 2 sign of that guy is equal to the of two over two so what I can do here is set up a little equation so that I can solve for beta right so let me do this over here I'm going to say that beta / 2 is going to be equal to pi over 4 because again s of this guy produces this and sign of this guy produces this so I can set these two equal to each other in order to get a solution for beta okay so that's why that's going on okay let's scroll down just a little bit to get some room I'm going to put or here going do the same thing with this so beta over 2 is equal to the 3 pi over 4 okay and all I need to do here it's very simple to get a solution I'm just going to end up multiplying both sides by two okay so this is basically going to be what this is going to cancel I'll have beta is equal 2 the two would cancel with the four and give me a two down here so pi over two and then over here basically I'm multiplying both sides by two and so this is going to cancel and this is going to cancel give me two here so we'll say or beta is equal to my 3 pi over 2 okay so these are our two solutions in terms of radians and they're within our domain or our restrictions so I'm going to do this in solution set notation so I'm just going to put P Pi / 2 and then comma 3 Pi / 2 okay so we're not asked for the general solution here so we don't need to worry about that if you were asked for the general solution remember your period here is different because you're dividing by two so basically the period here is 4 Pi okay not 2 pi okay so you got to remember that all right so one thing I want you to do is pause the video go ahead and plug pi over 2 in for beta in each case and verify that basically these Solutions do work you can just do one of them I'm going to just say trust me that they both work but just pause the video and see that for yourself that if you plug in Pi / 2 right here okay so you have pi over 2 divid by two and right here pi over 2 / two that basically you would get a true statement the left side would be equal to the right side okay let's take a look at another example so this one is slightly more complex not more complex but more tedious because you do need to use an identity and then there's some squaring involved so we have the negative cosine of theta is equal to -2 + 3 * the s of theta over 2 again when you think about these things if you end up with cosine and then s you're going to be thinking about an identity right so if I think about s of theta over two if I go back to my handout on the identities and we come down here let's go all the way down to where we get to the half angle identities and was the last one we talked about so here's the one for sign so you have S of a over2 in our case we're working with Theta so let's say s of theta / 2 is equal to Plus plus or minus your < TK of 1us the cosine of th/ 2 Okay so let's go back and let's just make a little replacement okay so I'm going to put the negative of cosine of theta is equal to I'm going to have -2 and then I'm going to have plus you've got your three here that's multiplying that so keep in mind that basically you have that plus or minus I'm going to put this in front so I'm going to do plus or minus like this and then the three and then times that little guy which was the square Ro T of 1- the cosine OFA and this was over two okay so all I did was I plugged in for S of theta over 2 just using the half angle identity okay it's all I did now what I want to do here remember if you have a radical somewhere you're going to end up having the square okay and in order to do that you first want to isolate this guy okay so what I would do here is I would first add two to both sides okay and let's scroll down just a little bit so I'm just going to say that I have 2 minus the cosine of Thea is equal to this is canceled over here so I have Plus or - 3 * the < TK of 1 - cosine of theta over two now I can square both sides okay so I can get rid of this radical so I'm going to square this side and I'm going to square this on okay when you square the plus or minus you don't have to worry about it because basically you can think about this as negative 1 being squared that's going to be positive one and a positive you square it's still positive okay so you can basically forget about that so over here I'm going to use my special product formula and say it's the first one squared so 2 s is four and then minus it's 2 * the first guy time the second guy so 2 * 2 is four and then times cosine of theta and then plus the last guy squared so this is just cosine squar Theta okay and we'll rearrange this in a moment I know this is not how we usually write this we'll put equals again don't worry about the plus or minus that's gone 3^ squ is going to be nine and then it's multiplied by this guy right here squared I'm just removing the radical okay so basically you have this 1 minus the cosine of theta / 2 now be careful here because 9 is multiplying this whole thing okay if you end up Distributing this to the numerator make sure you use parentheses okay very important what I'm going to do first is rearrange things so I'm going to put this as cosine s Theta okay then Min - 4 * cosine of theta and then basically plus 4 okay so I've got all my signs right I want to make sure on that and then this equals let's do 9 * 1 - cosine of theta like this okay because the 9 is m the one and also the negative cosine OFA then this is over two okay so now what I'm going to do is I'm going to multiply both sides by two so that I can clear this denominator okay so basically this comes here and cancels here and here I'm just going to put a two okay being multiplied by everything there so you'd have 2 * the cosine 2 Theta and then I would basically have minus 8 * the cosine of theta and then plus I would have eight and this equals the two's canceled so just let's go aead and distribute this so nine - 9 * the cosine of theta okay so so far so good let's scroll down just a little bit more okay so what I want to do here is basically subtract N9 away from each side of the equation and also add 9 * the cosine of theta to each side of the equation okay so basically what this is going to do is give me a zero on the right okay you can see where I'm going with this I'm going to end up factoring and using the zero product property so on the left I have my 2 * the cine 2 Theta and then basically I have8 cosine of theta + 9 cosine of theta so this is going to be plus cosine of theta or you could say plus 1 * cosine Theta whatever you want to do there and then basically 8 - 9 is 1 and this equals zero okay so the idea here is that of course if you wanted to you could use the quadratic formula but we have something that is factorable so let me go ahead and cut this away put that in there so I'm just going to factor the left side here again if you have trouble factoring this I see a lot of students having trouble write it as 2 U ^2 + u - 1 = 0 you can factor that no problem right you think about this as 2 U and U and then the last guy here is a negative one right so that tells me I need mixed signs here this guy is going to be positive so what I want to do is put a plus one here and a minus one here right because the outer would be 2 U and the inner would be minus U so that would give me the correct plus u there okay now once you factored it like this you just put your cosine of theta in for you okay so just's come back in here and place this with cosine of theta and your road block is gone okay so 2 * cosine of theta there okay so a little trick for you now I'm going to set each one of these equal to zero and solve let me just get rid of this we don't need this anymore so I'll say 2 * the cosine of theta minus 1 equal 0 or I'll say the cosine of theta + 1 equals 0 let me scroll down and get a little room going here so basically I'm going to add one to both sides and this is going to give me 2 * the cosine of theta is equal to one okay I'm going to now divide both sides by two and I get cine of theta is equal to2 okay over here let me go ahead and put or and I'll put or and then over here I'm just going to say minus one minus one so the cosine of theta is equal to 1 and I guess I don't really need another one here so let's put this over here well I guess I could write it down here so cosine of theta equals 1 okay so you see what we're doing here we basically have to solve this guy and then this guy so where's cosine going to be equal to a half remember cosine is positive in quadrants 1 and four and we know that cosine of 60° okay is a half so basically you would want a 60° reference angle in Quadrant 4 so that would be 300° right so you'd want 60° and then 300° again if you're still looking at your unit circle it's no big deal right this is a half here so you see it's 60° or pi over 3 in terms of radians and then over here you be in Quadrant 4 so you have a half here 300° or 5 Pi over3 in in terms of radians okay so again if you need to use the unit circle keep doing it you will memorize it eventually now while we're down here I might as well note that basically Pi okay cosine of pi is going to give you Nega 1 so let's go back up and let me erase the degrees and let me say that Theta here would be equal to your pi over 3 and then it would be 5 pi over 3 and then it would also be PI right so this covers these solutions from these guys now in order to make the video a little bit shorter a little bit more concise I'm not going to check these Solutions when you square both sides of an equation we know from basic algebra that there's the possibility of extraneous solutions right solutions that won't work in the original equation I have already checked this I know that these guys are going to work I can say that the solution set here is pi over 3 okay and then I'll say pi and then I'll say 5 pi over 3 like this okay so these Solutions do work in the original equation in the next example we'll see where it doesn't Okay so let's go to that now all right let's look at another example here so we have the > of 3 * cosine of th/ 2 is equal to 1 plus the cosine of thet okay again same interval from 0 to 2 pi Z is included again if you see something like this you have cosine of this and cosine of this well you can go ahead and use your identity so we come back here now cosine of in this case would be Theta Theta over 2 is plus or minus the square otk of 1 plus the cosine of theta over two so let's just make a little substitution here I'll say the square Ro TK of three okay and I'll put the plus or minus out in front of here then times my square root of 1 in this case it's plus the cosine of theta remember if it's s then it's minus if it's cosine it's plus then this is over two and this equals your 1 plus your cosine of theta okay so again all I need to do here is just Square both sides so let's Square this side and let's Square this side okay and over here the plus and minus forget about it it's gone square of 3 S is 3 this guy if I Square it I'm basically going to end up with the radicant so this is 1 + the cosine of/ 2 and to speed this up let me just wrap this in parentheses and put this like this and let's put equals over here I'm going to flip these I'm just going to put the cosine of theta plus one so it makes it easier so cosine of theta plus one like this so this would be the cosine squar Theta first guy squar plus 2 * the first guy * the second guy so it's basically just 2 * cosine of theta and then plus the last guy squar so that would be one okay so over here now all we need to do is basically multiply everything by two right so this would cancel let me show this visibly and let's go over here and do this so now I have 3 * 1 which is 3 plus 3 * the cosine of theta this equals 2 * the cosine 2 theta plus you'd have four cosine and I messed that up so cosine of theta and then plus 2 * 1 is 2 so from here because I have a squared term I want to think about either factoring or using the quadratic formula this one's going to be factorable so we'll be safe there basically let's move everything to the left because that's where I like it so let's subtract two so basically this is going to be one let's go ahead and subtract 4 * the cosine of theta so - 4 * the cosine of theta so this is going to be negative cosine of theta and then let subtract 2 * cosine s Theta so I'm running out of room over here so let's just kind of put this over here so minus 2 * cosine s of theta I know it's kind of getting a little messy but let's write it neatly now so we'll put this out in front we'll put -2 * cosine 2 thet and then we'll put minus cine of theta then we'll put plus one and of course everything's gone from over here so this equals zero if you want for factoring a lot of people do not like to factor with a leading coefficient that's negative you can multiply everything by negative 1 so this becomes positive this becomes positive this becomes negative and it has no effect on zero okay so basically if I want to factor this again you can use that trick with u if you want I don't I don't really need to do it so I'm just going to put 2 * cosine of theta and cosine of theta okay because this times this would give me this now the final guy is neg 1 so that can only come from positive 1 * negative 1 we know that this guy right here is positive so that means I want to put plus one here and minus one here right because the outer would be 2 * cosine of theta the inner would be minus cosine of theta so that would combine to give you this okay so let's go ahead and copy this and paste this in another sheet and again I'm just going to use my zero product property I'm going to say two time the cosine of theta - 1 = 0 or the cosine of theta + 1al 0 so I'm going to add one to both sides over here and I'll have 2 * the cosine of theta is equal to 1 divide both sides by two and again we get this cosine of theta is equal to a half so we already know that that's going to occur basically at 60° right or a reference angle of 60° in Quadrant 4 which is 300° we just drag this up here we'll say that Theta here is 5 pi over 3 or actually let's do this first so we'll do pi over 3 or then 5 pi over 3 so again this guy right here is your 60° this guy right here is going to be your 300 degrees okay so then or okay and I'll just put a comma here for this one it's really simple I just go minus one here minus one here so we have the cosine of theta is equal to1 we already know this is going to be Pi Okay so we've seen this before so this is exactly the same solution that we got in the last section or the last example but basically now you're going to see that one of them is not going to work so let's actually erase all of this we know it's a pain to check Solutions but it's absolutely necessary in this case so my original equation is the square < TK of 3 multiplied by the cosine of theta over 2 okay and then basically you have is equal to 1 + the cosine of theta so what do I have here to start let's go Square < TK of three times you would have the cosine of basically we' have pi over 3 / 2 so it's like multiplying by half so it's the cosine of pi/ 6 and the cosine of pi over 6 remember pi over 6 is 30° so that's going to beunk 3 over2 so let me just write this over here for reference so cosine of Pi / 6 is < TK of 3 over two okay so let me erase this and I'm going to put Square < TK 3 over two this equals 1 + the cosine of theta in this case we're working with our pi over three let me write this in so the cosine of pi over 3 and we already know that this is 60° here so this would be 1/2 okay so what is 1 plus a half most of you know this would be 2 over2 here so 2 over2 + 1 over two would be 3 over2 or basically three halves now if I do square of 3 * square of 3 that is three okay so this does give me three halves so the first solution is valid so this one you can put a big fat check okay the next one you're going to see a problem with so let's go ahead and erase this and try again so we have theare < TK of 3 times I'm putting 5 piun over 3 in here remember if I do 5 piun over 3 let me erase this we're not going to need this anymore so if I do 5 pi over 3 and then divid by two it's like having cosine of 5 piun over 6 now 5 pi over 6 is 150° so that has a 30° reference angle and it's in quadrant 2 so it would be the negative of the < TK of 3 over2 okay so this is the negative of the < TK of 3 / 2 okay over here I have 1 + the cosine of 5 piun over 3 and we already know this is going to be a half right so we know this is a half this ends up being three halves and you can see this is not going to work out because of the sign right squ 3 * S 3 is still three but now it's going to be basically negative of three Hales equals three Hales so this is false right this does not work so this is what I mean by checking the solutions if you don't check them there's a chance it's not going to happen every time but there's a chance chance that you report a wrong answer okay something that doesn't work in the original equation now this last one's going to work if I put a pi/ 2 in here remember Pi over2 is 90° and cosine of 90° is 0o right so if I did Square < TK 3 * 0 well that's 0o right so you can basically say this is zero over here and then on this part I have 1 plus the cosine of pi and the cosine of pi or the cosine of 180° is going to be1 so what is 1 +1 that's going to be zero so you get 0 equals 0 so this one checks out as well okay so let's go ahead and erase this and we can finally write our solution set I know this is super tedious but it's just something you have to do so we end up with pi over 3 okay and then we end up with pi these are the only two solutions in the interval were given from 0 degrees or from actually zero in terms of radians out to 2 pi in terms of radians so Zer is included 2 pi is not in this lesson we want to talk about solving trigonometric equations with multiple angles all right let's take a look at the first problem so here we have the cosine of 2 Beta is equal to -10 cine ^ 2 Beta + 8 now we just want to solve this from 0 to 2 pi where 0 is included and 2 pi is excluded now when you look at this if I'm trying to get cosine of beta equals some number here you have cosine of 2 Beta here you have cosine squ beta so this one right here is going to be the immediate problem for us because you want to write this in terms of the cosine of beta again equals some number in the end so what you would want to do is use a little identity to plug in for this guy and coming down here this is the one I'm going to work with if you have cosine of 2 Theta or two beta whatever you're working with you have three different formulas you can choose from I'm going to work with this one so what I'm going to do is say the cosine of 2 Beta is equal to 2 cosine 2 beta minus one so I'm just matching this formula here I'm just taking the beta in the place of theta and basically now you have cosine of 2 Beta And cosine of 2 Beta so those match and what you can do is take this guy right here and you could just plug it in for this guy right here and that's going to give me cosine squ beta over here and cosine squ beta over here so that's what I want so let's write this as two cosine squar beta minus one again just taking this and plugging it in for this so that's how I get this left side now so this equals on the right nothing's going to change so this is just -10 cosine SAR beta and then Plus 8 okay at this point I want to get cosine of beta equals some number if you have a squared guy involved then you should be thinking about factoring or in some cases the quadratic formula but for this particular problem you could actually solve it using the square root property and it's going to be a lot faster you could solve this with factoring if you want to but it's going to take a little bit more time so what I'm going to do is isolate cosine squ beta on one side and then have a number on the other side so I can use the square root property so I'm going to add 10 cine squ beta to both sides of the equation just like that and so right here this is going to cancel and then the other thing I'm going to do is add one to both sides of the equation so this is going to cancel so on the left you have 2 cosine 2 Beta then plus 10 cosine s beta that is 12 cine s beta and this equals 8 + 1 is going to be 9 now what you would want to do since 12 is multiplying the cosine squ beta there you would divide both sides of the equation by 12 so you want cosine squ beta by itself to use the square root property okay so this right here is going to cancel and you get cosine squar beta is equal to for 9 over 12 again each is divisible by three so 9 divid 3 is going to be three and 12 divided by 3 is going to be 4 so this is cosine 2 Beta is equal to 34s and now we can use our square root property again if we have something like let's say x^2 = 34s then by the square root property we can say x is equal to plus or minus the square < TK of 3/4s so that's theare < TK of 3 over the < TK of 4 which is 2 so X would be equal to plus orus the of 3 over2 well here it's the same thing I'm just going to get that cosine of beta is equal to plus or minus the Square < t of 34s and again this becomes the square root here of three over the square < TK of 4 and the square root of three really can't do anything with that but the square Ro of four is going to be two so the cosine of beta is equal to plus or minus the > 3 / 2 okay let's come to the units Circle and let's actually write this as the cosine of beta is equal to the 3 / 2 and then I'm going to write or the cosine of beta is equal to the negative of the of 3 over 2 so for beta here we're going to look in quadrants 1 and 4 where cosine is positive and I'm looking for an x coordinate of 3 over2 so right here at pi/ 6 and then coming down here it's going to be at 11 piun / 6 so notice it's pi over 6 here and then a pi over 6 reference angle here at 11 > 6 then if I'm looking for cosine of beta to be equal to 32 well I'm looking in quadrants 2 and 3 where cosine is negative so you can just think about a pi over 6 reference angle or again you could just look for that x coordinate of3 over2 so that's going to be right here at 5 piun over 6 again pi over 6 reference angle and then right here at 7 piun / 6 again Pi / 6 reference angle so because we're working in the interval where beta is greater than or equal to Z and less than 2 pi we can just list these four Solutions and be done with it so we're going to say that beta is equal to we have pi/ 6 we have 5 pi over 6 and then we have 7 pi over 6 and then lastly we have 11 pi over 6 all right let's take a look at the next problem so here we have 4 S of beta cosine of beta equal the of 3 we want to solve this over the interval from 0 to 2 pi where 0 is included and 2 pi is excluded so the first thing is when you look at this equation it might not be immediately obvious what you can do but if we look at our identities again you have something like this double angle identity the S of 2 theta equals 2 sin of theta cosine of theta which you can use to solve this equation so I'm going to write this as the S of 2 Beta is equal to 2 sin of beta cosine of beta come back up here and paste this in now this might not be immediately obvious but we're going to take this right here and match match it with this right here now right now you have this four right here and you have this two right here so you've got to rewrite this to where it's going to match so this is just straight multiplication so for four I could really write that as 2 * 2 so I could say that this is 2 * and then it would be 2 sin of beta cosine of beta and this equals the square < TK of 3 so this 2 * 2 sin of beta cosine of beta givs me back to 4 S of beta cosine of beta so I haven't done anything illegal if you look at this guy right here now it matches this guy right here perfectly two s of beta cosine of beta 2 s of beta cosine of beta so that means I can legally take this right here and I can plug it in for this right here so what that's going to do for me is I'm going to have 2 s of 2 Beta equal the of 3 so I'll be able to get it to S of 2 Beta equals some number and I'll be able to get a solution so let's go down and make a little substitution here here so I'm going to have my two don't forget about that and then what is it multiplying again for this I'm plugging in this so I'm going to have my S of 2 Beta And this equals my Square < t of three all right so just like I said we want to get S of 2 Beta equals some number right now you have two that's multiplying the sign of 2 Beta so let's divide both sides of the equation by two so this is going to cancel and we're going to get the S of 2 Beta is equal to the < TK of 3 / 2 all right there are quite a few ways you could solve this what I'm going to do is stick to the method that the book uses because we're solving over a specific interval we're not giving the general solution so I'm going to start with the interval we were given so we have beta is greater than or equal to zero and less than 2 pi well right now I have two beta so I'm going to multiply each part of this inequality here by two so 0 * 2 is still 0 and this is less than or equal to Beta * 2 is 2 Beta so now this matches this and then this is less than 2 piun * 2 is 4 piun so right now I'm going to work in this interval because again I have the S of 2 Beta not the sign of beta so that's why I'm adjusting this interval so let me move this up here and we're going to think about where are the S values 3 over 2 in this interval right here so let's start with the standard interval so just going from 0 to 2 pi Z included 2 pi excluded and that's going to give you a pi over3 solution so there's your 3 over2 for the yv value and then here's your 3 over2 for the y-v value as well so that's at 2 pi over 3 but again to get all of the solutions in this interval what you have to do is take those Solutions and you have to add 2 pi to them in order to get all the Solutions in this interval so I'm going to take and list pi over 3 and then let me list 2 pi over three and then what I would have to do is again take pi over 3 so pi over 3 and I've got to swing around by another 2 pi so let me go plus 2 pi so that's 6 pi over 3 if I get a common denominator let me make that three a little bit better so that's 7 pi over 3 so let's put that up there so that's 7 pi over 3 so again starting here just swinging around by another 2 pi you've got to do the same thing here so you're going to swing around by another 2 pi so let's put this as 2 pi over 3 plus again 6 pi over 3 so 6 pi over 3 is again 2 pi written with a common denominator so this would be 8 pi over 3 so 8 pi over 3 so that's basically all you have to do for this part the only thing is you have not solved for beta you've solved for two beta so in other words you could say that two beta is equal to this right here let me just drag this down and I don't know if that's going to fit and let me just grab this real quick and paste this in and go to another page so basically if you want to solve for beta you would divide everything by two or you could multiply everything by a half so in other words you could say that beta is equal to you'd have pi over 3 * 12 so that would be pi over 6 so piun / 6 and then you'd have 2 piun over 3 * 12 so the twos would cancel there so that would be pi over 3 so pi over 3 and then you'd have 7 piun over 3 * 12 that would give you 7 piun / 6 and then lastly you'd have 8 < / 3 * 12 so here 8 / 2 would give you 4 so this would end up being 4 piun so 4 pi over 3 so these are your Solutions in terms of beta so if you plug those back into the original equation they will work and they are in our interval where beta is greater than or equal to Zer and less than 2 pi so beta is piun / 6 piun over 3 7 piun over 6 and then 4 piun over 3 all right let's take a look at another problem so we have 9 is equal to4 plus the cosine of beta + 7K 4 We're solving this over the interval from 0 to 2 piun where 0 is included and 2 pi is excluded so I'm going to get cosine of beta + 7 piun 4 by itself so that'll be on one side there'll be a single number on the other side so that's the first goal so let me write this as -4 plus cosine of beta + 7 piun / 4 and then this would be equal to you have this netive 9 now if I want cosine of beta + 7 piun 4 on one side by itself I've got to get rid of this -4 here that I'm adding to this so I'm going to just add four to both sides of the equation this will cancel and on the left I have what I want so this is the cosine of this beta + 7 piun over 4 and this equals over here I would get a common denominator so let's write four as 8 over2 I'm just going to multiply by 2 over2 to do that so again this would be 8 over2 so 8 over 2 and 9 + 8 that's -1 so this would be over the common denominator of two so this is equal to12 now this is the part where people tend to get confused essentially what you're going to do is adjust your interval you don't have beta so if you were solving cosine of beta equals -2 well then I could go through and I could say well where is the x coordinate -2 where you're going to have right here at 2 pi over 3 and then you come over here and that's going to be at 4 pi over 3 we don't have that though we have this beta + 7 piun 4 we don't have beta well I can adjust that interval we had beta is greater than or equal to 0 and less than 2 pi well instead of beta I have beta + 7 pi over 4 okay well I'm just going to add 7 pi over 4 to each part so 0 + 7 piun over 4 is 7 piun over 4 this is less than or equal to Beta + 7 piun over 4 is beta + 7 piun over 4 so this matches this and this is less than 2 piun + 7 piun 4 so if we do 7 piun over 4 plus let's go 8 piun over 4 just writing 2 pi with a common denominator this would be 15 Pi so 15 pi over 4 so 15 pi over 4 so if we had cosine of beta = -2 again I would have this 2 piun over 3 and this 4 pi over 3 but in this case we have the cosine of beta + 7 piun over 4 so when I adjust this interval to this one right here let me get rid of this one right here so this is Crystal Clear what I'm going to do is look for where the cosine value is - one2 in this interval so you would start at 7 pi over 4 so in other words you've already swung around by 7 pi over 4 and then you want to swing around by another 2 pi to end up at 15 Pi over4 even though that's not included so you're going to catch this guy right here and this guy right here again but it's not going to be 2 pi over 3 or 4 pi over 3 it's going to be you adding 2 pi to those angles so what you would do is you would say that beta + 7 piun over 4 would be equal to you're going to take this 2 pi over 3 so 2 pi over 3 but now you've got to add 2 pi to that so that's going to be me adding 6 piun over 3 so that would give me 8 piun over 3 so you can get rid of these middle steps here and just say this is equal to 8 pi over 3 we'll solve this for beta in a moment let's just do the other one so then again you have this beta + 7 piun over 4 is equal to again you're going to take this 4 pi over 3 so this 4 pi over 3 and then you're going to add 2 pi so you're going to say this is plus 6 piun over 3 so this gives you 10 pi over 3 let me move this down here so this is more clear now I know the first time you do this this is super confusing but just think about the fact that again you started here at 7 pi over 4 that's where the interval is beginning so if I'm starting here here by the time I swing around this is not 2 pi over 3 this is 2 pi over 3 + 2 pi so that gives me that 8 piun over 3 and then as I continue this is not 4 pi over 3 this is going to be 4 piun over 3 + 2 pi so that's that 10 piun over 3 so you've got to adjust things because we're not solving cosine of betaal -2 we are solving cosine of beta plus 7 piun over 4al one2 so everything got adjusted so let's go ahead and grab this and I'm just going to come down here and we can solve these pretty easily so let's go or there I'm going to subtract 7 pi over 4 away from each side of the equation over here so this is going to cancel you're going to have beta is equal to 8 pi over 3 and then minus 7 piun over 4 you need a common denominator so let's multiply by 4 over 4 let's multiply this by 3 over 3 so this would be 32 piun over 12 so 32 piun over 12 21 Pi so 21 piun over 12 so this would give me 11 Pi so 11 pi over 12 so that's one solution let me get rid of this and just write this as beta is equal to 11 pi over 12 and let me solve this one now so I'm going to subtract 7 pi over 4 away from each side of the equation so that cancels you get beta is equal to let's go 10 piun over 3 I'm going to just multiply this by 4 now to get a common denominator and then - 7 piun / 4 * 33 so we will say that beta is equal to 10 piun * 4 is going to be 40 piun and this is over 12 and then minus 7 piun * 3 is 21 piun so 21 pi and this is over 4 * 3 is 12 so 40 Pi - 21 Pi is 19 Pi so this is 19 piun over 12 let me just put a comma here and put 19 pi over 12 so those are your Solutions in that given inter interval again where beta is greater than or equal to Z and less than 2 pi okay let's take a look at another problem so here we have the cosine of 6X is equal to sin of 4X plus the cosine of 2x so here we're going to solve over this interval from 0 to Pi where zero is included and Pi is excluded when you solve a problem like this that involves these sum to product identities in most cases you're not going to solve that over an interval because there's just too many solutions you will give the general solution let me start this problem off as saying we have the cosine of 6X I'm going to add the S of 4X to both sides and I'm going to subtract the cosine of 2x away from each side so let me subtract away the cosine of 2x first so I have this and then I'm going to add sine of 4X and then we're going to say this equals zero okay let's grab this real quick and let's come down here so I have our little identities set up for us so these are the again sum to product identities so we have this cosine of a minus cosine of B here you have cosine of 6X so 6X is your a minus cosine of 2x so 2x is your B let me actually slide this down and I'll write this out first and then we'll plug in so if you had the cosine of 6X minus the cosine of 2x so I'm just going to do this part right here this is equal to -2 and then sine of you have a plus b over2 so this plus this so 6X 6X + 2x/ 2 and then you have your s of a - b over 2 so 6x - 2x so 6x - 2x again this is over 2 6X + 2x is 8x so this right here would be 8X over 2 so this 8 would cancel with this two and give me a four so this would be 4X right here so let's write that like that I'm just going to wrap this for clarity and then here 6x x - 2x is 4X and so this would be 4X over 2 which is 2x so let's write the S of 2x now since this right here matches this right here we can take this guy right here and plug it in for this right here now why are we doing that notice you'd have a sign and a sign and then you have a sign and more specifically you have this 4X here and this 4X here so we're going to be able to actually this with factoring so let's come down a little bit and let's plug in so this goes in here so -2 and then s of you have your 4X and then sine of you have your 2X and then plus your sine of 4X and this equals zero so you have your s of 4X and your s of 4X so we are going to factor that out so let's factor that out so out in front of some parentheses you would have the S of 4 4 x and then inside you would have -2 sin of 2x so -2 sin of 2x and then plus if this gets pulled out you're going to have a one and then this equals zero now at this point you can use your zero products property so you can say that the S of 4X is equal to Zer or we're going to do this one so -2 sin of 2x + 1 = 0 in this particular case you have sign of something it doesn't matter what it is in this case it's 4X equals z or some number this is okay for this when you want to clean it up you would want s of 2x equals some number so let's go ahead and subtract one away from each side this cancels so actually let me just erase it so we can save some space here and then I would divide both sides by -2 so this would cancel so you're going to get the S of 2x is equal to 1/2 so positive2 now here's where you need to make make a decision again in most cases you're going to solve for the general solution and when I solve for the general solution I like to use a substitution technique so you can do it this way if you want if you're solving over an interval you can do it the normal way that we've been doing it by adjusting our interval so instead of s of 4X equal 0 for this particular one I'm going to let U be equal to 4X now this has nothing to do with this so actually let me cut this away and put this on another page so nobody's lost because people will get confused so now I can solve this let me get rid of this or by saying that I have the S of U is equal to Z so if I look at the S of U equals 0 using my unit circle let me write this in so the S of U equals 0 so if you look at this guy is zero so that's at zero and then this guy is zero that's at Pi so you could write that u in terms of the general solution is equal to 0 + 2 pi n and then you could say Pi + 2 pi n or n is any integer because things are going to repeat every 2 pi in terms of U now what you could actually do is write this as 2 pi n and then Pi + 2 pi n but you would notice that actually you can combine those Solutions and make it a little bit more concise because things are happening every Pi so in other words I could just write this as Pi n so I could write this as Pi n because if I plugged in a zero for n I'd have zero if I plugged in a one I'd have Pi if I plugged in a two I'd have 2 pi a three I'd have 3 pi so on and so forth so this is the general solution in terms of U so let's come back up and let's say that U is equal to Pi n now we let U be equal to 4X so all I have to do is plug in a 4X for U so 4X is equal to Pi n and then just solve for x how do I do that I divide both sides by four so this would cancel and you're going to get that X is equal to pi over 4 * n so let me get rid of this and bring this over here so X is equal to < / 4 * n now we will get our solution in the interval in a moment let me get rid of this and let me come down here and solve this so I'm going to use you again but it's only in terms of this it has nothing to do with what we did before and actually so nobody gets lost let me just pick something else let me let W so nobody is confused be equal to 2x so now I want to solve the S of w is equal to2 come back to the units Circle and let's think about this again s of wal2 well we know how to solve that we know that Solutions are going to repeat every 2 pi so where is the y coordinate 1/2 so right here at pi/ 6 and then that's going to be right here at 5 pi over 6 so we can say that W is equal to pi/ 6 Plus again these Solutions repeat every 2 pi again this is in terms of w so we're going to say 2 pi n and then we're going to have 5 piun / 6 + 2 piun n again things are happening every 2 pi in terms of w okay so in terms of w so let's grab this all right so let's come back here let's get rid of this and now W is 2x so all you have to do is go back and say okay well 2x is equal to we have pi over 6 and then plus you have 2 piun N the other one is 2x is equal to 5 pi/ 6 + 2 piun N let me get rid of this and actually I probably should slide these over a little bit because I have to do some multiplication here let me make that border a little bit better and let me slide this down just a little bit and slide this up just a little bit now we have 2X not X so if you want X by itself you can divide everything by two but because we have a fraction involved it's better to multiply by a half it's just a little bit easier to do it that way let me get rid of this border here so multiply by a half and then multiply by half so we know that this would cancel and just give me X is equal to < 6 * a half is > / 12 and then plus 1/2 * 2 piun N the twos would cancel you just have Pi n so get rid of this so that's one guy there let me move this over and then this guy I'm going to do the same thing so let me move this up I'm going to multiply this by 1/2 multiply this by 1/2 and multiply this by 1/2 so we're going to see that the twos are going to cancel you're going to have X is equal to if you have 5 piun over 6 * a half well we would have 5 pi i/ 12 and then plus again the twos would cancel so you're going to get Pi n so let me get rid of this and actually this color really doesn't show up very well so let me change this and say that X is equal to pk/ 12+ < n and then we have 5 piun / 12 + Pi n okay let me bring this back here and now let's put our solution together so right now I'm giving you the general solution solution and then I'll give you the solution over an interval so for the general solution you're going to take this one so we're going to say that we have < over 4 * n then if we think about this one we have < / 12 + piun n and then this one is 5 piun / 12 + piun n that's our general solution now if we want to go with the solution where X is greater than or equal to Z and less than Pi well again you're going to start here and plug in a zero for n so that would give you let's go X is equal to 0 and if I plugged in a one I'd have pi over 4 if I plugged in a two I'd have 2 pi over 4 which is pi over 2 if I plugged in a three I'd have 3 pi over 4 if I plugged in a four I'd have 4 pi over 4 which is pi and that's not in this interval so I'm going to stop there then these you would have pi over 12 if you plugged in a zero there and then if you add Pi you're not going to be in this interval so then then this one I would have 5 piun over 12 so 5 piun over 12 again if I add Pi I'm not going to be in this interval so these are your Solutions in this interval so 0 < over 4 piun over 2 3 piun over 4 < over 12 and then 5 piun over 12 okay so I want to think about this graphically with you if you're struggling with finding the general solution this can be extremely helpful although it's extremely timeconsuming to think about this with a graph so we have y = the cosine of 6X minus the cosine of 2x plus the S of 4X so this was my equation it's just that instead of Y we had zero over here so if I plugged in a zero for y so in other words if I had 0 is equal to the cosine of 6X minus the cosine of 2x plus the sin of 4X all I'm doing is I'm saying for which X values am I going to get a yvalue of zero well where is y0 when I'm thinking about this on the coordinate plane well that's on the x- axis right so that's where Y is zero so anywhere where I cross the x- Axis or have an X intercept I'm going to have a solution for this equation so thinking about this where X is greater than or equal to Z and less than Pi I get this graph right here now in a moment we're going to talk about the fact that the period for this guy is going to be Pi we discussed this in the last practice test maybe didn't watch the video for that but I'll explain where this is coming from in a moment so that means that I can go from zero to Pi and I'm going to include Zer and exclude pi and any solutions I find will actually repeat in that next cycle so I can just take that solution and add Pi but you're going to see that you can condense some of those solutions that make it a little bit simpler and let's just start with what we have so notice that you have this zero here so that matches this one I'm going to skip this one over for a moment so we have this pi over 4 so that's there I'm going to skip this one so this is pi over 2 so that's there and then I'm going to go to this one which is 3 pi over 4 now if you think about those Solutions you have zero and then you have pi over 4 and then you have pi over 2 and then you have 3 3 pi over 4 again you could go through and say plus pi n and then plus pi n and then plus pi n and then plus pi n that is valid but what's happening is if you add pi to this let me just show if I added Pi this would get me to Pi if I added pi to this I would be adding 4 pi over 4 so this would get me to 5 pi over 4 if I added pi to this that would get me+ 2 piun / 2 so that would be 3 piun / 2 and then if I added pi to this I would add 4 pi over 4 so that would get me to 7 pi over 4 well if you go through here and you just look at these first guys here so it would be a jump of pi over 4 each time so 0 plus pi over 4 is pi over 4 P pi over 4 plus pi 4 is pi over 2 pi over 2 plus pi over 4 is 3 piun over 4 3 piun over 4 + piun over 4 is piun Pi + piun over 4 is 5 piun over 4 5 piun over 4 + piun 4 is 3 piun 2 3 piun 2 plus p piun over 4 is 7 piun over 4 and so on and so forth so that's why we we got that < over 4 * n with the general solution so we had this piun over 4 * n then the other ones you really can't do anything with them other than to say that you have pi/ 12 so that's this one right here and then just plus pi in so it's just going to repeat in the next cycle so this is Pi / 12 + Pi n let me slide this down and then the other one we have 5 pi over 12 so that's this here so again that's just going to repeat in the next cycle so 5 piun over 12 plus piun and all right let's come to the graph now these Solutions are extremely hard to read so you probably need to follow me on Desmos or something like that in order to fully see everything but I'm just going to first talk about the period for this guy I'm going to let you visually see that and explain where it's coming from so in the practice test for the last lesson we talked about the fact that if you had something like y = the cosine of 6X minus the cosine of 2x plus the sign of 4X if you wanted the period for this then basically you would think about the least common multiple of the periods so how do you get the least common multiple of the periods so this right here you think about it as let's just say you had y equal the cosine of 6X what's the period Well it's going to be 2 pi over 6 right 2 pi over 6 which is pi over 3 so pi over 3 again that's coming from that formula if you have y is equal to the cosine of BX well it's 2 pi which is the standard period for cosine / B so here it's 2 Pi / 6 so that gives me pi over 3 it's the same thing if you run into to S right so s and cosine is the same thing so let's do this one so this is the cosine of 2x so that's 2 piun / 2 so that gives me just pi and then for this one it's the S of 4X so that's 2 pi over 4 so that's pi/ 2 so we have not discussed how to find the least common multiple of fractions so in other words let's say I said what is the least common multiple of 1/3 1 and 1/2 well some of you know that it's going to be one but others don't now there's a formula you can use for it but it's a little bit complicated actually if you ask me I think it's a little bit faster to just list things so if you had something like pi over 3 all you're doing is you're multiplying it by 1 then two then three then four you're generating some multiples so times 1 it would just be pi over 3 and then time 2 it would be 2 pi over 3 and then time 3 it would be 3 pi over 3 which of course is just pi and then we could just do one more so * 4 it would be 4 pi over three let me put a border for this one if it's Pi well that's real easy multiply by one you get Pi then 2 pi Pi then 3 Pi let's stop at 4 Pi then for the other one you have Pi / 2 so let's go * 1 it's pi over 2 * 2 it's 2 Pi / 2 which of course is going to just be pi and then * 3 it's 3 Pi / 2 * 4 it's going to be 2 pi and you could stop there if you look at the least common multiple here well looking through what do I have that's common to everything you have Pi you have pi and you have Pi so in other words if you had 1/3 if you had one and you had 1/2 the LCM would be 1 but in this case we have pi over 3 which can be thought of as 1/3 * pi you have Pi which can be thought of as 1 pi and then pi/ 2 which you could think of as 1/2 * pi so the least common multiple there is pi so that tells me that the period for this graph here is going to be Pi let me get rid of this and let me actually trace this out so you can see it if you want to follow me type in X is equal to Pi like this in Desmos it's going to fix it for you and give you a pi like that you're going to get a vertical line and you can go xal 2 pi and then x = 3 Pi so on and so forth you're going to get these vertical lines and it's going to really allow you to see the period of this guy so I'm just going to trace from 0o to Pi and then we'll see that this is going to repeat from PI to 2 pi so this right here this is going to be one cycle for this graph so this is one cycle so from Zer to Pi and then again if I went from PI to 2 pi the same thing is going to repeat so basically I come up come down come back back up come down come back up so this is way up here this is exactly matching what happened in the last cycle so you can see that these two match in the period is going to be Pi so this tells me that things that happened from Z to Pi Z included Pi excluded would happen again from PI to 2 Pi Pi included 2 pi excluded so that's how we ended up on the last page where I did this where you could see the solutions more clearly and looks like I erased my 5 pi over 12 but as I was saying it just shows you that solutions that happen in this cycle will repeat in the next cycle so you can take your 0 plus pi n you could take your Pi / 12 plus pi n you could take your pi over 4 plus pi n so on and so forth It's just that the Zer pi over 4 pi over 2 3 pi over 4 those guys can be condensed into this form right here so the pi over 4 * n where n is any integer and then for your pi over 12 well you really can't do any better than just saying plus pi n so you have your pi over 12 you add Pi that gets you to add 13 PI 12 I know you can't see it it but it's right there and then 5 pi2 again if you add Pi you're going to be at that 17 Pi / 12 again I know it's hard to see but just follow it on Desmos if you want to look at it so this is our general solution again if you want to play around with the graph on Desmos you can visually see this and it'll help you to understand the concept of the general solution and how things are repeating in this lesson we want to talk about solving trigonometric equations with inverse trigonometric functions all right before we start looking at any equations I want to review some Concepts that are going to be important for today's lesson so these are going to come from our lesson on inverse trigonometric functions so I'm going to let F be a one: one function what does it mean to have a one: one function again if you go back to the definition of a function you have a relation where for each X there's one y so you would look at the graph of the relation and it needs to pass the vertical line test so no vertical line would impact the graph in more than one location and if it passes that you have a function so for each X there's one y now when you think about about a one to one function you start with a function and you apply another rule where you say now also for each y there's one X so for that you would check it with the horizontal line test so no horizontal line would impact the graph of a one: one function in more than one location if it does that's not a one: one function if it doesn't then you're good to go and you have a one: one function and if you have a one: one function you can generate an inverse so let's say f is your one:1 function the way you would notate your inverse you would say it was F inverse like this so again this is not an exponent of-1 this is the inverse of my function f the way you generate your inverse is you swap your x's and your y's so the domain or the set of allowable X values for f becomes the range or the set of allowable yv values for f inverse and then the range for f becomes the domain for f inverse so in other words if I look at a point on F and let's say it's 3 comma 2 when I go to the inverse these guys are going to be swapped so the X will become the Y and the Y will become the X so this would be 2 comma 3 so your x value of 3 becomes your y value of 3 and that could be a lot better there and then your y value of two will become your x value of two in the inverse so following this thought process here where the domain becomes the range and the range becomes the domain what happens is when you look at your function and you compose it with its inverse you're going to have this little cancellation that happens or you can say these guys are going to undo each other so F of f inverse of x is equal to X for every X in the domain of f inverse and then F inverse of f ofx is equal to X for every X in the domain of f so this is something we're going to use today when we're solving our trigonometric equations so as a practical example just using what we have here let's say you had F of f inverse of let's say 2 so using this right here it should be equal to two so I can just give an answer of two right away but if you wanted to prove this you can go about it the long way so you could say what is f inverse of 2 well I come up here and I say okay here's an x value of two if I plug that in I know that my output or my y-value is going to be three so what I want to do is say that this is now F of three okay well what is f of three well again if I look at my x value of three the associated y value is going to be two now so this is going to be two so again F of f inverse of two is going to give me two at the end of the day I can shortcut this and not go through this inside part now when I look at the other one in that we have F inverse of let's say F of 3 well again I already know that this is going to give me three if I go about this the long way again if I start on the inside of this guy right here well what is f of three F of three is two so this would be F inverse of two and F inverse of 2 is going to be three so these guys are just undoing each other again F of f inverse of X is equal to X as long as X is in the domain of f inverse and then F inverse of f ofx is equal to X again as long as X is in the domain of f all right let's continue reviewing our lesson on inverse trigonometric functions so if we look at the graph of yal the S of X we can see the domain is from negative Infinity to positive infinity and the range is from negative - 1 to positive 1 with both being included we can also see that it's not a one toone function it would fail the horizontal line test so if we want to find the inverse sign function we can't start with just yal the S of X we have to do something called imposing a domain restriction and we're going to come up with this what we call restricted sign function so y equal the S of X where X is going to be greater than or equal Tok / 2 and less than or equal to Pi / 2 so I have this listed as the domain from NE piun over 2 to pi over 2 with both being included so when you restrict the domain you end up with a one: one function with the same range so the range is still the same from negative 1 to positive one with both being included when you start with a restricted sign function and you swap the x's and the Y's you're going to generate your inverse sign function so y equals the inverse s of X so you could say y equal the ark sign of X and now the domain which is from Nega 1 to positive 1 both being included is coming from the range of your restricted sign function and then the range from negative piun / 2 to Pi / 2 again both being included is coming from your domain of the restricted sign function now the reason it's important to know all that when you start using the cancellation properties with s you need to understand when you're composing sign with its inverse when are you going to be able to cancel things so in other words we like to cancel this with this and say okay well the result of this is just going to be equal to X but this is only true if x is in the domain of the inverse sign function and the domain of the inverse sign function is from negative 1 to positive 1 with both being included so the sign of the inverse sign of X is equal to X for X being greater than or equal to 1 and less than or equal to postive 1 if you pull out your calculator and you try to do s of the inverse sign of two that is not going to give you two as a result it's going to give you an error because two is not in the domain of the inverse sign function all right the next one is a little bit more challenging to understand you can explain it with two different approaches so if you have something like the inverse sign of the S of X is equal to X this is only valid for X being greater than or equal to piun over2 and less than or equal to Pi / 2 so this is actually pretty important to understand the first way to think about it is let's say you look at the sign of X that is not the restricted sign function so we know that for this guy the domain is all real numbers and plugging something in from Infinity to positive Infinity is going to give you a result that's going to be from negative 1 to positive 1 so I'm just going to mark that out and say that what's going to be plugged in for the inverse sign function is going to be from negative 1 to positive 1 well you know that that's the domain for the inverse sign function and I know that the range is going to be from negative pi/ 2 to pi over 2 so basically I'm always going to get something out of this guy right here that's from negative piun over 2 to pi over 2 but what's going to happen is if you don't plug in something for X in this interval right here you can't just cancel you'd have to evaluate the S of X first and then take the inverse sign of that if you think about it from the other approach you could say that these guys are only going to undo each other if you're working with the domain from the restricted sign function which is where X is greater than or equal to piun 2 and less than or equal to pi2 if you're within that domain well then the inverse sign of the S of X is going to give you X these guys are going to undo each other okay using a similar thought process again when we start with yal the cosine of x it's not a one:1 function so we restrict the main to where X is greater than or equal to Z and less than or equal to Pi you're going to have the same range so from Nega 1 to positive 1 both being included so again if you start with that restricted cosine function and you swap the x's and the Y's you're going to get your inverse cosine function so y equals inverse cosine of x or y equal Arc cosine of x if you want to say it that way your domain is from negative 1 to positive 1 both being included coming from the range of the restricted cosine function and then the range is from Z to Pi both being included that's coming from the domain of the restricted cosine function so here's something like the cosine of the inverse cosine of x is equal to X for X being greater than or equal to 1 and less than or equal to POS 1 again you've got to be within the domain of the inverse cosine function for this to work and then for this one the inverse cosine of the cosine of x is going to be equal to X for X being greater than or equal to zero and less than or equal to Pi so again you've got to be within the domain of the restricted cosine function for this to work again the other way to think about this you could say that the cosine of x the domain for that is all real numbers so I can plug in whatever I want but the issue is if it's not in this interval here I've got to evaluate that first before I take the inverse cosine of it because the inverse cosine is only going to give me something from Z to Pi with both guys being included so these guys will undo each other as long as X is in the domain of the restricted cosine function so again being greater than or equal to zero and less than or equal to Pi and for the last one we look at the tangent of X so y equal the tangent of X again not a one:1 function you restricted the domain to where X is strictly greater than piun / 2 and strictly less than piun / 2 so NE piun / 2 and P Pi / 2 are not included and then when you look at the range it's going to be all real numbers so when you swap the x's and the Y's you get your inverse tangent function so y equals the inverse tangent of x or y equals the arc tangent of X so the domain is all real numbers again coming from the range of the restricted tangent function and then the range is going to be from piun / 2 to Pi / 2 again these guys are excluded here remember that the tangent of Pi / 2 or the tangent of negative Pi / 2 those are undefined because the cosine of Pi / 2 is zero and you'd have division by 0 which is undefined so when we think about the range here this is coming from the domain of the restricted tangent function all right so this first one's really easy to understand so the tangent of the inverse tangent of X is equal to X for any X that is a real number so that's what this means this means X is an element of the set of real numbers and basically if you think about the domain of the inverse tangent function again it's all real numbers so that's why we can plug in whatever we want here and these guys will undo each other when we think about this one again this is the one that's more tricky to understand so the inverse tangent of the T of X this is equal to X for X being greater thank / 2 and less thank / 2 so this x is going to be in the domain of the restricted tangent function if you want these guys to undo each other all right let's take a look at a few sample problems so suppose you're given the inverse cosine of the cosine of Pi / 6 the first question you would ask is is this angle pi/ 6 in the domain of the restricted cosine function if it is these guys are going to undo each other so in other words is pi/ 6 greater than or equal to zero well yeah that part's true and less than or equal to Pi well yeah that's true as well so I can go ahead and say that this would cancel away with this and I'm just left with that Pi / 6 so that's going to be my answer if you wanted to look at this on the unit circle again you would start let's say we had the inverse cosine of the cosine of pi/ 6 and we're going to do this the long way so we're going to start on the inside and basically we would say what is the cosine of Pi / 6 so I come to Pi 6 that's right here the X coordinate is my cosine value so that's of 3 over2 so now we want the inverse cosine of the < TK of 3/ 2 when I think about the inverse cosine function again I'm going to work from 0 to Pi where 0 and Pi are both included so I'm just going to ask the question what angle has a cosine value of 3 over2 in this interval from 0 to Pi where 0 and Pi are both included well we already know that that's going to be < / 6 here's your cosine value of 3 over2 so you're just reversing what you just did you found the cosine of piun / 6 and that was 32 well now the inverse cosine of3 32 is going to get you back to P pi over 6 so this right here is your answer and again if you see something like the inverse cosine of the cosine of pi/ 6 because this pi/ 6 is in the domain of the restricted cosine function or you could also say the range of the inverse cosine function you can go ahead and just shortcut this process and say this is going to cancel with with this so they'll undo each other and you're going to be left with this pi/ 6 as your answer okay let's look at another example so here we have the inverse cosine of the cosine of 5 pi over 4 this is the one where a lot of people would make a mistake so if you give the answer as 5i over 4 that is incorrect In other words you can't just cancel here because this 5 pi over 4 if you go to the unit circle that's going to be right here so that is not within the domain of the restricted cosine function or the range of the inverse Co sign function again you're thinking about this from 0 to Pi where both 0 and Pi are included so what you have to do for this particular problem you have to go about it the long way so you have to say this is the inverse cosine of the cosine of 5 piun over 4 okay so what is the cosine of 5 pi over 4 if I come over here that's going to be the negative of the 2 over2 okay so now I have the inverse cosine of the negative of thek of2 over 2 okay so what is the inverse cosine of the negative of the2 over 2 so if this guy is negative I want to be in quadrant 2 where cosine is negative and I'm looking for where I have a cosine value of2 over2 so that's going to be right here and that's going to be at 3 pi over 4 so this would be equal to 3 pi over 4 now if you had gotten the inverse cosine of the cosine of 3 pi over 4 again because this 3i 4 is in the domain of the restricted cosine function or you could say the range of the inverse cosine function in this particular situation you can cancel this with this those will undo each other and then basically you're left with that angle of 3 pi over 4 when you get a situation like this because this is not within the domain of the restricted cosine function or the range of the inverse cosine function you have to do more work you have to first evaluate the cosine of 5i 4 which is2 over2 then you take the inverse cosine of2 over2 and that's going to give you this angle in quadrant 2 which is 3 pi over 4 all right let's take a look at one more of these so here we have the inverse sign of the S of 6 pi over 5 now when we get this problem again I'm always asking is this guy going to undo this guy in other words are these guys going to cancel away and just give me 6 pi over 5 as my answer and here you're going to say no because for this to work that way you have to have this 6 pi/ 5 being in the domain of the restricted sign function or the range of the inverse sign function and it's not if you think about let's say the range of the inverse sign function again you're working from piun / 2 to piun / 2 so you've got to be there when you think about 6 piun over 5 that's not directly on the unit circle I've sketched it in so let me put 6 pi over 5 and you might be scratching your head saying where's that angle at and you can do some analysis and think about okay well it's going to end up being in quadrant 3 so I'm just going to write this one out and I'll give you an easier way to do that by converting it into degrees in a moment so this would be greater than pi and less than 3 Pi / 2 so that that tells me I have an angle in quadrant 3 so let me just put that this is in quadrant 3 now an easier way to do that if you're struggling with that you can always convert it over into degrees way easier to do so I'm just going to say it's 6 pi over 5 times I'm going to go 180° over pi and so basically the pies are going to cancel and so 180 divid 5 is 36 and so you'd have 6 * 36 that's 26 and now this is in terms of degrees so this is 216° so I think a lot of people will find this easy easier to work with and also immediately you know that this is going to be in quadrant 3 whereas with 6i 5 you really have to think about it because it's in terms of radians I've already sketched this in to help us out so this guy in Orange which I'm not sure if it comes through so well so I'm going to try to do a rough Trace over what I did so this would be your angle right here so this is what I'm looking at so that's my 6 pi over 5 now I know that the S of 6 pi over five is going to be negative because it's in quadrant 3 I also know that the reference angle is going to be pi over 5 and the way I figure that out is if I'm in quadrant 3 I take my angle which is 6 pi over 5 and I subtract away Pi so I'll do that as 5 pi over 5 and that gives me pi over 5 so this guy right here this reference angle so this piece right here that's going to be pi over 5 now I know that sign is negative again in quadrants three and four so all I have to do is come over into quadrant four and find an angle with a reference angle of pi over five but again I don't want to do this I don't want this angle right here because I want to rotate clockwise in Quadrant 4 when I have a negative argument for my inverse sign function so I want this guy right here right here so I'm going to draw over this maybe use a different color so this guy right here that is going to be netive pi over 5 so I've got my pi over five reference angle so now I want to convert my problem over to where I use something that's going to be in the range of the inverse sign function so let me get rid of all of this I think we're ready to go and all we have to do now is say that the inverse sign of the S of 6 piun over 5 would be equal to the inverse sign of again I'm going to convert this over into an angle that's in the range of the inverse sign function and so I'm going to say it's the sign of < over 5 because again the S of 6 piun over 5 and the S of Nega Pi 5 those are both going to be the same value they're both in quadrants where sign is negative so they're negative in quadrants 3 and four and the reference angle is pi over 5 in each case so all I have to do now is give my answer as these guys are going to cancel and I'm left with that angle which is pi over 5 all right let's take a look at a few examples where we're going to solve some trigonometric equations using inverse trigonometric functions and the cancellation properties that we talked about earlier so suppose you had something like -5 +2 tangent of -2x + 5 piun 4 and this equals 11 now we want to solve this over the interval from 0 to 2 pi where Zer is included and 2 pi is excluded now earlier in the section we talked about how we could solve equations of this form but here we're going to use a different approach so let me copy the problem so -5 + 12 and then we have our tangent of this is -2x + 5 piun / 4 and this equals -1 so let me go ahead and grab this real quick just remember the interval that we're solving over I'm going to come here to a fresh sheet so we have a little bit of room to work so the idea here is still the same you want to isolate the tangent of something in this case it's -2x + 5 pi over 4 so you want to isolate that on one side you want a number on the other so what I'm going to do is add five to both sides of the equation so over here this is going to cancel away and I'm going to have 1/2 tangent of -2x + 5 piun over 4 and this would be equal to over here I have1 hves and then Plus for five let's write that with a common denominator so let's multiply by 2 over2 so this would be 10 Hales and then now I can add-1 + 10 that's -1 put that over the common denominator of two so this would give me2 now again if my goal is to isolate this guy right here I have to think about the fact that I have that 1/2 that's multiplying that so to get rid of that I'm going to multiply both sides of the equation by two and so this is going to cancel and this is going to cancel and so at this point what I have is the tangent of this -2x + 5 piun over 4 and this equals over here I just have my -1 so here's where we're going to use our inverse tangent function and the cancellation property that we talked about earlier so I'm going to take the inverse tangent of each side so let's start over here I'm going to say that I have the inverse tangent of this guy right here so this would be the tangent of -2x + 5 piun over 4 and then this guy over here on the right you're going to do the same thing so I'm going to take the inverse tangent of this1 now again we have our property where we said that the inverse tangent of the tangent of X in this case X is just this guy right here so this is what's on the inside we would say that this is equal to X as long as X is greater thank / 2 and less than pi/ 2 so we're going to use this property here and basically we're going to say that this guy and this guy are going to undo each other or cancel each other however you want to think about that and so I'm just going to be left with this part right here so I'm going to say this is -2X let me move this down a little bit more and then plus 5 piun / 4 and this equals the inverse tangent of 1 now most of you know that's piun over 4 but just in case you don't again you can always go to the unit circle and think about these things so for the inverse tangent function again you're working from piun / 2 to piun / 2 with those values being excluded so I'm going to make sure that I stop a little bit short there so those guys are not included again the negative Pi / 2 and the Pi / 2 and you're just asking the question where in this interval here am I going to have a tangent value that's going to be -1 well a lot of you can see that at 7 piun over 4 you have a s value that's ne2 over2 and a cosine value that's 2 over2 so if you had the tangent of 7 pi over 4 you're dividing opposites so that's equal to -1 now we can't put 7 pi over 4 because again we have to rotate clockwise here so you've got to go like this so this would be netive piun over 4 and of course that's coterminal with 7 pi over 4 so the tangent of 7 piun 4 is equal to the tangent ofun / 4 and that equals 1 so if I took the inverse tangent so the inverse tangent of1 that's going to give me thisk over 4 okay so coming back to this point what we want to do is solve for x so now I have this guy that I want to isolate so let's subtract 5 pi over 4 away from each side of the equation so of course this is going to cancel and now I'm going to have -2X is equal to let's write this as piun - 5 piun over the common denominator of 4 so this right here piun - 5 piun is - 6 piun so this is going to be -6 piun over 4 now to simplify this if you think about 6 and four there's a common factor of two so 6 ID 2 is 3 and 4 ID 2 is 2 so let's go ahead and write this as -3 piun over 2 so this is -3 piun over 2 now as the last step again I want to isolate X so what I want to do is think about the fact that -2 is multiplying X so you can divide both sides by -2 or since you have a fraction here it's probably going to be a little bit easier here if you multiply both sides by - one2 so multiply both sides by -2 so of course you could write this as -2 over 1 and this right here would cancel and become a one and so you would have X is equal to Nega * negative is positive and then basically you would have 3 pi over 4 so that is only one solution and that's part of the issue with this technique here so you have x = 3 pi over 4 so let's go back up to the top and let me write that X is equal to 3 piun over 4 so I know that this guy right here is in quadrant 2 you're going to find that you actually have a solution in quadrant 1 2 3 and four with this problem and the way I know that is I would look at the period of this guy so I know how often these Solutions are going to repeat again the standard period for tangent is going to be Pi so when you look at Y is equal to the tangent of X well the period for that guy the period for that guy is going to be Pi but if you have y is equal to something like the tangent of BX well then the period so the period is now going to be pi over B but that's when B is positive in this case you're going to have a negative here so you can adjust the formula and say pi over the absolute value of B so for this case you have this -2 that's multiplying X so the period for this guy would be pi over the absolute value of -2 which would be pi over 2 so that's going to be the period let me get rid of this so let me put this like this and let me put that the period period is Pi / 2 and actually let me grab this real quick just going to come down here to the unit circle and rewrite things so X is equal to 3 piun over 4 again this is in quadrant 2 and I'm going to put that the period so the period is pi/ 2 I think you can do it on the unit circle the first time just so you can get a visualization of what's going on so I know that Solutions repeat every pi/ 2 so here I'm going to be at 3 pi over 4 so what I want to do is actually subtract away Pi / 2 and find my solution in quadrant 1 a lot of you know that's going to be pi 4 well let's go ahead and do the work here so we have 3 < / 4 - < / 2 let's go ahead and write this as 2 piun over 4 so we can have a common denominator so - 2 piun over 4 and so this would give me again piun over 4 so that's a solution in quadrant 1 now if I want the solution in quadrant 3 then what I would do is actually take my 3 pi over 4 so 3 pi over 4 and I'm going to add 2 pi over 4 and that's going to give me 5 piun over 4 so that's going to be right here and then in Quadrant 4 of course it's going to be 7 pi over 4 but let's just go through it we would have the 5 piun over 4 this guy right here and we're adding our pi over 2 or 2 piun over 4 so we have 5 piun over 4+ 2 piun over 4 so that's 7 piun over 4 so those are your Solutions in that interval that I gave if we have X being greater than or equal to zero and less than 2 pi well then X is going to be equal to < over 4 and then you have 3 piun over 4 and then you you have 5 pi over 4 let me scooch this down a little bit and then you have 7 pi over 4 then in terms of the general solution if you did want that although it didn't ask for it you could start with pi over 4 so start with something like pi over 4 and then you could just add pi over two to get to the next solution and then add pi over two to get to the next solution and then add pi over two to get to the next solution and add pi over two to get to the next solution and that will continue forever and ever and ever so pi over4 plus your pi over 2 time n where n is any integer all right one thing that I find helpful especially for beginner students is to look at a graph of an equation after you solve it so you can go to Desmos so I would say that you have 12 and then tangent of -2x + 5 piun over 4 and then you have this ne5 here so let me write minus 5 I'm going to add this to both sides so I'm going to add 11 halves over here and then this would be equal to zero because I added it over here as well so this right here this5 that's -10 /2 so this is -10 / 2 and -10 + 11 is going to be POS 1 so this right here would be +2 so the way you would key this into Desmos is you would say that Y is equal to this and of course if I plug in a zero for y now I'm just finding my X intercepts or the X values that make the Y value zero so I have the graph set up for us again this is y = 12 tangent of -2x + 5 piun over 4 + 12 so if you want to key that into Desmos again if I plugged in a zero for y now I'm looking for the X intercepts so where are the X values that make the yv value zero and of course you're going to see this is consistent with what we found so here you have your pi over 4 your 3 pi over 4 your 5 pi over 4 your 7 pi over 4 so that's in the interval that we solved in from 0 to 2 pi where 0 is included and 2 pi is excluded and then you would continue with this pattern forever so you can keep just adding Pi / 2 to your solution to get to the next one so 7 < 4 plus < / 2 or you could say 2 piun over 4 that's 9 piun over 4 then 11 piun over 4 then 13 piun over 4 then 15 piun over 4 then 17 piun over 4 so on and so forth all right let's go ahead and look at another example so here we have 40 s of beta - 3 then + 1 is equal to 6 again we want to solve this over the interval from 0 to 2 pi where 0 is included and 2 pi is excluded so we're going to have 40 and then sine of this beta minus 3 and then + 1 is equal to 6 so let me go ahead and grab this let's come down to a fresh sheet here and again my goal to start is to get the sign of this beta minus 3 on one side by itself so I want to isolate that so in order to do that I'm going to start by subtracting one away from each side of the equation so this would cancel and so now we're going to have 40 and I'm going to put times the sign of this beta - 3 and this would be equal to 6 - 1 is 5 again I want to isolate this right here so you have 40 That's multiplying that so all I'm going to do is divide both sides of the equation by 40 so of course this is going to cancel and now you're going to have the S of this beta minus 3 is equal to when you think about 5id 5 that's 1 and 40 ID 5 is 8 so this is 1 18 now I've got to get this guy by itself and in order to do that I've got to get rid of this guy right here so I'm going to use my property from earlier remember that if you have the inverse sign so the inverse sign of the S of X this equals X as long as X is within the domain of the restricted sign function so we would say for X being greater than or equal to piun / 2 and less than or equal to Pi / 2 so here I'm just going to take the inverse sign of each side and use this rule here so let me actually get rid of this and I'm going to say that I have the inverse sign of this guy right here so the sign of my Beta minus 3 and this equals again I'm going to do that over here as well so the inverse sign of 1/8 so at this point we know that this will undo this so those are going to cancel and I'm left with my Beta minus 3 and this equals my inverse sign so my inverse sign of 1/8 now at this point you can give one solution for beta you could just add three to both sides of the equation and so this right here is going to cancel and you're going to say that beta is equal to the inverse sign of 1/8 and then plus 3 so this is only one solution to get the other solution what you're going to have to do is think about your angle in quadrant 2 that has a sign value of 1/8 so what do I mean by that remember that if you have the inverse sign of something that's positive so this is a positive argument you're going to get something in quadrant one and actually you're going to get this angle right here let me try to trace over this as best as I can so that angle right there if you're working with radians that's going to be let me put the inverse sign L of 1/8 is about 0.125 in terms of radians if you want to work with degrees it would be about 7.18 de so that's going to be for that angle right there let me just put this in so everybody knows what that is so this is going to be about 7.18 de or 0.125 radians now where is the sign value also going to be 1/8 well again in quadrant 2 two where I have a reference angle of 7.18 De or 0.125 radians so you want this angle right here as well so this angle right here as well which again would have that same reference angle so if I go down like this this would be about about 7.18 de or 0.125 or radians now how do I get this angle right here so this is the angle if I look at it well all I have to do is take the straight angle which is pi and subtract away this part right here so you could do that by saying that you have your beta is equal to Pi minus the inverse sign of 1/8 so again that's taking this piece out and then plus three like that so let me go ahead and grab this real quick and so we have these two guys here and basically all I'm going to do is combine them into one little nice solution so I'll just put a comma here you can also use solution set notation if you want now if you wanted to give the general solution then basically all you'd have to do is say this guy plus 2 pi where n is any integer and then this guy plus 2 pi where n is any integer because your Solutions are going to repeat every 2 pi now if you want let me go ahead and grab this real quick let me paste this in if you want to verify these Solutions again you can sketch the graph of this guy the way I would do it is I would say that you have y is equal to 40 sin of beta minus 3 and then I would subtract six away from each side so I would say -5 so basically if you subtract six away from each side here you would have this part right here is equal to zero I'm replacing the zero with Y so I can graph this guy and again if Y is zero you're looking for the value for beta or you could put an X here if you want for graphing purposes so you're looking for the value for x such that the Y value is zero so you're looking for the X intercepts now in terms of approximating this again the inverse sign of 1/8 that's going to be about 0.125 and then if you add three to that you're going to get about 3 .125 so let's say that beta is about 3.125 so if you key this into your calculator you're going to get about 6.01 let's just say six so 6.16 like that let me just grab this real quick so coming down to the graph again you can sketch this on Desmos this is one of the solutions we found so about 3.125 this is another one so this is about 6.16 so that's all in the interval from 0 to 2 pi where Z is included and 2 pi is excluded and of course if you wanted other Solutions if you wanted to give the general solution everything everything is going to repeat every 2 pi so if you take this solution and you add 2 pi you're going to get to this solution here if you take this solution and you add 2 pi you're going to get to this solution here all right let's change things up and now we're going to look at some problems that involve solving equations where you have an inverse trigonometric function involved so here we have 3/4s * the inverse cosine of x / 2 and this equals < 4 now for this type of problem what you want to do is isolate your inverse cosine function so I want inverse cosine of in this case X over2 on one side and I want a number on the other so let's go ahead and just copy the problem so 3/4s I'm going to put times the inverse cosine of this X over 2 and this equals pi over 4 if I want to isolate this guy right here well I have three4 that's multiplying this guy so what I'm going to do is multiply both sides of the equation by the reciprocal of 3/4s which is 4/3 remember that a number times its reciprocal will always give you one so when I multiply 4/3 * 3/4s those guys are going to cancel and give me a big fat one so now the left side is just what I want it's the inverse cosine of this X over 2 and this is equal to over here the fours would cancel and so you would have pi over 3 now at this point you could probably solve this by inspection you can just look at this and say well the inverse cosine of2 would be pi over 3 so I know X is going to end up being one it's a really easy equation but if you didn't know that what could you do to get your solution well there's two strategies you could use the first thing is you can go back to what we talked about earlier with the cancellation property so if you have something like the cosine of the inverse cosine of let's just say x right now we have X over2 but let's just say x this would be equal to X as long as this guy is within the domain of the inverse cosine function so as long as X is greater than or equal to NE 1 and less than or equal to postive 1 now we're just going to use that here and then I'll show you an easier way to do it so I'm going to take the cosine of each side so the cosine of the inverse cosine of x/ 2 is equal to I'm going to take the cosine of this side as well so the cosine of pi over 3 so at this point I'm just going to cancel this with this these guys will undo each other and you're going to have X over 2 is equal to the cosine of pi over 3 we know is2 and at this point you know that X is 1 right the denominators are equal so you could if you wanted to just multiply both sides by two let's just go ahead and clear the denominators away to make this official so this guy would cancel with this guy and this guy would cancel with this guy and you get X is equal to 1 an alternative approach if you get something like this let me get rid of this right here is to do this in a simpler way so you'll probably see this technique in your textbook you know that if you have the cosine of pi over 3 you get 1/2 so this is the angle this is the cosine value then you also know that the inverse cosine of2 so that's the cosine value this is going to give you the angle again between 0 and Pi where 0 and Pi are both included where that cosine value is 1/2 so this is pi over 3 so if I look at this the inverse cosine of x / 2 is equal to piun 3 well this is your cosine value and this is the angle so all I really have to do is say well that means that the cosine of the angle pi over 3 is equal to again this is the cosine value so this is X over2 so you can kind of shortcut that process and just set it up this way just using the basic definition for cosine and inverse cosine so once I look at this I know that the cosine of pi over 3 is again 1 12 so this becomes 12 is equal to X /2 and that just becomes X is equal to 1 now with these problems I always suggest checking them you'll get a lot of problems that they throw at you with no solution here or they get really tricky so always check your problems here so what I would do is plug in a one for x and see if I get a true statement so you get 3/4s times the inverse cosine of 12 is that equal to < over 4 so the inverse cosine of 1/2 is going to be piun over 3 so you would have 34s * piun over 3 is that equal to pi over 4 yes it is if you cancel this with this you would have < over 4 is equal to < over 4 so x = 1 is a valid solution all right so again I'm going to recommend that in this particular section you look at the graph of everything just so you can confirm things so I went ahead and just rearranged the equation so I could get y = 34s * the inverse cosine of x / 2 and then minus < over 4 so the equation was 3/4s and then cosine inverse of x/ 2 and that was equal to pi over 4 so I just subtracted pi over 4 away from each side so would give you minus piun over 4 like this and that would be equal to zero so again all I'm doing is I'm plugging in a zero for y and I'm saying for which x value am I going to get a yvalue of zero so where's the X intercept if you look at the graph what you're going to see that occurs right here where you have a one for your x value and a zero for your yvalue so the ordered pair 1 comma 0 and again that just confirms our solution that X is going to be equal to 1 all right let's take a look at another one so here we have2 * the inverse cosine of x over 4 this is equal to Pi so in this particular case again I'm trying to isolate the inverse cosine function first so here the inverse cosine of x over 4 so let me just write2 * the inverse cosine of x over 4 and this would be equal to Pi now if I want to isolate this guy I need to get rid of this 1/2 that's multiplying it so I'm just going to multiply both sides by two and so we're going to see that this right here is going to cancel with this right here so now I have my inverse cosine of x over 4 that's on the left it's I ACC ated and this equals < * 2 is just 2 piun now let's stop for a moment and think about this is there going to be a solution for X is there something that I can plug in there divide by four and then take the inverse cosine of that and get 2 pi no because the range of the inverse cosine function is from 0 to Pi with 0 and Pi both being included so this right here is impossible so we would say that there's no solution right so there's no solution again if you want to look at this graphically if you started with * the inverse cosine of x over 4 and again this was equal to Pi so let's put equals Pi well I'm going to subtract Pi away from each side so now I'll have minus pi equals 0 well again I'm just plugging in a zero for y and I'm saying for which x value am I going to get a yv value of zero so where would I cross the x- Axis or get an X intercept and you can see that this graph does not cross the x-axis basically it goes from here to here it's never going to cross the xaxis so there is no solution all right let's take a look at a problem that's a little bit more tedious so here we have the inverse cosine of 35ths minus the inverse s of X in this equals zero so if you get a problem like this what I would do is I would isolate the inverse sign of X so in other words right now you have minus the inverse sign of X so I would add the inverse sign of x to both sides of the equation so let me just show that real quick so I'm going to add the inverse sign of X on the left and I'm also going to add the inverse sign of X on the right so this would cancel away so what I would have here is the inverse cosine of 35ths is equal to the inverse sign of X so I'm actually going to write this one on the left that's just my preference so I'm going to write the inverse sign of X is equal to the inverse cosine of 35ths okay I'm going to come down to a fresh sheet so we have a lot of room to work so what I'm going to do is actually take the sign of each side so I'm going to say that I have the sign sign of the inverse sign of X and this equals I'm going to take the sign of the inverse cosine of 3 FS when I think about taking the S of again the inverse sign of X these guys will undo each other as long as X is within the domain of the inverse sign function so from negative 1 to positive one with both being included so X is equal to the S of the inverse Co of 3 FS what is the inverse cosine of 3 fths again that's an angle in quadrant one because this guy is positive where the cosine value is going to be 35ths so what I can do is say that I'm going to let something like Theta be equal to the inverse cosine of 3 FS so again this angle Theta is going to be in quadrant 1 because the inverse cosine of something that's positive is going to give me something in quadrant one so I'm going to say that Theta is is in Quadrant One additionally we know that this right here is the cosine value so I can say then that the cosine of this Theta is going to be 35s so what am I really trying to find for X here well it's the sign of again the inverse cosine of 3 fths well the inverse cosine of 3 fths is Theta so really I'm just trying to find the S of theta so you could really write this as X is equal to the S of theta how do we figure out what the sign of theta is again if I know the cosine of theta is 35ths and I know Theta is in quadrant 1 this is really easy to figure out so this right here is your adjacent and this right here is your hypotenuse remember for cosine it's adjacent over hypotenuse again you could also say that this is your X over your R for the S of theta this is going to be your opposite over your hypotenuse or you could write that as your y over your R well I know that the r or the hypotenuse in this case is five so I can just put that in there and again you can just use your Pythagorean theorem to figure out what Y is or the opposite a lot of you already know it's going to be four because this is a 345 right triangle but again you can always make a sketch now for this one it's very easy so maybe you don't need a sketch but I'll just go ahead and make one for completeness so the adjacent is going to be three so you could say this is your X here so X is three your hypotenuse is going to be five so I'm going to say this is your r r is five and of course the opposite or your Y is going to end up being four but when we first solve the problem we don't know that so what you would do is say that you have x^2 + y^2 is equal to R 2 just coming from the Pythagorean theorem and so we know X is going to be 3 so 3^ 2 is 9 and then plus y^2 again at this point Y is unknown and this equals we know R is 5 so 5^ s is 25 so what I would do at this point is subtract 9 away from each side of the equation this cancels you get y^2 is equal to 25 - 9 is going to be 16 and at this point I would say that Y is equal to the principal sare root of 16 which is 4 you would use the principal square root here because you're in quadrant one and Y values are positive so coming back here this would be a four so x equals the S of theta and the S of theta is 4 fths so you can just say that X is going to be equal to 4 fths here now how would you check something like this remember that Theta is an angle whose cosine value is 3 fths and whose s value is 4 FS so coming back up here again if you plugged in of 4 fths there you'd have the inverse cosine of 3 FS minus the inverse sign of 4 fths does that equal zero so essentially what you have is let's just say this is Theta well this would also be Theta it's the same angle right so the cosine of this angle is 3 fths and the sign of this angle is 4 fths so it's just theeta - Theta is equal to 0 so you get 0 equal 0 so this 45ths is a valid solution for X again you can always go to Desmos and verify this solution graphically so if you do something like yal the inverse cosine of 35ths minus the inverse s of X again if you plugged in a zero for y I'm asking for which x value am I going to get a y value of zero and that's going to occur at 4 FS comma 0 so this right here is your x intercept again where Y is z so that confirms our solution that X is going to be 4 FS all right now let's look at something that's very tedious overall I'm just going to tell you that you're definitely going to get problems like this in your textbook and they take quite a while to solve especially if you're doing this the long way I'll also tell you that for most of these problems there's an Associated identity that you can use to greatly speed up your work I would suggest for the first few times that you solve this type of problem that you do it the long way so you can get a full understanding of what's going on and then later on when you come across these problems again you can just use the identities and go a lot faster all right so let's look at the inverse s of x minus the inverse cosine of x and this equals Pi / 6 so what should we do first if we get a problem like this pick one of these guys so it could be the inverse s of X or the inverse cosine of x and you're going to isolate that on one side so in this case I'm going to isolate the inverse s of X and to do that I'm just going to add the inverse cosine of x to both sides of this equation so I'm going to go ahead and show this as cancelling so now I'm going to write this as the inverse sign of X is equal to over here I'm going to have the inverse cine of X and then plus < / 6 now I'm going to end up taking the sign of each side but before I do that let me just grab this real quick and let me paste this in I want to write this in away to where it's Crystal Clear what we're doing and what everything means so I'm going to make a little substitution you've seen me do that already so the inverse cosine of x we know that that's an angle from 0 to Pi where 0 and Pi are both included where the cosine value is X so in other words I can let something like Theta or in this case let's just use beta so let's let beta be equal to the inverse cosine of x so then what I can say is that the cosine of beta is going to be equal to X I know that in terms of beta right now now just looking at this beta is going to be from 0 to Pi where 0 and Pi are both included that's coming from the range of the inverse cosine function so let me say that beta is greater than or equal to zero and less than or equal to Pi so I'm just going to change this from the inverse cosine of x to Beta so I'm going to erase this and just put this right now as beta okay let's put this over here like this now there's something else that you have to consider let me actually grab this and slide it down right here if I have the inverse sign of X that's going to give me something from piun 2 to piun / 2 where both of those guys are included so that tells me that this right here this beta + pi/ 6 has to be greater than or equal to piun 2 and less than or equal to piun / 2 I'm going to put this in terms of beta being in the middle of that guy so to do that I'm just going to subtract pi over 6 away from each part of this guy over here let me just add piun over 6 that would be the same thing so basically this would cancel and over here I'd want to get a common denominator so multiply this by 3 over3 and over here I'd multiply this by 3 over3 as well so I would end up with piun plus -3 piun over the common denominator of 6 so that's -4 piun over 6 so this would be -4 piun / 6 and of course each guy here is divisible by two so 4 ID 2 is 2 so this would be -2 piun over 6 / 2 is 3 so let me erase all of this and put this as -2 piun over 3 and let me erase this and over here let's think about what we have so this would be 3 Pi minus pi over the common denominator of 6 so this would be 2 piun over 6 so 2 piun / 6 each is divisible by 2 so 2id 2 is 1 6 ID 2 is 3 so this is pi over 3 so this is pi over 3 let me actually write this up here and I'm going to write this a little bit more concisely and let me come back up here and let me paste this in here and I'm just going to Arrow to that guy right there so we know that both of these would actually have to be true because beta is equal to the inverse cosine of x I know that beta has to be greater than or equal to Z and less than or equal to Pi but then it would also have to be true that beta is greater than or equal to to -2 piun 3 and less than or equal to piun 3 because of this right here so the inverse sign of x equals this beta +k 6 so what you actually have to do is put these two guys together using the keyword and so you're trying to find the intersection of the two so if you consider these guys together so I'm going to use the keyword and here so beta is greater than or equal to Z and less than or equal to Pi so that would look like this graphically so here's zero and here's Pi again that's about 3145 let's just say it's about right there so this would be about pi right there so this is what that looks like if you were to graph that inequality then for this one beta is greater than or equal to -2 piun 3 and less or equal to pi over 3 well let's say that this right here is about -2 piun over 3 and let's say that this right here is about pi over 3 just in case you're curious -2 piun over 3 is about -2 let's say 09 and Pi over3 that's about let's say 1.05 so graphically this would look like this right here so this right here if you want to look at that inequality so something like that so if I want the intersection of the two again I'm looking for values that are going to work in both in other words it would have to satisfy this one and this one so when I set that up I'm running out of room but let me just do it over here I would say that beta is greater than or equal to what well consider that over here you have a zero and over here you have a negative -2 Pi over3 well it's got to be zero it's got to be the one that's more restrictive or further to the right because it's got to work in both if I had a value that was let's say here that was to the left of zero but to the right of -2 pi over 3 well it would work in this one but it wouldn't work in this one so I'm going to have to choose zero here and then I say it's less than or equal to well for this one I want the one that's further to the left right because it's going to be more restrictive and include fewer values so I want the pi over 3 in other words if I had something like let's say two that was to the right of pi over 3 well it works in this one but doesn't work in this one so it's got to work in both both so I'm going to choose pi over 3 okay let me come down here and I'm going to paste this in and I'm coming back to this in a second and let me grab this and let me come down here and paste this in so we're going to need this identity that's why I have that set up already so what I'm going to do now is take the sign of each side so I'm going to take the S of the inverse sign of X and this is equal to the S of beta + pi/ 6 when I think about this guy right here again from from the cancellation property that we talked about earlier this will undo this as long as X is in the domain of the inverse sign function so from -1 to positive one with both being included so let's go ahead and say that X is equal to this guy right here so the sign of beta + pk/ 6 now I'm going to erase this and now we're going to use this sum identity for sign to simplify this so what I'll do is say that this is X is equal to again if you have the S of a plus b it's equal to the S of a * the cosine of B plus the cosine of a * the S of B so this would be the S of the first guy which is beta and then times the cosine of the second guy which is pi/ 6 and then plus the cosine of the first guy which is beta and then times the S of the second guy which is pi over 6 now I know most of you know this but if you go to pi/ 6 on the unit circle the x coordinate is the cosine value so that's 32 and the y coordinate is the S value so that's 1/2 so coming back up here I can replace the cosine of piun / 6 with > 3 over 2 and I can replace the S of piun / 6 with 12 so this right here is 1/2 now we also know what the cosine of beta is so the cosine of beta again is going to be X so this right here is going to be X all right so we're just going to plug some things in so we have X is equal to I'm going to go < TK 3 / 2 time your s of beta and then plus here you're going to have 12 time the cosine of beta which is X so this is just X over2 so this right here is X over 2 now you have a common denominator so what I'm going to do is actually say x is equal to the < TK of 3 * the S of beta + x over the common denominator of two so the question now is what is the sign of beta so that's the to hold up so what is the S of beta so again this is the opposite over the hypotenuse do we know what the hypotenuse is remember the cosine of beta the cosine of beta is equal to your X over one right because X can be written as X over one and this right here is your adjacent and this right here is your hypotenuse so I know the hypotenuse is one so let me put that there now what about the opposite can I figure that out remember that we said that beta is greater than or equal to0 and less than or equal to < over 3 so I'm just going to sketch a triangle in quadrant one and it would look something like this so your adjacent is just going to be X your hypotenuse is going to be 1 and you're going to find through the Pythagorean theorem that the opposite is going to be the square root of 1 - x^2 so again if you want to go through this let's say that we label this right now generically as y so we would say that we have x² so I'm just calling this length right here x so it's X2 plus I'm going to have this guy right here let's call that y so y^2 is equal to this guy right here your hypotenuse we know that's one so 1^ sared you're going to solve this for y so you're just going to subtract x^2 away from each side of the equation I'm just going to erase this and then to solve for y you want the principal square root of this guy right here so 1 2 is just 1 - x^2 and the reason for that is again your y- value is never going to be negative because beta is greater than or equal to Z and less than or equal to P pi over 3 so you use the principal square root here so this right here is what you're going to get for your opposite the square root of 1 - x^2 let me actually move this over I don't think it'll fit I'm just going to say that this is the square root of 1 - x^2 let me make this a little bit longer here so this is over one so what I would do now is plug this right here in for that s of beta and then we would just solve for x what I want to do is say that X is equal to the square < TK of 3 and then times again for the S of beta I have the square Ro T of 1 - x^2 if you want you can combine those now I think it'll make it a little bit easier to understand again if you have the square root of a Time the sare RO TK of B we know this is the square root of a b so the square < TK of 3 * the square < TK of 1 - x^2 would be the Square t of 3 * the quantity 1 - x^ 2 and if you distribute that three into each guy you would have the square root of 3us 3x 2 so again 3 * 1 is 3 and then - 3 * x^ 2 that's 3x^2 let me make this a little bit better here and let me erase this and also erase this so then we would say plus you have this x here and then this is over two now I want to grab this guy right here and let me come down to this sheet here the reason I'm keeping this here is because we're we're going to end up squaring both sides of our equation and we want to remember that our angle beta which represents the inverse cosine of x that is greater than or equal to Z and less than or equal to pi over 3 so I know that my X values here are greater than zero so just keep that in mind so for this one right here I'm going to multiply both sides by two me put a little border and section this off so multiply over here by two and of course this is going to cancel away and so now we'd say that we have 2X is equal to theare < TK of 3 - 3 x^2 and then plus X at this point I would subtract X away from each side of the equation and so this right here is going to cancel so 2x - x is X so X is equal to the square < TK of 3 - 3x^ 2 now I want to square both sides so what that would give me let me come down here a little bit is x^2 is equal to the square root of 3 - 3x^2 and again I'm squaring this as well so just squaring both sides of our equation and that's going to give me x^2 is equal to so you're left with that radican there so 3 - 3x^2 me make this three a little bit better there so at this point I just want to add 3x^2 to both sides of the equation this is going to cancel and so you have x^2 + 3x^2 that's going to give me 4 x^2 and this equals over here I just have three now if you want to solve for x you divide both sides of the equation by 4 this would cancel you're going to get x^2 is equal to 34s now here's where we need to think remember with the square root property we would say that X is equal to plus or minus the sare < TK of 34s so this turns into plus or minus the square < TK of 3 so the square < TK of 3 over two do I need the plus or minus here no because again I cut it off but we said that beta which was our inverse cosine of x is greater than than or equal to 0 and less than or equal to < over 3 which told me that X is going to be greater than Z so what this tells me that I don't need the negative of the of 3 over2 that guy is not going to work it's going to be an extraneous solution it's something that was produced because we squared both sides of our equation so you just get X = the 3 over2 all right so for these problems again I always recommend you check them so let's put X is equal to the square < TK of 3 /2 let's just make sure that it works again if you were to try to check x equals the negative of the of 3 over2 the extraneous solution there it's not going to work remember that X values would be positive where we're dealing with this angle beta so what I would think about here is what is the inverse sign of this Square < TK of 3 over2 a lot of you know that it's going to be pi over 3 we'll go to the unit circle and grab it and then what is the inverse cosine of < TK of 3/ 2 most of you know that's pi/ 6 so let's put this as pi over 3 let's put this as pi/ 6 let's go to the unit circle real quick just in case you need it so again for the inverse sign of 32 I'm working from piun / 2 to P piun / 2 and I'm looking for a s value of 32 so that's going to be right here right here and the associated angle again in this interval from negative pi/ 2 to Pi / 2 is going to be pi over 3 so that tells me that the inverse sign of the < TK 3/ 2 again is < over 3 and then for the inverse cosine of the < TK 3 / 2 again with this one now I'm working from 0 to Pi both of those are included so from 0 to Pi and again where do I have a cosine value of of 3 over 2 well that's going to be right here at pi/ 6 so this would end up being pi over 6 all right so now let's just plug in so the inverse sign of 3 over 2 is piun / 3 so that is < over 3 and then you'd have minus the inverse cosine of 3 2 that's < / 6 does this equal < / 6 well if I get a common denominator here I'd multiply this by 2 over 2 and so that would give me 2 pi minus piun over the common denominator of 6 does that equal pi over 6 yeah 2 pi minus Pi is Pi so this would be pi/ 6 is equal to pk/ 6 which is a true statement so that solution x = 3 2 is a valid solution again you can always verify this graphically so you can set this up as y equals the inverse sign of x minus the inverse cosine of xus pi/ 6 where did that come from again if you have your inverse sign of x minus your inverse cosine of x is equal to pi/ 6 well if you subtract pi over 6 away from each side of the equation this would become minus < / 6 is equal to 0 so all I'm really doing here is plugging in a0 for y and I'm asking for which x value am I going to get a yv value of zero so again that's going to be your x intercept so here that occurs atun of 3 over2 comma 0 so that tells me the x value is3 over2 so this confirms my solution that x = 3 over 2 in this lesson we want to talk about solving trigonometric inequalities all right let's go ahead and look at the first problem so this will be a really simple example just to get started so we want to solve for x is greater than or equal to Z and less than 2 pi and so we have 4 - the sin of X is greater than 7 so how do we solve something like this in this particular case it's a very simple example so you could isolate the S of X on one side and then have a number on the other and then you can go to the unit circle and get your solution that way I'm not going to actually do that because I want to build up the concept that we're going to use for the harder examples so what I'm going to actually do is make the right side zero to start so I'm going to say that I have let's just reorder this the negative of the sign of X and then plus 4 I'm just going to leave a space here and put this is greater than 7 now if I want the right side to be zero I just need to subtract s halves away from each side of the inequality I would want a common denominator so let me multiply 4 over 1 * 2 over 2 and let me put this minus sign down here so at this point I'm going to write that I have the negative of the S of X and then basically this 4 * 2 is 8 so this would be 8 over this 2 so 8 over 2 that's the same thing as four we're just getting a common denominator and 8 over 2 - 7/2 we do 8 - 7 which is 1 over the common denominator of 2 so that's just 1/2 so this is +2 and then this is greater than 7 half - 7 is obviously zero at this point what I would do is get rid of this negative that's in front of the S of X I'm actually going to write this as -1 * the S of X so it's Crystal Clear what we're doing I'm just going to multiply both sides of the inequality by1 remember if you do that you have to flip the direction of the inequality symbol so let me write -1 * this -1 * the S of X and then plus you're going to have your 12 * -1 this is going to get flipped so right now it's a greater than because we're multiplying both sides by a negative it's going to become a less then and then you have 0 * -1 so 1 * 1 is POS 1 so this is just the S of X now and then you have 1/ 12 * 1 that's -2 and then this is less than 0 * anything is 0 so I'm going to grab this real quick okay so at this point I'm just going to come to the unit circle and put a little border here now this is an easier example so of course you could just figure out well where is the S of X less than 1/2 just by adding 1/2 to both sides I'm not going to do that because again I want to build up the method that we're going to use with the more challenging scenarios but once you have everything down in terms of what methods to use when you can go ahead and solve it that way it's no problem so I'm going to write down that we want to find the critical values here and how do we do that again you replace your inequality symbol with equals so you have the S of x - 12 is equal to zero add 1/2 to both sides and you'll get the S of X is equal to 1 half so basically the solutions there if we're working in that interval let me write that out so we know we're working from zero and Zer is included out to 2 pi and 2 pi is excluded where your Solutions there you're going to have pi/ 6 you have a sign value of a half and then also 5 pi over 6 you have a sign value of 1/2 so when I write down that the critical values are going to be where X is going to be pi over 6 and then 5 pi over 6 the reason we want to know that is we want to know where this guy right here the left side of this guy is going to be zero so basically that acts as a boundary for us and we're going to be able to set up some intervals on our number line or you can also do it with a table or you can do it on the unit circle you could split this up and say okay well one interval is going to be from 0 to pi/ 6 and I'm going to exclude everything for right now so basically you'd have an interval from 0 to pi/ 6 so that's that's one of them and then another interval you would take from pi/ 6 let me do this in a different color so from pi/ 6 and I'm going to exclude everything again so piun over 6 to 5 piun over 6 both are excluded so let me write another interval is pi/ 6 to 5 piun over 6 both being excluded and then a final interval here you would take from 5 pi/ 6 going to 2 pi again with both being excluded there so let's say a final interval would be 5 pi/ 6 going out to 2 pi both excluded and I have that written more neatly with a little table again if you want to use a number line it's really up to you so all you have to do is pick something in each interval and test it and see if it works so you can say true false something like that you can also just figure out the sign so basically is it positive or is it negative so coming back up that's really easy to do in this first guy here so between 0 and pi/ 6 is this guy going to be positive or negative well basically it's going to be negative so let me put a negative sign there and then between pi/ 6 and 5 Pi 6 would this guy be positive or negative well it would be positive and then in this interval right here between 5 Pi 6 and 2 pi is it positive or negative well it's going to be negative so coming back we want to put that it's negative here it's positive here and it's negative here so if you want to put a solution together you would say that it's going to work here it's not going to work here and it's going to work here in other words we're trying to figure out where is the S of x - 12 less than zero so I'm trying to figure out where it's negative because that would mean it's less than zero so here it works so let me put this in as X being greater than or equal to0 I'm going to go ahead and include 0 because that's in the interval we're solving over again when I set up my table I exclude things just to make it simple so X would be greater than or equal to zero and you would have less than pi over 6 and then this part right here it's positive so we don't want that over here you're going to have this interval so 2 pi is excluded and 5 pi over 6 they're both excluded so you're going to say x is greater than 5 pi pi/ 6 and less than 2 pi so just put this together let me slide down a little bit so let me actually just grab this and move it here and then grab this and move it here and I'm just going to use the keyword or that's all you have to do another way you can do this is you can put it in interval notation so you can put a bracket with zero and then you're going to go out to pi over 6 and that's going to be excluded so you use a parenthesis and then the union with you're going to use a parenthesis with 5 pi/ 6 because it's not included and you're going to go out to 2 pi and of course that's excluded so you're going to use a parenthesis so this would be your solution X can be anything that's greater than or equal to0 and less than pi/ 6 or X can be anything that's greater than 5 pi/ 6 and less than 2 pi okay you can always look at these things on Desmos if you're a little bit confused or you want a visual confirmation of your solution so we have y = the sin of x -2 remember we were solving the sin of x - 12 is less than zero so all I'm doing here is asking where are the y- values going to be less than zero again in that interval that I'm solving over so if I look at zero which is the start of the interval I'm solving over you can see that you have a yv value that's less than zero so that's going to continue up to the point where you're not at pi over 6 so anything less than pi over 6 so again that's where you get the Zer included up to an excluding pi over 6 then the union width if you look at anything greater than 5 pi over 6 5 Pi 6 itself doesn't work because that's where you're at zero but anything greater than 5 pi/ 6 going up to an excluding 2 pi because that's excluded from the interval we're solving over that's going to work as well so you put a parenthesis and then again 5 pi over 6 and then out to 2 pi with a parenthesis okay let's take a look at a problem that's a little bit more challenging so again we want to solve for x is greater than or equal to zero and less than 2 pi we have -2 cosine ^ 2 is greater than or equal to 1 + 3 cosine of x so again I want to make the right side zero in this case I'm going to start off by actually making the left side zero because it's easier and then I'm just going to switch it to right so I'm just going to add 2 cosine 2 x to both sides of my inequality here so this right here would cancel and give me Zer on the left so you would have 0 is greater than or equal to let's write this out in front so 2 cosine 2 x and then + 3 cosine of x and then + 1 okay let me slide down here a little bit now what I'm going to do is just rewrite this so you can basically say that 2 cosine ^ 2 x + 3 cosine of x + 1 is less than or equal to zero so that's how I'm going to set it up if you want to solve it like this that's fine as well so I'm going to say this is 2 cosine 2 x + 3 cosine of x + 1 is less than or equal to 0 all right at this point you want to find your critical values so let me put a little border here and just say that we want the critical values and again how do we figure out what the critical values are well basically you replace this inequality symbol with an equal symbol so I'm going to solve the equation two cine 2 x + 3 cosine of x + 1 is equal to 0 now this guy is quadratic in form so when you get something like this generally speaking in a trigonometry class you can Factor it sometimes you do have to use a quadratic formula but generally speaking you're going to be able to factor so let's go ahead and set this up because we can solve it with factoring here so this is a two which is a prime number that tells me this is 2 cosine of x and this is cosine of x so that's easy the last guy here is a one and this guy is positive so one is either 1 * 1 or negative 1 * netive 1 but because this guy is positive I know this is a plus one and this is a plus one if you want to check it you do your outer so that's plus 2 cosine of x and then your inner is plus cosine of x so 2 cosine of x plus cosine of x is 3 cosine of x so you're good to go with the factorization so at this point we would use the zero product property so I would say that 2 cosine of x + 1 so this Factor here set that equal to zero let's go ahead and solve that real quick I'm going to subtract one away from each side of the equation that cancels you get 2 cosine of x is equal to1 divide both sides by 2 and you're going to get that this cancels so the cosine of x equals - one2 so let me write that the cosine of x is equal to -2 and then the other guy is the cosine of x + 1 equal 0 so let me put or we have the cosine of x + 1 is equal to zero you can subtract one away from each side of the equation that cancels so you get the cosine of x is equal to ne1 so let me grab this real quick and let me come down to the unit circle so we should know these off the top of our head at this point but just in case you don't we know that cosine is negative in quadrants 2 and 3 and I'm just looking for where I have a cosine value of -2 so that's going to be right here so there's your cosine value of -2 and that's 2 piun over 3 so then also here's 4i over 3 cosine value of -2 then for cosine of x being equal to1 well I have a cosine value of1 only at Pi so let me put here that we're solving over the interval from 0 to 2 pi so that's clear and then basically your Solutions are X is equal to you have 2 pi over 3 you have pi and then you have 4 pi over 3 now these Solutions are the critical values and basically we're using this to split the number line up into intervals so basically I'm going to exclude everything here I know a lot of people like to include zero because it's included in the interval that we're solving over but just to keep things simple I am going to exclude everything and then we'll put things together at the end so for one interval here I'm going to go between zero and let's say 2 pi over 3 so that's one interval then another interval so we're going to go between 2 pi over 3 and pi and then another interval would be between pi and 4 pi over 3 and then for the final interval here we'll say between 4 pi over 3 and 2 pi so again I'm not going to include any of those guys okay so once again I'm going to come to a little table that's what I like to do to organize my information so you saw that I set these intervals up on on the unit circle so I have between 0 and 2 pi over 3 then between 2 pi over 3 and Pi between pi and 4 pi over 3 and then between 4 pi over 3 and 2 pi so when you're testing things don't ever pick the critical values themselves just leave those alone when we think about zero and 2 pi that's coming from our interval so again we're solving over the interval from 0 to 2 pi where 0 is included and 2 pi is excluded and again a lot of you are going to ask well why did I not include zero here you can if you want this is personal preference when I set this up I just exclude everything and then once I'm done and I'm putting the solution together that's where I'm going to think about what's included and what's not included so again this is up to you let me drag this up here out of the way and then basically I'm just going to write my inequality so we have two different forms we can use so the unfactored form is 2 cine 2 x and then plus 3 cosine of x and then + 1 is less than or equal to zero so this is the form I would work with if I was using a calculator again you could just pick something in each interval plug it in and see if it works and then the form I would use if I'm not using a calculator is the the factored form because it's just easier to work with that if you're not using a calculator so this was 2 cine of x + 1 and then times the quantity cosine of x + 1 okay let me get rid of this unfactored form again you can use that if you want I'm just going to stick with this one so we know that a positive time a positive or A negative * a negative is a positive and we know that a positive * a negative and a negative * a positive would be a negative I'm going to tell you that this right here so this right here this Factor based on what we can plug in is always going to be positive so I'm just going to put a plus right here so basically if this guy right here is negative then the whole thing will be negative and if it's positive then the whole thing would be positive now let me stop for a minute because some of you will say well why is this Factor always positive I want you to think about the cosine function in general so if you have the cosine of x or you could write y equals the cosine of x the range for that guy as we know is going to be from -1 to postive 1 and so you get this -1 here when you plug in a pi for X well we can't plug in pi here because it's a critical value so you're not going to be able to use that so basically when I think about the cosine of x well the smallest I can make this is if I plug in a pi for x and I get a Nega - 1 -1 + one would be zero but it's not going to be zero because I can't actually choose Pi so it's going to have to be positive here so I know that when I look at this Factor over here now I have let me put this as a little border we have 2 cosine of x + 1 so just ask yourself the question where would it be negative you can solve a very simple inequality for that so just say where is this less than zero subtract one away from each side of the inequality that cancels you're going to get 2 cosine of x is strictly less than NE 1 so this is less than1 divide both sides by two and so basically the cosine of x if that's less than - one2 then this right here would be negative and so the whole thing would be negative and this guy would be satisfied because if this side right here is negative or if it's zero we'll have a true statement so let's go back up and see where the cosine value is less than - one2 so that looks like this interval right here so that would make this negative and then this interval right here so right here so that would make this negative all right so coming back to the table we know that this guy right here would be negative and this interval right here so between 2 pi over 3 and pi and then this interval right here so between pi and 4 pi over 3 so let me put that in there so we would satisfy our inequality there because this side right here would be negative and that's always going to be less than zero so in this interval here and this interval here this Factor would be positive and so the whole thing would be positive and so that's never going to be less than zero or equal to zero you can put a plus here and a plus here to show that okay so to put the solution together you would think about where this works so because it's a non-strict inequality and we allow for the possibility of this guy to be equal to zero well 2 piun over 3 pi and 4 pi P or three those are all going to work so if you started off by just listing these in terms of the table you could start with X is greater than or equal to 2 piun over 3 and less than or equal to Pi and then say or you could look at this one so you have X is greater than or equal to Pi and less than or equal to 4 pi over 3 now of course you wouldn't write it like this because of the keyword or what's going to happen is this is just going to be X is greater than or equal to 2 piun 3 and less than or equal to 4 piun 3 so let me tighten this down like this me get rid of that and put a less than or equal to here and just drag this over there like that so this would be your solution and if you want to list the interval notation you're going to put a bracket next to 2 pi over 3 and you're going to go out to 4 pi over 3 and you're going to use a bracket there as well because both of these are included so once again if you want to look at this on Desmos just to get a visual idea of what's going on you can type in y = 2 cosine ^ 2 X Plus 3 cosine of x + 1 and essentially we had the inequality 2 cosine 2 x + 3 cine of x + 1 is less than or equal to 0 so you're basically asking where are the Y values either 0 so at your X intercepts or less than zero so if you look for your X intercepts again we already know where that's going to be you have your 2 pi over 3 you have your pi and your 4 pi over 3 and then basically you're saying where is it either there or less than zero so can sketch between these guys and you're going to see it's going to be less than zero so between those guys so if you're looking for where it's either zero or less than zero you're going to take this section right here so from 2 pi over 3 going up to and including this 4 pi over 3 so that's why we said the solution was again from 2 pi over 3 going up to and including 4 pi over 3 so that's going to be your solution in the interval that we're solving over from 0 to 2 pi where Z's included and 2 Pi's excluded all right let's take a look at one more of these so we want to solve for x is greater than or equal to Zer and less than 2 pi we're given c^ 2 x - 3 cose of x + 3 is greater than zero all right so for this one we're going to use a little identity to clean things up you'll notice that the right side is already zero so we're good to go there one thing that I would note if you're working with something that's not s or cosine you have to think about where your guy is undefined so if you think about the cang squ X and the cosecant of x that you have let's come to down to the unit circle and remind ourselves if you have something like y equal the cosecant of X where is that undefined remember that the cosecant of X is equal to 1 over the S of X so basically this guy right here cannot be zero so where is your sign value zero it's going to be right here at zero and then right here at Pi so in the interval that we're solving over the X cannot be the angle zero or Pi because again that would would be undefined because you'd have division by zero so let me put here that the domain and typically we'll list this by saying that basically it's the set of all X such that X is not equal to Pi n where n is an integer so 0 Pi 2 pi 3 Pi something like that because again you get division by zero then if you think about the cotangent of X so y equal the cent of X and I know we have cotangent squ X but basically you can extend this to that well when we think about the domain here so what is the domain here well basically it's going to be the same right because we have the cosine divided by the sign and so the sign value cannot be zero so that's going to happen at zero then Pi then 2 pi then 3 Pi so on and so forth so basically it's the same thing so it's the set of all X such that X is not equal to Pi n or n is any integer okay now if you had again cotangent squ X the same thing would apply because you're just squaring this so you still can't have division by zero so that would still apply there if you had yal cang 2 x all right let's get rid of this we're going to say that at zero and Pi this guy is going to be undefined so I'm going to keep this highlighted for now and actually we might do better by just saying undefined so undefined and just putting a little note here so I'm going to put zero and pi and that's going to be for in the interval that we're solving over I'm just going to paste this in here and I have a little identity set up for us already so we know that the cotangent X is equal to cose 2 X - 1 so essentially this right here matches this right here so I'm just going to take this right here and I'm going to plug it in for this right here so that's all I'm going to do so let's say that we have cose 2 X - 1 again this gets plugged in for this then - 3 cose of X and then + 3 is greater than 0 so you have this -1 and then + 3 so 1 + 3 is pos2 so let's go ahead and write this as cose 2 x and then - 3 cose of X and then basically + 2 and this is greater than zero so at this point you would want to find your critical values so the critical values now this is quadratic inform and you can solve it with factoring so let's go ahead and replace this with equals so you have cose 2 x - 3 cant of x + 2 is equal to 0 now I would note that again with a critical values you're also finding places where it's undefined and we know that it's zero and Pi like we talked about earlier so you can throw that in the mix now or you can put it in later whatever you want to do but I'm just going to put it in there now so we don't forget so here what I want to do is Factor on the left so let me go ahead and set this up with some parentheses very easy to factor this so this is cosecant 2 x so it's cosecant of x * cosecant of X so very easy this is 2 which is only going to come from either 1 * 2 or -1 * -2 this is negative right here so this has to be -1 and -2 let me make that a little bit better if you check the outer here is -2 cose of X and the inner here is minus cosecant of X well -2 cosecant of xus cose of X is-3 cose of X so this is the correct factorization okay let me slide down here a little bit and essentially you want to find cosecant of x - 1 = 0 add one to each side you get the cosecant of X is equal to 1 most of you know that's going to be at pi/ 2 but we'll go to the unit circle in a moment then we have or we have the cose of x - 2 equals 0 add two to both sides so the cosecant of X is equal to 2 so that's going to be at piun 6 and 5 piun over 6 again let's just go to the unit circle and let me paste this in here so I have my undefined right there of 0 and Pi just so we don't forget so the cosecant of x equals 1 again if you don't know this you can just convert it over into a sign problem so we know that the S of X is equal to 1/ the cosecant of X so basically the cosecant of X is 1 so this would be 1 over 1 so where is the S of x equal to one well basically you're going to be at Pi / 2 that's where your sign value was one so for that one you can go ahead and say that basically your X is going to be equal to pi/ 2 now for the other one you have cosecant of xal 2 again you can convert that over into a sign problem so you can say that the sine of x is equal to 1 over the cosecant of X and the cosecant of X is 2 so the S of X would be equal to 12 and where does that happen well that's going to happen right here at pi over 6 and then right here at 5 pi over 6 let me get rid of this and actually we don't need this anymore we can actually get rid of this information we're just going to list everything out so from solving our equation we have and I can put this in order so let's go pi/ 6 and then we have pi/ 2 and then we have 5 piun 6 so this right here comes from solving the related equation and then this right here this comes from us determining where things are undefined again in that interval so basically you're going to split your you could say number line or unit circle again however you want to think about that up into different intervals so you're going to have one that's between 0 and pi/ 6 so that's going to be this part right here so you think about something in there and then the next guy would be between pi over 6 and pi over 2 so between pi over 6 and pi over 2 and then you're going to do between Pi / 2 and 5 piun / 6 and then you're going to do between 5 piun over 6 and pi and then you're going to do between pi and 2 pi so that's the big one right there in the previous example you saw me use the factored form but I'm just going to tell you for something like this this is going to be very slow if you do that so you can use the factor form and determine whether it's going to be positive or negative and do it that way if I get this type of problem personally I'm just going to use a calculator and so I'm just going to go through and set up a little table this is how I like to do it again nothing will be included here so between 0 and piun over 6 between piun 6 and piun over 2 between p piun over 2 and 5 piun over 6 between 5 piun 6 and and pi and then between pi and 2 pi now I've already picked something to make this a little bit faster that's in that interval you can pick anything in that interval you want just don't pick the end points themselves just pick something between so I pick pi over 12 here and you punch that into your calculator you'll get about 5.34 again I'm rounding this it doesn't matter you're just trying to figure out if it's positive or negative and so this is positive so this guy if you plug in pi over 3 you're going to get something that's negative so about .13 so this is negative here if you plug in the 3 pi over 4 4 you're going to get something that's about .24 so again this is negative and if you look at 11 piun 12 which would have a piun 12 reference angle you're going to see that you get that 5.34 again so approximately and this is positive and then lastly I'm going to pick this 5 pi over 4 if you plug that in you're going to get about 8.24 so this is positive so we want to know where this guy is greater than zero so anywhere where it's positive we're good to go we have to be very careful about our end points here so we know that they're not going to satisfy the inequality in terms of where this guy is zero but you also have to worry about where things are undefined so that's where this gets really really tricky so let me come down here and let me god you through this so that you don't make a mistake so you could say between 0o and pi over 6 this guy is going to work because it's positive do I want to include Pi / 6 no because that would make it zero so that doesn't work because it's a strict inequality does zero work no because that's where it's undefined so again normally you're solving from Z to 2 pi and basically you would include zero and exclude two pi but here zero is where you're going to be undefined so you have to exclude that there that's why I never include zero when I do these types of problems and people always complain but then you run into a situation like that and you maybe have zero included or something like that you're not thinking about it and then you get the wrong answer so that's why I exclude everything all the time so basically for this one you would say that X is strictly greater than zero and less than pi over 6 so neither would be included skip this one and skip this one now you have between 5 Pi 6 and Pi would 5 Pi 6 work work no because it makes it zero and that wouldn't be included it's a strict inequality would Pi work no because in this case it's undefined at Pi so it's undefined at 0 Pi 2 pi all those things so basically this is X is greater than 5 pi over 6 and less than Pi so it can't be Pi itself so then over here you're going to have a positive again but once again Pi cannot be included and 2 pi cannot be included so you're going to say that X is greater than pi and it's going to be less than 2 pi now let me put the keyword or in here I'm going to try to squeeze that in everywhere and then let me come down here and write this more neatly in case you're trying to put this down for your notes so basically X is greater than zero and less than pi/ 6 then you're going to say or so you're going to say that X is greater than 5 pi over 6 and less than Pi now this is strictly less than here so you cannot combine those two like a lot of people want to do PI cannot be included because it's undefined at Pi so don't put that in there let me come down here a little bit and say or so we have another one of those we'll say x is greater than pi and less than 2 pi so that's kind of a lot but when you solve an inequality like this sometimes it gets ridiculous and if you solve one that's harder sometimes it gets really ridiculous if you want to do the interval notation it's not that bad so you're going to say from 0 to pi over 6 again both are excluded and then the union width you're going to do from 5 piun over 6 to Pi again that's both going to be excluded and then the union width looks like maybe I can't fit this let me drag it down down here so maybe I can fit it so the union width down here you would go from PI to 2 pi again both are excluded all right so once again if you want to go to Desmos and see this visually you can type in y = cose 2 x - 3 cose of x + 2 and you'll visually be able to see all the things that I was talking about so basically you're asking where is this guy going to be greater than zero so where are the yv values greater than zero well if you start here at zero notice that this guy can't be zero because again if you plugged in a zero here this is going to be undefined so anything that's greater than zero is going to work and this is going to be going up to and then excluding this Pi / 6 so when you hit pi over 6 this guy is actually going to be zero so that would be excluded so that's where you're getting that 0 to Pi / 6 with both excluded part and then the union width if you look at it here I'm going to be below zero at zero below Zer at zero so once I get to 5 pi over 6 I know that's hard to read that's another x intercept that we found and basically anything greater than that going up to an excluding pi we're going to be above zero again so we're going to say 5 pi/ 6 going out to Pi but Pi of course will not be included because you can see that this guy is going to have a vertical ASM toote at x equals PI right these guys are going to approach an x value of pi but it won't be able to actually touch it because it's undefined there so then you have the union width you would say anything that's larger than Pi so anything that's larger than Pi well this is going to work again you're going to be on this green graph right here and this is going to work up into the point where you get to anything less than 2 pi so again that's where you're undefined again so from PI to 2 pi with both being excluded so if you want to go to Desmos and check this out you can visually see our solution so anything between 0 and pi/ 6 and then also you could say anything between 5 pi over 6 and pi and then lastly anything between pi and 2 pi so all of those areas would work in the interval that we're solving over which was from 0 to 2 pi with 0 being included and 2 pi being excluded in this lesson we want to talk about the law of signs and specifically here we'll look at the easier case where we have angle side angle and then also if we have the side angle angle okay so before we jump in and look at problems I want to talk a little bit about where the law of signs comes from and so if you're not really interested in how this guy gets derived you can just skip ahead to the examples so the first thing is when we talk about solving triangles earlier in the course we saw we solved right triangles with the SOA TOA right this just means we're finding all the missing sides and all the missing angles now when we work with an oblique triangle this is a non right triangle then basically we have to use something like the law of signs or the law of cosiness because we don't have a right triangle okay so basically what I'm going to do is I'm going to start off with this acute triangle okay which let's just go ahead and call it ABC okay and I want to just draw your attention to a few things in terms of the naming conventions so you see that you have capital letter a cap capital letter b and capital letter c those are your angles okay and then when you look at the lowercase letters those are the sides okay that are across from their corresponding capital letter angle right so what do I mean by that in other words this is angle a if I go across I have lowercase a this is for the length of this side here okay that's all that is if you look at B here this angle right here if I go across this is B right so it's the length of this side here right so on and so forth you can do c as well this is angle C if if I go across okay here's lowercase C here so it's the length of this side so you just need to understand how this is named because pretty much in every book this is how they're going to do it now what I'm going to do because we have an oblique triangle we're going to split this triangle ABC up into two different right triangles okay so you see you have this perpendicular here that's been drawn and we've labeled this as H for the height okay you've seen this in Geometry where you basically have the height so you can get the area okay so basically what we would do here is think about these two separate triangles so I'm going to say I have triangle triangle and I'll call the first one a DB so a d b okay so if I think about that that's a d b like this so that's a right triangle now and then I also have another triangle that I'm going to form inside so let's call this triangle okay and let's go ahead and say this one's BDC so let's say b d c so b d c like this okay so you see the two right triangles now let's think about let's say the sign of this angle here to start so what is the sign of this angle a well basically I'm going to think about the opposite here which in this case is H right if I'm thinking about this right triangle so H over what's the hypotenuse well in this case it's going to be C right again you might have to highlight this to really see so this is a DB like this if you think about this right triangle this is your opposite and this is your hypotenuse so this is H over C okay if you use a similar thought process if you looked at C okay this angle here what is the sign of this C guy well now it's going to be the opposite which is H over the hypotenuse which is a here okay so let's grab these two results and let's just write this over here so I have the sign of this angle a here was H over C and then I also have the sign of this angle C is H over a okay what you're going to do is you're going to solve each one for H so in other words I would multiply this side by C and this side by C and over here let me make sure this is lowercase so I don't confuse anybody over here this is going to be a and this is going to be a so you multiply both sides here by a multiply both sides here by C what's going to happen is this over here is going to cancel okay and this over here is going to cancel so really I can just erase this and say that H here is equal to C * the S of A and H here is equal to a * the sign of C so since each one is set equal to H I can set them equal to each other legally so I can just erase this and I can just drag this up here or actually I'll put this over here to match the convention in the book so let me write this a little bit bigger so a Time the S of C okay is equal to you're going to have C time the S of a like this okay so what I want to do now is just divide both sides by the S of a times the S of C so I'm going to divide this side by the S of a times the S of C okay and this side by the S of a a Time the S of C so from here I'm going to cancel this with this and I'm also going to cancel this with this and so what I get is I get this a over the S of a let me make that better so let me make that better so s of a like this is equal to the C okay the lowercase C is over the S of C okay like this so what this is telling you is that basically the lengths of the sides in a triangle are proportional to the signs of the measures of the angles opposite them we could also do a similar process to show that a over the sign of a okay is equal to B over the sign of B so I could say a over the S let me make this better trying to write this as small as I can so s of a is equal to B over the S of B and you could also show that b over the sign of B is equal to C over the sign of C so I could also put C over the S of C okay so you can go through and prove that to yourself if you want to go back to the triangle but this is basically what we're going to be using throughout this lesson it's very very easy okay overall you're just going to use one part so in other words you'll either use this guy right here or you'll use this guy right here or you'll use the original we had which is a over the sign of a is equal to C over the S of C okay so let's go ahead and jump in and look at an example so we're given that b which again if you get a capital letter this is an angle is equal to 46° C is equal to 69 degre and a is 34 CM so I'm going to tell you like most people will when you start these problems just make sure you draw a little sketch it doesn't matter if it's accurate at all you just want something to put numbers down all right so I've already pre-drawn this little sketch for us so you see that your angle B here is 46° your angle C here is 69° and then this guy if you go opposite of a this is your a here it's 34 CM okay so now if I think about what I need to find I need to find the measure of angle a so let's go ahead and do that first remember your Triangle Sum property tells you that the angles inside of a triangle are going to sum to 180° so in a calculator you could go 180° - 69° - 46° and this is going to give you 65° okay so you can go ahead and put a 65 degrees in there like that okay so that's what a is going to be equal to some teachers want you to draw this out and some will say Okay I want your answer so we'll say a is equal to 65° here like that so we'll do both okay in terms of B how are we going to figure this out well you've got to work with what you have so in other words I have an angle measure a and I have lowercase a the side length Okay so I can say that a over the S of a is equal to B over the S of B okay so I can just plug in for a here this is 34 CM so this is 34 cm and then for the the sign of a you're taking the sign of 65° so the sign of 65° then for this guy right here B over the sign of B well B the capital letter B is 46 deg so we're going to put 46 degrees there and basically if you want to solve this guy for B right lowercase b you would say that lowercase b is equal to you're just going to multiply both sides by the sign of 46° so you would say 34 CM times the S of 46° over the sign of 65° now if I punch that up on my calculator I get about 26.98 59 okay let's just go ahead and round this to the nearest 10th okay so I'm going to say that would be about 27 so let's just approximate this I'm just going to erase all this and I'm going to use the approximately symbol okay so I'm going to put approximately 27 okay and this should be in terms of centimeters so let's put this up here so B is approx Pro imately 27 cm and in terms of where you need a round just get with your teacher or your assignment it should tell you now for C the mistake that a lot of people make is they use this rounded value you want to stick with the information you're given you're given a hardcoded value here of 34 CM just use that again so I'm going to want to do and let me do this on another page I'll do a which is 34 CM over the sign okay of a let me make that better so the sign of a which was again again it's 65° is equal to you're going to have lowercase C over the sign of what was our angle C that was 69° okay so again very easy to do overall you're just going to multiply both sides by S of 69° so you're going to have 34 CM times the S of 69° okay over the S of 65° okay so this is what C is going to be so let's erase this so this would give me about 35.2 3 let's say 1 again if you want to go to the nearest 10th we'll just say 35 so let's just erase this we don't need it anymore and we'll go back up and we'll say that this is about 35 okay centimeters so you can put this here C is about 35 CM so just depending on what your teacher asks for you might want to list things like this you might be asked to draw a triangle and then fill in the missing parts whatever it is just make sure you have all the missing sides okay here we got B is about 27 CM C is about 35 cm and then you're missing angle which here is 65 degrees okay let's look at another one again these are very very easy so a here is 52° B here is 67° and a lower case a is 18 cm so let's go down and so again a is 52° B is 67° and basically C is unknown but again we can find that really quickly just go ahead and do 180° - 52° - 67° and that's going to give you 61° okay so let's put that in there so this is 61° and again you might want to list this up here so let's say C is equal to 61° and then we want to find B and C here okay so I have my angle measure 52° here for angle a and then side a is 18 cm so again I can just set this up I can say 18 cm 18 cm over the sign of this angle a which is 52° is equal to let's just go ahead and start with B so I'll say B over the S of 67 degrees okay and if you haven't noticed basically this guy right here always ends up being multiplied by this so a shorter way to do this would be B is equal to you know basically the sign of whatever this angle is times this ratio here okay that's all you need to do so let's go ahead and say B is equal to 18 cm times the S of 67° over the sign of 52° let me just erase this so I get about 21026 let's say five okay and let's round us to the nearest 10th so that's going to give us about 21 so let me erase this and I'll say that this is about approximately 21 and don't forget your units so we'll say centimeters and again I'll answer over here so B is approximately okay approximately 21 CM okay for C let's use a little shortcut okay let's go ahead and say that c is equal to again I want this guy right here 18 cm okay over this angle here the sign of this guy so the sign of 52° but I'm multiplying this by again remember you would have had C over the S of C which is the S of 61° but we multiply both sides by that guy okay so we can clear it so this is times the sign of this angle here which is 61° okay no need to do two steps when you can do it in one so if we go ahead and crank this out you're going to get about 19.97% here is 61° let's just look at one more so we have a is equal to 83° C is equal to 53° and lowercase C here the side length is eight yards again we have our lowercase a this side here we need to solve for lowercase b this side we need to solve for and we don't know the measure of this angle B I always start with the measure of the angle because it's the easiest so you want 180° minus 83° - 53° and that gives me 1 well I would get 44° okay so this is 44 4° so let's erase this let's put B is equal to 44° okay so in this particular case let's just go ahead and set up both of them really quickly so we know we need to find this guy right here so I'm going to use what's given here I have this side C here which is eight yards and I have the angle measure C which is 53° so in each case I'm going to use this ratio where I say eight yards okay over let's go ahead and say the sign of 53° okay so if I wanted a so a would be equal to what it would be this times the sign of this angle here okay so the sign of this 83° okay so you set that one up and then for b b would be equal to again I'm going to keep this part the same so eight yards over the S of 53° let me make this better so s of 53° and then times the sign now of of in this case the angle B is going to be 44° okay so 44° there all right so let's start with a let's punch that up on our calculator so if I round this to the nearest 10th I'll just go Ahad and say it's about 9.9 so 9.9 and this is yards so I'll say approximately let's answer it twice so here I'll put approximately 9.9 yards and I'll do it up here as well so a is approximately 9.9 yards and then for this one let's just go ahead and say that's about seven right again rounding to the nearest 10th so it's about seven so let me get rid of this and I'll say this is approximately seven yard and then up here B is approximately 7 yards okay so again we found that a is about 9.9 yards B is about seven yards and then this missing angle measure of capital letter B here is 44° okay so another thing you're going to see in this section is about the area of a triangle so this is going to involve side angle side so you're going to have two sides okay and the angle between them so the method used to derive the law of signs can also be used to derive a little formula to find the area of a triangle so I want you to recall from basic geometry the area is equal to 12 time the base time the height okay so here our base would be this lowercase b right because that's this side here right and then our height would be this H here okay this guy right here so basically I want you to think about our right triangle ADB again again if we did s of a so s of a this is equal to again H C so H over C now if I solve this for H I'm going to get what I'm going to get S of a * C is equal to H so s of a s of a * C is equal to H so I'm going to plug this in right here into this formula okay so I'm plugging this in right there so what I'm going to get is that the area is equal to2 time the base times instead of H I'm going to have basically C * the S of a so let's erase this and put C time the S of a okay now similarly you could go through and calculate other formulas okay using a similar thought process so you could say the area is equal to2 times this now let's go a * B okay then times the sign of c and then lastly let's go area is equal to 12 times let's see what didn't we use so it's going to be a * C and then times the sign of B okay so notice that this capital letter that's involved okay this angle is always the one that's not involved here I know these are lowercase but if you look you have lowercase b and lower case C this is uppercase a lower case a lower case b uppercase C lowercase a lowercase C uppercase B okay and that's because if you look at the way this is set up if you have side B and side C so this is side B and this is side C so where's the angle between them well that's going to be this guy right here it's going to be a right if I looked at a and b so those sides so this is side A and this is side B okay what's the angle between them it's going to be C if I look at the last one which is a and C so if I look at a and I look at C what's the angle between them it's going to be B okay so this is a little formula you can use okay because in some cases you're not going to know the height you're not going to know what H is okay so you can basically use this instead all right let's look at this example here so B here is 5 km a here is 10 km and C here is 121° okay so again you sketch yourself a little diagram and basically we have again the side B here is going to be 5 km if I look at the side a here it's 10 km and we know that c is 121 degrees so again you have this side here you have this angle you have this side here again notice that the angle is between them okay it's very important so how do we do this well again we'll say the area the area is equal to I'm going to go 1/2 times again you take these sides so 5 kilm so 5 kilm times you're going to go 10 kilm so 10 kilm and then times let me kind of this is going to run off the screen so let me drag this over a little bit so then times the sign of the angle between them so the sign of 121 degrees okay and so let's just go ahead and crank this out and actually since this doesn't fit on the screen very well let me kind of simplify this here so 5 * 10 is 50 50 ID 2 is 25 kilm * kilm is going to be kilm s so let me get rid of all of this and say that the area the area is going to be 25 kilm squared okay and then basically times the S of 121° so let's just go ahead and punch this up into our calculator just do 25 * the S of 121° and I'll go ahead and say it's about 21.4 again if you're rounding to the nearest tenth so the area is about okay approximately 21.4 and it's going to be kilometers again this is squared very important that you square your units there because again you're end up multiplying the kilometers times the kilometers so you get kilm squared okay let me throw a wrench at you so this is kind of a harder problem a harder variation of this problem so you have a is 11 ft C is 69° and B is 76° so you've seen the pattern before with side angle side right where the angle is between them but this one you have two angles in a side so if I go down to my little sketch here you see you have these two angles in this side here okay so basically to solve this guy to find the area I either need to figure out C or I need to figure out B okay it doesn't matter which one you do you're need the same answer but basically if you had C you have a side an angle on a side if you had B You' have a side and angle on a side so let's just go ahead and do B okay so we're given this 11 ft here so to to use the law of signs I've got to First work with a find the angle measure here so that's going to be 180° - 69° - 76° which is what that's going to be 35° okay so this right here is going to be 35° so let's erase this okay so now again if I'm trying to find this guy right here I say that basically a which is 11t 11t over the S okay of a in this case is 35 degrees so the S of 35° is equal to you'll have this lowercase b again it would be over the S of 76° but we know we're just going to mulp this by the sign of this angle measure here which is 76° okay so basically I'd have the sign of 76° time 11 ft over the sign of 35° so that would be about 18.6 okay so let me erase all of this and I'm just going to say and I'll just put this up here that this guy right here is about 18.6 and this is in terms of feet so now it's really easy to go back to our problem we're going to say that the area Okay is going to be equal to2 times again I'm taking this guy right here which is the 18.6 okay feet and then times this guy right here which is the 11 ft okay and then times the sign of the angle okay between them so in this case it's 69° so times let me do this down here the S of 69° okay so let's just punch this into a calculator so if I round this to the nearest 10th I'm going to say this is about 95.5 ft squared okay or square feet however you want to say that so let's go ahead and say this is about about 95.5 ft squared or again square feet however you want to put that in this lesson we want to talk about the law of signs where we have the ambiguous case what's given is the SSA or the side side angle all right so in the last lesson we talked about how to basically solve these oblique triangles these are non-right triangles when you're given two angles in one one side so this could be angle side angle which is Asa or it could be side angle angle right SAA now that's the easy case you just go immediately into using the law of signs okay you can just basically set up these proportions and you can solve for what you need to solve for when you get to the ambiguous case you're going to see side side angle and now you have to dig into the details of the problem so before I get started into that I just want to recap some things that we're going to need for this lesson some of you might find this quite boring so you can just skip ahead to the examples but I think you might find it quite beneficial okay so first and foremost let's revisit the law of signs we have this a over s of a is equal to B over s of B which is equal to C over s of C so we just write this in a compact form you're only going to use two of these at once right so you might set these two equal to each other like this okay if you're trying to solve for something in there or you might set these two equal to each other or you might set this one and this one equal to each other whatever you need to do okay so you're basically going to use two at one time now alternative you could also write this like this so you could say the S of a over a is equal to the S of B over B which is equal to the S of C over C now really quickly because I know somebody's going to be confused remember we name the triangles okay generally speaking you'll see triangle ABC in your textbook so capital letter a capital letter B capital letter C remember those are angles okay the lowercase counterparts in other words if I think about capital letter a and lowercase a the relationship is basically the lowercase letter is going to be opposite of the angle okay that's named with that capital letter so you have Capal letter A if I look at lowercase a that's going to be the side opposite of that angle okay it's they going to be the length of that if I look at lowercase B it's going to be the side opposite of this angle okay with capital letter B that's the name of then if I look at lowercase C it's going to be the side opposite of this angle named with an uppercase C or a capital i c however you want to say that now let's think about some of the properties for an angle let's call it beta of a triangle okay so this is really important when you think about the sum of the angles in a triangle we know it's 180 degrees okay we know we can't have an angle in there that's going to be 0 degrees it's going to be not possible and also we can't have an angle that's 180 degrees that's not possible as well so we know that the sign of beta has to be greater than zero can't can't be zero because again you can't have a Zer degree angle okay and you also can't have an angle that's 180 degrees if we go back to the unit circle remember if you think about a sign value of zero it's going to happen basically if your angle is 0 degrees or if it's 180 degrees and I just said that's not possible okay so if we go back we think about the next part of this the sign of beta is again greater than zero and less than or equal to one well remember when you think about the sign okay we know the range of sign in general is from negative 1 to one but again if you're working between 0 and 180° not including 0 or 180 well then basically you're going to be between Z and one where Zer is not included and one is right if you have a 90° angle you can have a sign value of one okay that means you have a right triangle so let's go back here I said s of beta equals 1 if this is the case then beta is 90° again you have a right triangle lastly we need to understand this because when we have two solutions here this is going to come in so remember the concept of a reference angle if you have sign of some angle beta let's call it 30 de so s of 30° this is equal to S of 180° - 30 okay 30° here if I take that away from 180° this would be 150° so this is s of 150° we know this is true because of the concept of a reference angle if you go back and you think about this let me erase this highlighting here if I had a 30° angle the sign value is a half if I had a 150° angle okay and I didn't draw that really well so let me try that one more time the reference angle there is 30° so the sign value is also a half okay again this is going to come up in a moment because we're basically going to be using the inverse to find find our missing angle measure and it might be possible that we have one that's let's say 30° and another that's 150° okay so we need to keep that in mind now just a few other quick things about triangles that you may or may not know assuming the triangle has sides that are all of different lengths the smallest angle is going to be opposite the shortest side let me highlight that the smallest angle is opposite the shortest side the middle valued angle is opposite the intermediate side the largest angle is opposite H the largest side okay so you need to make sure you understand that because again it's going to come up today all right let's jump in now and look at some examples I'm going to go through all the different scenarios and give you an example of each so the first thing we're going to look at would be where there's no such triangle and you're given an acute angle so we're given some acute angle a and remember opposite of a we're going to have this side here which is named with lowercase a so we're given some angle a and then opposite of that we have some side a these are both going to be given to us we're also getting given this side B here okay so the first thing you can think about if you wanted to show that there was no such triangle you could say that H is going to be greater than a and it's going to be less than b okay so you could prove it this way you could also show that the sign okay of this guy B if you try to set this up with the law of signs you're going to end up with something that's greater than one okay so I'll show you both for this example that's coming up in a minute now you might be thinking how in the world am I going to get H so let's talk about that real quick let's get rid of this okay let's think about if we formed a right triangle here okay and we thought about this angle a well the sign of a is equal to what it's the opposite over the hypotenuse so in this case because I formed a right triangle here the opposite of it is H and then the hypotenuse is B okay if you wanted to solve this for H you would say that H is equal to B okay whatever this side is here times the sign of a the angle that you're given okay this is pretty easy to remember just going to erase this real quick and just drag this up here let me go down and we'll read off our problem and we'll come back and we'll just kind of use this as a visualization of it so we're given here that a is 61 degre so notice this is an acute angle right then B is equal to 29 cm and a is equal to 20 cm so again this is your angle here and then this right here because it's the lowercase letter counterpart this is the side that's opposite of that okay so when we set this up we'll say this this is 61° okay that's the angle you're given opposite of that you have this side let's just go ahead and say again it's a so this is 20 cm and then B we're told this is 29 CM so again one way you can prove that there's no such triangle you can basically plug into this guy right here and you could say okay is H falling into this category is it greater than a and less than b if it is you can stop and say no such triangle okay so basically I'm going to have B here which is going to be 29 CM so H is is equal to 29 CM times the S of a so the S the S of 61° so this is going to be an approximation so H is about or approximately we'll say 25.4 CM okay so you can see here that if basically if I set up this inequality I could say that 20 cm which is a is less than 25.4 CM okay which is H is less than you could say 29 CM which is your value for B okay so we fall into this case here where we could say there's no such triangle okay it's not possible the other thing you can do and let me just erase this if you wanted to do it this way you could say basically that the sign okay the sign of this angle here 61 degrees okay so the S of a over a itself in this case is going to be 20 cm is equal to again the S of this angle B over this B measure here which is going to be 29 cm now you can multiply both sides by 29 centimeters let me just kind of slide this down a little bit so we'll say times 29 cmers over here so we'll say this is equal to the sign of B so the S of B is approximately let's just go ahead and say it's about 1.3 in terms of this guy and again the sign value is not going to be greater than one so that's another way you can prove that there's no such triangle so however you want to do it basically if you're going through the process of trying to solve the triangle and you get something that doesn't make any sense it's just nonsense you want to stop and say no such triangle okay so let me just put this over here and just say no such triangle all right let's look at another scenario this is one where we end up having a right triangle so basically here you're going to see that this a value here is exactly equal to H okay so if a is equal to H and then you could also say either a or H because they're equal let's just go ahead and say a is going to be less than b then we know that we have a right triangle now another thing here since you notice you have b as a right angle you could say that the sign of B remember this is going to be equal to one so two ways to kind of check this let's go ahead and look at our example real quick so we have a is 30° C is equal 2 * of 5 in and a is equal to thek of 5 in now instead of giving you a letter B I gave you a letter C just to throw a wrench at you this is not any harder if you want to work with this on this particular image you just need to cross this out let me use a lighter color you need to cross this out and put capital letter B cross this out and put capital letter C cross this out and put lowercase C cross this out and put lowercase b okay just adjust for what you're given they're not always going to give you things in the exact format it is in your book okay so the first thing is I want to prove that a is equal to H how do I do that well I'm given that a is equal to the square root of 5 in I'm given that and I messed that up I'm given that c is equal to 2 * the of 5 in okay and I'm given that this angle a is 30° now remember for the height for the height we can say the sign of the angle a the sign of the angle a is equal to again it's the opposite which you could use as H here okay over the hypotenuse which in this case we're going to use C okay so basically H is going to be C * the S of a so H you write this over here H is going to be C which we know is 2 * Square 5 2 * 5 in times the sign of in this case is 30° right the sign of a so what is the sign of 30° we all know from the unit circle it's a half so let me put a little border here H is going to be equal to 2 * theun 5 in time2 okay my writing is getting a little sloppy here so let me change that and put inches there so we know or actually I can move this out to the end so that we don't have to mess with it for right now we see this two we cancel with this two and basically H is equal to theun of 5 in what is a square of 5 in so you've already shown that a is equal to H and also you can see that a is less than b in this case we swapped B and C around so again you got to keep that straight in your mind so this value here okay is less than this value over here it's all you need to worry about now in this particular case you could also show that the sign of this angle in this case would be the angle C now is equal to one so let's go down and think about this so again you use your law of signs you could say that the sign of 30° over thisare < TK 5 is equal to you would have your s of C over you would have two * squ of 5 Okay so let's just go ahead and say this is a half so I have2 and then I'm dividing by square of five so I'll just say down here it's 2 * 5 okay so 1 over 2 * of 5 like this is equal to I'm going to multiply both sides by this so I'm going to multiply both sides by 2 * of 5 and so look this is going to cancel here with here and this is going to be equal to the S of C so the sign of C is equal to one okay so a little rough sketch of this if you want to look at it basically if you want to find the rest of this so you you already know that c is a right angle right the sign of that guy is one we already know that opposite of that we're given that this is 2 * squ of 5 in we know a is 30° opposite of that square of 5 in so we need to find this and this so where does that come from if I want to find this 60° here the measure of angle B again what I want to do if I come back up here let me erase this I would say that b is equal to 180° - 30° - 90° which is going to be 60° right remember the three angles in a triangle have to sum to 180° so I can just take 180° minus the measure of one angle minus the measure of the other angle that gives me the measure of the remaining angle right which in this case is B so B here is equal to 60° now how do I find this lowercase b how do I find this guy right right here and let me write in here that c is 90° so we can solve this completely well lowercase b is going to be found using again the law of ss so I'm just going to go in this case I'll go sare < t of 5 in over I'll go the S of 30° so I just did a over the S of a is equal to I'm going to say B okay B which is unknown over the sign of B so the sign of 60° so the sign of 60 deges we know the sign of 60 degrees is square of 3 over2 right so let me put this in here as < TK 3 over 2 and we know the sign of 30° is a half okay so this is a half okay so let's think about simplifying this a little bit we know this would basically be 2 * the of 5 in okay if I simplify that so this would be 2 * of 5 in over here this would be 2 B over the square < TK of 3 so what I'm going to do to get B by itself I'm going to multiply both sides by the square of 3 over 2 so this will cancel and this will cancel okay and over here square of 3 2 so this is going to cancel and basically you get that b is equal to 3 * 5 in which I could write as a 15 in right so basically B is the square otk of 15 okay inches so let me erase this and so now we've solved the triangle right we know all the sides and all the angles again if you go back to this little picture here or this little image whatever you to call it again this is 60° that's for B and then the side opposite is square Ro of 15 in all right let's look at another scenario so this will be our third scenario so here you're going to see that basically your a is going to be greater than or equal to your B and you're going to have exactly one triangle okay so you can also say that zero is going to be less than your sign of this angle here B okay which is going to be less than one okay it can't be equal to one because you don't have a right triangle in this particular case Okay so let's go ahead and look at an example and so we have C is equal to 39° B is equal to 28 in and C is equal to 30 in now again I switched the letters around on you don't freak out so this is your angle that you're given this side here lowercase C is the one opposite of that okay so you just plug back into the diagram that you're looking at but for this one you really don't need it if you notice that this guy right here okay this right here is 30 in it's opposite of the angle you're given if this measurement here which is 30 in is greater than or equal to this measure here you know you're in the scenario where you're can have exactly one solution you can just start pounding away in okay you don't need to do anything else so the first thing I'm going to do is I'm going to solve for my angle B so let's start with that so I'm going to say that the S of B over B okay is equal to the S of C over C okay we know we can multiply both sides by B here okay so you just get rid of this and then I can just plug in what is C it's 39° let me change my color here so this is 39° what is lowercase C here it's 30 in so this is 30 in okay and then what is B here it's going to be 28 in okay if you want to cancel the units visibly you can so this is going to cancel so basically let me make this a little bit better okay so this is 30 right there basically you punch this into a calculator getting an approximation so you say sine of B is approximately I'll say 5874 so 5874 and I'm using more decimal places here because I'm going to take the inverse okay so to find B here you want to take the inverse of this guy okay so you could wrap this entire thing in parenthesis here so this entire thing here plug it into the calculator Go Side inverse of that or you can do this this way you're going to be close enough so I'll go ahead and say s inverse of this guy right here 0 5874 is going to be approximately we round it to the nearest 10th let's just say it's 36 Dees so let's go ahead and say this is 36° and again this is going to be my B here okay so let me erase all of this and I will go ahead and say that b is going to be 36 Dees okay and this is approximately so let me make the approximately symbol here so then what is a I'm solving for a here again this is pretty easy I'm going to do 180° - 39° - 36° this is going to give me what so 180 - 39 is 141 take away 36 you get 105 again this is an approximation okay it's not going to be exact so let's say this is approximately 105° and if I want the lowercase a here again I can just use the law of signs again so again go with what you know so I'm going to go that 30 in which is this lower case C over the sign of 39 degrees so the S of 39 degrees again the sign of C let me make that better I'll say is equal to you're going to have your lowercase a over your s of 105 de again I'm going to multiply both sides by the sign of 105° so I might as well just do it over here so the S of 105° so again remember this is an approximation so you can show that here if you want to and say this is approximately a okay most teachers will allow you to put equals there as well and I'll go ahead and say this is about 46 in so let's say it's about 46 in okay now let me show you this sketch really quickly and then I'm going to come back and show you why there can't be two solutions okay how you could prove this so let me show you the sketch I made so here you have your angle a which is 105° opposite of that you have your side a which is 46 in again this is an approximation for B it's 36° opposite of that okay over here you got 28 in and then C is 39° opposite of that you've got 30 in okay so let's go back up let me show you why there can't be two solutions we saw let me just erase or actually let me just box this off we got an angle measure here of 36° for B okay approximately but let's just work with that if we look in quadrant 2 and we think about okay what angle has a reference angle of 36° again we talked about this earlier you would take 180° you would subtract away 36° and that would give you what that would give you 144° okay so remember when you did that inverse sign operation that's only going to give you something in quadrant one or four okay we're working with positive values here so you're always going to get something in quadrant one if you want to go into quadrant two and get something okay the something that has the same sign value in this case 144 de you've got to use that concept of a reference angle the problem here is that if I say that b could also be equal to this even if it was approximately well we're given that c here is 39 right degrees so 144 plus 39 is going to give me 183 remember the sum and let me make that better the sum of the three angles in a triangle is 180° so this is not possible I can't have something that's 144 degrees because I'm going to go over okay so you can go ahead and exclude that right away if you want to prove that to yourself you can but again if you get an angle in this case it was C it's 39° and then the side opposite there which is 30 in is bigger than this B that they're giving you okay the side in that position which is 28 in you can go ahead and conclude that you're going to have exactly one triangle and it's not going to be a right triangle in this case all right let's go ahead and take a look at one final scenario with given an acute angle and its side opposite so you'll see this diagram or this image here is pretty busy a lot of stuff going on because we're going to have two triangles here so let me just quickly recap a few of the things that we've already seen so we saw that if a was basically less than H and H was less than B that there was no triangle right so there is no triangle in this particular case but notice that a here is less than b so this is kind of one of the things where you have to really be careful about what you're doing because if you see a is less than b you could have no triangle or if we're going to see in a second you could have two triangles then we saw that if a was equal to H okay and you could say a or H whatever you want to do if a basically is less than b then you would have a right triangle and you would just have one obviously so one right triangle okay then we saw that if a was greater than or equal to B you would basically have and let me not put a comma there you would basically have one triangle so one solution one triangle here what you're going to see is that a is greater than H and it's less than b so when this happens you're going to have exactly two triangles okay so some students don't like to check things this way they don't like to involve H and you don't have to right you can always use your law of signs so if you're using the law of signs with this guy when you try to find the sign of the angle B it's going to be greater than one okay so that's how you know you have no triangle here when you try to find the sign of B it's going to be greater than zero so you can say it's going to be greater than zero and less than one so this guy is going to give you your two triangles this guy gives you your no triangle okay so it's important to understand that when we return here let me go ahead and write this again so a is greater than H and less than b okay when we return here a is going to be the same in each case it's just how it's drawn okay so you'll see this in your book this little swing here so basically the first way you could draw a triangle this way okay so basically you'd have this B sub one here that's your angle B you have your angle C here and then down here I put C sub one most books don't even put a c down there but basically C sub1 would be from A to B sub1 like that okay so the length the entire thing that would be your side C now if we swing it this way okay and we draw this a different way the length of a is the same but now my triangle is going to look like that okay basically and so when you think about C sub 2 now it's just going to be this short piece here right so that's my side C so the angle C changed okay from basically here to just a smaller version okay and the angle B changes as well so you're going to see the angle B change the angle C change and the side C change when you look at these guys and draw two triangles let's go ahead and look at an example here so we have C is equal to 25° we have b equal 14 km and Cal 8 km so again when I think about this I have the angle and it's opposite okay don't worry about the letters it's completely irrelevant okay don't get hung up on that if I want to find H again I'm using that same formula take this guy okay so this measure here that's not involved in the angle and it's opposite multiply by the sign of the angle you're given it's all you're doing so I'm going to take 14 kilm multiply by the sign of 25° and this is going to be equal to H okay and I like my variable to be on the left so let's just go ahead and put it like this so H equals this we'll say h is approximately let's go ahe and say 5.9 kilm okay so if I think about this is this guy less than this okay and greater than this yes right if I line this up I would say 5.9 kilm is less than 8 kilometers okay which is less than 14 kilomet so I am in a scenario where I have two triangles okay that proves that now again if you just looked at this piece of information and this piece of information you can only conclude that you either have two triangles or no triangles okay so you'd have to use the law of signs to get further into that so let's go ahead and do that right now so we can start solving this guy we know we're going to have two okay so I'm going to first start by finding my angle B okay so how do I do that again I would say that the sign of C okay so the sign of 25 degrees over C which is 8 kilm then basically is equal to the sign of B over okay I'll go ahead and say this 14 kilm here now I know I'm going to multiply both sides by 14 kilm so let me slide this down a little bit and let me do this in a different ink so I'll say 14 kilm we know that this is going to cancel here and let me just erase it from here so what would this give us we get about 739 let's just go ahead and say six so I'll say the s sign of B is approximately 7396 okay so you want to use your inverse sign function so I'll say s inverse of this guy right here so 7396 again if you wanted to wrap this guy right here okay inside the parentheses when you're doing s inverse that's fine too okay this is going to be more accurate now if I do this I'm going to get about 47.7 de so let me erase all this and I'll say B okay B and I'm going to put sub one is about I put equals so let me put approximately 47.7 de okay so how do I get B sub2 remember we talked about this earlier when we said we would use the concept of a reference angle all we'd have to do is go into quadrant 2 and say okay well 180° minus this reference angle 47.7 de would give me 132.5 okay degrees so I'll say this is approximately 132.5 degrees okay now I'm going to separate these two I'm just going to put you know uh maybe a border here or something like that because we do need to find a and you can put sub one and you want to put lowercase a here sub one right so this is all for the first triangle okay so I'm going to put approximately and approximately and let's just erase this real quick okay so another thing here too and I didn't do this basically you know this is going to work because again when you add this so 132.5 de plus this 25° here this is not going to be than 180° we saw earlier in the lesson we tried to add these guys together we got 183° so it was greater than 180° so it didn't work here you have room for another angle because this is going to be 157.5 de okay so you see here for sure this is going to work and you're going to have two triangles okay so let's erase this let me also add down here I'll put a sub 2 okay and I'll put lowercase a sub 2 like this so let's work on this first one here so if I want a sub one all I need to do is go 180° - 25° - 47.7 de so this is going to give me we'd say 107.3 Dees so this is going to be 107.3 de so let's erase this and again we're going to use our law of signs so I'll go ahead and say that I have 8 kilm over the S of 25° so this is basically C over the sign of C I'm going to multiply this by in this case I would have the sign of 107.3 de okay this will be equal to a again if I did a over the sign of a I would end up multiplying both sides by this that's how I got that okay so if you punch this up on a calculator we'll say this is about 18.1 and this is going to be in terms of kilometers so we'll say 18.1 kilm okay let's erase this and I'll quickly show you the sketch of this guy so you see here your angle a is 107.3 de opposite of that I've got 18.1 km your angle B here here is 47.7 De opposite of that I've got 14 km and then your angle C here is 25° opposite of that is 8 km okay so let's go ahead and look at the other guy and again we're going to solve things in a similar way so now for a sub 2 I'm going to have 180° - 25° minus now 132.5 de so this equals 22.7 Dees okay so let's put 22.7 Dees in like that and now I just need a sub 2 so again I'm just going to do the same thing so I'm going to go 8 km over the S of 25° times in this case I'm going to do the sign of this guy so 22.7 Dees okay and this is going to be equal to a so let's go ahead and punch this up on a calculator you're going to get a a is approximately 7.3 and this is in terms of kilm so 7.3 kilm like this so let's erase this again I've drawn this triangle angle for us so if we look at our sketch here we have our angle a which is 22.7 De then opposite of that we have this 7.3 km then we have our angle B which is 132.5 De opposite of that we have our 14 km then we have our angle C which is 25° opposite of that we have our 8 km So based on the information given we can see that we can draw two different triangles all right so let's move on to the final two scenarios these are pretty easy overall this is when the angle you're given is now an obtuse angle so basically you just have two situations that deal with this the first one is let's say you have some angle a it's obtuse and you're given that and you're given the side opposite of that so this a here if a is less than or equal to B so if a is less than or equal to B then you'll have zero triangles so zero triangles where you can say no triangle will exist let's look at an example so here we have a is 125° B is 18 mil and a is 14 mil so again this this a right here is opposite of this angle a here which is 125° by rule 125° angle in a triangle is going to be the largest angle so therefore it should be opposite the largest side and it's not right this right here this 14 miles this 14 miles okay is less than this 18 miles so there's no triangle right no triangle no triangle exists exists I guess I could clean that up a little bit so let's make that better so egist for the final scenario we're given some angle a again it's an obtuse angle and we're given the side opposite okay and then we can basically say this side opposite a in this case is greater than b so if a is greater than b here so if this is greater than this we will have exactly one triangle so let's look at an example so here we have a is4 de C is 8T and a is 35t so clearly this right here okay which is is opposite of this right here is larger than this other guy you've got so what I would do here start by finding my angle measure C so I would do the S of C is equal to let's go ahead and say the S of 114° so the sign of a over a which is 35t times it's going to be C which is 8T okay so remember you divide s of C by 8T and you multiply both sides by 8T so that's how we set this up so we know that this is going to cancel with this and so basically what you'd have is about20 let's say 88 okay so let's say the sign of C is approximately I'm going to say 02088 okay so we're going to do the inverse sign so s inverse of this guy 02088 this is going to be equal to let's say about 12.5 so maybe we'll go 12.1 so let's say that c just erase this and put C is approximately 12.1 de okay so let's erase this all right so now let's go ahead and find my B so this is approximately what well I'm just going to do 180 degrees - 114° - 12.1 de this is going to be equal to let's say 53.9 Okay degrees so we'll say this is approximately 53.9 de okay so we have that and lastly we just need this lowercase b or this side measure so let's go ahead and put this as approximately what again I'm going to use the information I'm given so I'm given a and then a here so I'm going to say since I want lowercase b I'll say B is equal to or put approximately whatever you want to do I'm going to say it's going to be this 35 ft okay so over we'll do the sign of 114° so a over the S of a and I'm going to multiply this by the sign of this B here so the sign of 53.9 De again we would have had B over the sign of B so I just multiplied both sides by the sign of B so that's how I got that set up okay so basically what I'm going to have here is let's say about 31 ft okay so let's say approximately 31 ft okay all right so we've solved this triangle when you get the obtuse situation it's much easier when you're dealing with an acute angle and the opposite side okay that's when you really have to dig into the details and really think if you're get in up two angle and its opposite side that's pretty easy right you're basically going to say okay well I have this one and this one so is this one right here larger than this one right here that they gave me if it is you have one solution if it's not meaning they're equal or it's less then there's no solution okay no triangle so let's go down and look at a little sketch here so I have angle a is 114° and again opposite of that you have the 35 ft angle B is 53.9 De opposite of that you have 31 ft angle C is 12.1 de opposite of that you have 8 ft in this lesson we want to talk about the law of cosiness with the side angle side and the side side side and also we'll look at Heron's formula for the area of a triangle all right so over the course of the last two lessons we've been learning about how to solve these oblique triangles using the law of signs so we saw the easy case with the two angles in the side given angle side angle or side angle angle okay and then we saw the harder case the ambiguous case of the law of signs where we had to think about things right this is where we had two sides in a non-included angle so side side angle now we're going to look at something that is a little bit different right so we're going to have side angle side so two sides and an included angle and then also side side side right so all three sides no angles when these two cases present themselves we're going to be using the law of cosiness now okay and we're going to say that a unique triangle can be determined so we don't have to go through all this stuff that we saw with the ambiguous case of of the law of signs all right before we get going on some problems I want to give you some insight in terms of where the law of cosiness comes from I'm not going to go through deriving each one because there's going to be three of them but I'm going to show you one of them and you can get the other ones on your own if you want now some of you might not find this very interesting some of you might find it extremely boring just skip ahead to the problems if that's you okay so the first thing is you'll notice that we have this triangle here and it's an oblique triangle and it's named ABC okay typical now you'll notice that we've put this guy on the coordinate plane okay where B basically is at the origin right so we have this 0 comma 0 you'll see for C you have this a comma 0 so where does this come from well opposite of angle a you have side a right labeled with this lowercase a right so this is a length you could think about this as the line segment BC okay right here okay well basically this has a length of a so the x coordinate there would be a the y-coordinate would be zero because you're still on the X a axis right so that's where we get the a comma 0 there okay now up here this one might be a little bit more confusing but basically to get this guy right here and this guy right here you could start off by calling them X comma y okay just start with that generically then what you would do is you would take the sign of this angle B okay so how would we do that remember this is an obtuse angle here so what you do is you'd form this right triangle here okay like this and you'd say okay the sign of B is going to be equal what it's going to be equal to the opposite which in this case is going to be y over the hypotenuse which in this case is C so y over C okay and then let's go ahead and do the cosine of this angle B also here you have the adjacent over the hypotenuse so this is X over C okay so so far so good now where does this come from this C * cosine of B and C * s of B well basically you're just going to solve for y and then solve for x and then plug in in each case okay that's all you're doing so I'm going to multiply here both both sides by C so by C by C and basically over here this is going to cancel this is going to cancel so y equals c * the sign of B and see here you see C * the S of B okay and here x equals c * the cosine of B and again C * the cosine of B so you're just taking what x is and you're plugging it in there okay you're taking what Y is or what Y is equal to and you're plugging it in there so that's where all this stuff comes from so let me erase everything and now let's talk about a formula to find basically the length this guy right here so this would be using the distance formula we know how to do this from basic algebra but again I'm going to call this side here B right lowercase b because it's opposite of angle B so how would we do this with the distance formula we would say what let's just write it here real quick so we have D the distance between these two points on the coordinate plane is equal to the square root of you have your x sub2 - your x sub 1^ 2 plus your y sub 2us your y sub1 squ okay so from here now what I want to do is I want to replace D with B because the distance from here to here it has a name lowercase b so let's put that in there okay and this is the square root of again it doesn't matter how you pick these points I'm going to label this guy right here as x sub 2 y sub 2 and I'll label this one as X sub1 Y sub1 just to make it easy so I'm going to plug in C * cosine of B so C * the cosine of B and then basically minus a this quantity will be squared and then plus we're going to do this C * s of B so C * sine of B and then you would do minus 0 I mean I can put that in there but it doesn't really do anything so I'm just going to close this and square okay so let me copy this real quick and let's paste this here so we have a lot of room to work okay so basically the first thing you would do here I mean you could Square everything inside first but I'm just going to square both sides and I'm gonna say b^2 is equal to remove the square root symbol so you're gonna have C * cosine of B minus a this quantity is squared plus I'm G to have c times the S of B this quantity is squared okay so basically what we want to do here is just expand everything right so this one right here this is just multiplication so you can just apply your exponent each case right so this will be c^ 2 time sin^2 B okay but for this one right here you have two terms right so you have to expand that or basically foil it out right in this case it's a special product formula so I'm going to have this as c^ s okay first guy squar times cosine squ B okay so this whole thing is squared right C is squared cosine squ B right because cosine of B is squared okay then you'd have minus 2 * this guy times this guy so 2 * C * cosine of B then times a okay if you want to write AC there we can kind of rearrange that make that a little bit cleaner and then after that you'd have plus the last last guy which is a s okay so let me put all this together and let's go ahead and say don't forget this B out here so I'm going to say this is b^2 okay is equal to this is going to be c^ 2 * the cosine 2 of B and then minus we have 2 a c times the cosine of B and then plus a 2 and then don't forget about this one so then plus C SAR time s^ SAR B okay so I can erase this at this point I'm just doing that for kind of side work or scratch work so I don't want that to confuse you so I'm just gonna get rid of it okay let's move this up here now one thing that might not be completely obvious but you remember from probably earlier in the course we saw something where we said sin sare theta plus cosine squ Theta equal 1 you have a cosine squ B and you have a sin squ B here okay so if I can get those two together they're going to be equal to one okay so this is what we're going to do let's go ahead and say we have b^2 is equal to I'm going to go c^ 2 * cine s b okay and then I have this as a term here this is a term and then this whole thing is a term so I'm going to put this over here so plus c^ 2 * sin 2 B okay and then what didn't I use I didn't use this and I didn't use this so let me put those after so let me go plus a^ 2 and then minus 2 a c times the cosine of B okay so some of you will see where I'm going with this other will be lost and that's fine I'll show you right now basically what I'm going to do here is factor out the c^ S okay so I'm going to have the b^2 is equal to we'll have this c^2 out here and then inside you'll have cine 2ar B plus s^ 2 B okay so that's taken care of and then plus you have your a^ 2 minus 2 a c time the cosine of B okay so this right here we know is equal to one right by definition so you can can erase this and just put a one in there or just erase it all together because you don't need it anymore so let's move this down here okay slide this down here and basically you're done so this is the law of cosine so b^2 is equal to c^2 then plus a^ 2 then minus 2 a c multipli by the cosine of B now in your book it's going to be rearranged they'll put a^ s here first okay and they'll put c^2 here in the second position but again because this is just addition here it wouldn't matter okay but this is how it's going to be listed in your book so I want to be consistent so where would this be useful let's go back up I have things in terms of B squ right so if I knew my side a and my side C and my angle here B okay my included angle well then I could find the length of this guy be here or the distance between these two points using that formula okay now additionally we have two other forms and I'm not going to derive this obviously it would take forever but you can do a similar process so I'm going to do a^ 2 is equal to in this particular case you're going to have B and C now so b^2 plus c^ 2 and then minus you'll have 2 * B * C time the cosine of a okay and I think you see the pattern right so basically you think about the side that you are solving for it's the other two sides that you're going to need right so B and C in this case you get B and C here and it's the cosine of the angle a opposite of this side a and then the last one you probably guessed it so c^2 is equal to you're going to have a and B now so a^ 2 + b^ 2 - 2 * a b time the cosine of again opposite of C would be Big C cosine of capital letter C however you want to think about that so to give a little summary of this little law of cosiness we're going to say that the square of a side of a triangle is equal to the sum of the squares of the other two sides then you could go less twice the product of the two sides and the coine of the included angle okay so this is a very easy pattern to memorize just go ahead and write it down your notes we're going to use it today okay let's look at some examples here I'm going to do two with side angle side and then two with the side side side and then we'll look at some Heron's formula area problems so the first thing you'll notice here is you have side angle side right so I have side B is 21 in I have angle C is 100° and I have side a is 27 in so let's go ahead and write this down so we had B was equal to 21 in okay we had a was equal to it was 27 in and then we had angle C was 100° okay so with this guy again what you're trying to do in this particular case you're trying to figure out when you start what is the missing side right so what is this C over here what is this so equals what so let's come back okay we're just going to plug into it it's very simple so C squ is equal to hopefully you didn't forget this so it's the two other guys so it would be a 2 + b 2 okay then minus it's 2 * this guy time this guy so 2 * a * B and then times the cosine of this angle here that you have so the cosine of C okay once you work with this a few times it becomes very very easy to remember it and you're just going to plug in and basically hit this on a calculator and you're done okay so let me show you what you're going to plug in here and we'll approximate it together you know that you want C by itself okay and C is going to be positive right so we don't need to consider plus or minus square root so I'm just going to say c is equal to the square root of all of this okay so let me grab that and I meant to just grab this part and it doesn't seem to be doing that so let me write it a s plus b ^2 minus 2 a and then times the cosine of C okay so C is equal to the square root of let's plug in here so for a I'm going to plug in 27 okay you can put the units in if you want or not in the end it's going to come out to just be inches so I'm just going to leave them off okay then Plus for your B it's going to be and I forgot to square this for your B it's going to be 21 so 21^ 2ar and then - 2 * your 27 * your 21 and then times your cosine of 100 okay degrees now you're just going to punch this into a calculator and I get about 36 let's say 97 and I'm going to do 37 okay so I'm going to say my side C okay this is a lowercase C is equal to 37 and then the units are inches okay all right so let's erase this and let's put here that c is approximately 37 in now we need to find our two missing angles here okay and let's talk for a minute about using the law of signs or using the law of cosiness if you go back and you think about it you can always in any of these formulas you can solve for this guy okay and you can end up using your inverse cosine function to solve okay so that's fine if you want to use the law of signs you have to be very very careful why remember if you have an obtuse angle and a triangle you can only have one of them okay and it's going to be opposite the largest side if we go back here remember assuming the triangle has sides that are all of different lengths the smallest angle is opposite the shortest side the middle Val valued angle is opposite the intermediate side and the largest angle opposite the largest side okay so what am I getting at here well basically what's going to happen is in this particular case you have an obtuse angle already okay so it is opposite of the largest side and we can clearly see that but let's say that you didn't have an obtuse angle let's say you had an angle that was like 30° for example well when you go to find your other missing angles if you're using the law of signs you might run into a problem okay because when you think about the inverse sign function you can't get an obtuse angle out of that okay you can you can only get something in terms of from Quadrant One between 0 and 90 deg okay that's all you can do so if you had something larger than 90° you would get the wrong answer right so for example if the angle measure should be 110° if you plug in the inverse sign function there you would end up with 70° okay that's just how it works so what we do to fix this is we find the smaller of the two remaining angles using the law of signs okay because if I find the smaller one then it's not possible for it to be the obtuse angle because again a triangle can only have one obtuse angle so if I'm finding the smaller one it can't be that one so I'm good to go to use the law of signs okay so let's go ahead and find the angle measure B right because it's opposite the smaller side right this one's 21 this one's 27 okay so let's go back and let's go ahead and say that we have the sign of in this case C which is 100° okay over C which is 37 in is equal to the sign of B over B is 21 in okay so we already know how this is going to work we're basically going to multiply both sides by 21 inches so we'll say that the sign of B the S of B is equal to will have 21 inches times you're going to have the sign of 100 degrees this is all over your 37 in okay and you can go Ahad and cancel the units inches out if you want visibly just punch this into your calculator and you're going to find that the sign of B is approximately let's just say it's 0.55 I'll just say 89 okay so if you use your inverse sign function on this I'll say s inverse of this 0.55 89 this is approximately we'll say 33.9803409 de okay so that would be 80° and then minus your 34° which would give me 46° okay so a is approximately 46° Okay so we've solved this guy let's go back up and I'll just say this is 34° and I'll say this is 46° all right let's look at another example with side angle side I'll go faster on this one so basically here you have a side C that's 16 cm you have an angle B that's 89° and you have a side a that's 28 CM so let's say C is 16 cm let's say that angle B sorry I was writing a but angle B is 89° and then let's say that side a is going to be 28 CM okay so again the first thing I want to do is find the missing side right so we're going to say B is equal to what so again just use your law of cosin very easy so we're going to say b^2 so b^2 is equal to again which ones are not involved so it's going to be a and C so a 2 + c^ 2 and then minus 2 * a * C time the cosine of this angle here which is B okay so we're just going to plug in so let's just go ahead and say this is B is equal to the square root of I'm going to plug in for a that's going to be 28 and forget about the units you don't really need them because it's going to end up just being centimeters so then Plus for the C it's going to be 16 and then for the minus two * again it's going to be and this needs to be squared two times for this is going to be 28 and then times 16 and then times the cosine of in this case it's 89° okay let's punch this into a calculator so I get about 32 let's just call it that so I'll say it's about 32 CM so we'll come here and we'll say that our side B is approximately 32 cm and again if we go back up here we can just label this I'll say this is approximately 32 cm and again you just want to take out of these two sides here you want to take the smaller side okay okay and then go opposite you want to find that angle measure so I'm looking for angle C here because again it's not going to be obtuse it's not possible so I can use my law of signs if you want to use the law of cosines you can do that as well it's up to you okay so let's go back and let's figure out angle C so I'm just going to say the S of C is equal to so I'm going to write 16 cm here again you can leave the units off because you know you're going to end up canceling them out but I'll just show it here and then times it's going to be basically the sign of 89° which is basically B okay over B which is in this case 32 CM okay I'm going to visibly cancel these units punch this into a calculator you'll see that the sign of C is approximately I would say 4999 let's just call it that but if you wanted to make this a little quicker you just take this whole thing plug it into your calculator for the inverse sign function and you'll have C okay I'm just doing this in two steps to make it a little bit easier to comprehend so0 49 999 and then I'll say sin inverse of this guy5 4999 would be approximately let's say it's about 30° okay so that would be for C okay so let's erase this so we'll say C is about 30° okay and then again if you want to find a you're going to do 180° - 89° - 30° which is 61 degrees so I'll say that a is approximately okay I'll say 61 de so we'll say this is about 61° and we'll say this one is about 30° all right let's change things up just a little bit and look at the case where we have the side side side so we have all three sides and to solve the triangle you need to find the three missing angles so in this particular case let me just read this off we have our side a which is going to be 17 miles let me make this equal sign a little bit better okay and then we have our side B our side B which is going to be 15 miles and then we have our side C lastly which is going to be 28 miles okay so based on the information about a triangle that we know we know that this angle C here would be the largest angle okay because it's opposite of this largest side okay so when you have all three sides you're going to first find the largest angle using your law of cosiness okay so let me grab this real quick let me cut it away let's come to a fresh sheet so we have a lot of room to work okay and I'm going to be focusing on finding our C so what is this guy what is the measure so let's turn back to our law of cosign and I'm going to use the one with the cosine of C involved so in other words let me just erase this real quick so I have some room let me box this off I'm going to say or actually I can actually do this up here I'm going to say that c^2 is equal to remember take the other ones that aren't involved so you're going to have a 2 plus your b ^ 2 and minus 2 * a * B okay * the co of C so the idea here is that you're going to isolate the cosine of C okay so it's going to be equal to something okay you're going to be able to plug in for a b and c here okay and so then you can use your inverse cosine function and find out what C is approximately now when you talk about the law of signs again with the inverse sign function you run into problems because the inverse sign function is not going to be able to give you something in quadrant 2 right so you're not going to be able to get an obtuse angle out of that with the inverse cosine function you can right it returns something from 0 to Pi in terms of radians or from 0 to 180° so if you get a negative cosine value you know you're in quadrant 2 and you have an obtuse angle if you get a positive cosine value okay you're in quadrant one and so you have an acute angle okay let's quickly solve this guy for the cosine of C in the next example I'll just give you the formula but I just want you to see where this comes from in your book so the first thing I would do is I would subtract a^ s and I subtract b^2 away from each side okay so Min - a^ 2 and then minus b^ 2 so it's going to cancel over here let me write this down here I'll say we have c^ 2 - a^ 2 - b^ 2 is equal to the ne of 2 a * the cosine of C okay this is multiplication here so what I'm going to do to isolate this I'm going to divide both sides by -2ab okay so divide both sides by -2ab let me erase this from up here so we can write this formula up here this is of course going to cancel so that's gone so you'll have the cosine of C is equal to you'll have your c^ 2 - a^ 2 - b^2 over the negative of 2 a b now because this is a fraction I can keep my negative down here or I can get rid of it and I can bring it up to the top like this right so I could distribute it to each part so I could say this is c^2 plus a 2 plus b^ 2 okay like this and then I'm going to rearrange it so I'm going to say and this is how it's going to be in your book so that's why I'm doing this so the cosine of C is equal to I'll say this is a 2 + b^ 2us c^ 2 all over 2 a b and I know it takes a while to get these things and in tutorials we're used to just G being given the formula but I want you to see where things are coming from in your book so you're not always just getting things hand okay so let me grab this up here and basically now I'm just going to plug in okay and to make this quicker if you wanted to you could just do the inverse cosine of whatever this is to give you C right so you could say C is equal to the cosine inverse of so I'm just going to plug in for a b and c here you can leave your units off you know they're going to cancel so a is going to be 17 so that's squared plus b is going to be 15 that's squared minus C is going to be 28 that's squared all over 2 times you're going to have your 17 and then times your 15 okay so just punch this into a calculator use your cosine inverse function and let's say that's about 122° okay if we round to the nearest 10th so I'll say here that c is approximately 122° okay and if you look at this value right here you'll notice that this guy is negative okay so if the cosine value is negative you're going to get an obtuse angle right because you're getting something in quadrant two if this guy ends up being positive you're going to be in quadrant one and you're going to have an acute angle okay but you see how you avoid that problem versus when you're working with the law of signs right there you really have to be careful about what you're doing so now that we have the largest angle when I think about the angle a or the angle B they have to be acute angles right if you always find the first angle as the largest angle then you don't have to worry about the next two angles you can just use the law of signs okay so let me get rid of this and let's just use our law of signs here real quick so I'm going to say in each case that I have the sign of in this case 122° over whatever C is which is 28 miles again you can write the units if you want it's going to end up canceling so I'm going to multiply this by let's just start with a we'll say it's approximately what so let me go the S of a is equal to this so I'm going to multiply this by what I would have divided by here which is 17 miles okay so you can see that this is going to cancel here then to make this a little bit quicker let's just go ahead and say a is going to be equal to we're going to do our inverse sign function so sin inverse of this whole thing so the S of 122° over 28 then times 17 okay like this so if we punch that into a calculator you're going to get about 31 degrees so let's say a is about 31° okay and then once you have those two again you could just use the angle sum property of a triangle so again 180° is the sum of all the angles me make that zero a little bit better then minus you've got this one as 122° then minus you got this one as 31° you can just say this is 27° okay so this is about 27° so let's go back back up and I'll say C is 122° again this is an approximation a is about 31° and B is about 27° okay let's look at one more example I'll go a little bit quicker through this one so we'll say a is 13 ft okay and then you could say B is going to be 29 ft and then C is going to be 21 feet okay so let me grab this and let's paste this in again you're going to work with your largest angle first so look for your largest side so that's going to be here right so we're going to be looking for B so I'm just going to give you this formula again if you wrote out the b^2 is equal to a 2 + c^ 2 - 2 * a * C * the cosine of B okay and you solved it for the cosine of B you're going to end up with the cosine of B is equal to we'll say a 2+ c^ 2 minus b^ 2 all over your 2 a c okay so let's get rid of this and again if you really wanted to take this forward you could say that b is equal to the cosine inverse of you can just grab this right here so a 2 + c^ 2 - b^ 2 over 2 a c okay so you can just plug in like this and basically find out what B is approximately all right so let's plug in for a we have 13 ft so we have let me change colors here we have 13t I'm just going to put 13 forget about the units so for C I'm going to put 21 so put 21 and then for B I'm going to put 29 so let's put 29 there and then 2 * a * C so a is 13 and C is going to be 21 okay so let's punch this into the calculator and I get about 115° so B is approximately 115 degrees okay let's erase this and again I've got my obtuse angle okay so once I have my obuse angle the other ones are going to be acute for sure okay so I don't have to worry about the law of signs right it's not going to get give me an error if I use that now all right so let's solve for a so we have the S of a is equal to we go a which is basically 13 ft times let me make that three a little bit better times we're going to go the sign of B so the S of 115° and this is all over B itself which is 29 ft now we're going to end up taking the inverse of this so you might as well say that a is equal to let me slide this down a is equal to the inverse sign of this guy right here okay so let's punch this into the calculator and let me visibly cancel these units and you'll get about 24° so I'll say a is approximately 24° so let's erase this and then again for C all you have to do is do 180° - 115° - 24° and that's going to give me 41° okay so this is about 41° so let's go back up and we'll say B is about 115° a is about 24 degrees and C is about 41 degrees all right let's talk about this Heron's formula and how to find the area of a triangle given the three sides so side side side you can also use this with side angle side okay and I'll show you that in a moment but it's much slower than the formula we saw with the law of signs okay so with this guy right here I'm given that basically let's say side a is going to be so a is 10.7 M then let's do side B that's going to be 9 M and then side C is 10 let me put the C there so C is going to be 10.5 M okay so let me grab this cut this away bring this over here let me paste this in here okay let me give you a little formula okay so for Heron's formula basically the first thing is you're going to find the perimeter of the triangle and divide it by two okay in your book you're going to see this called s so s is equal to you could say 1/2 times you sum the sides so a plus b plus C okay you know how to get the perimeter of a triangle so in this particular case s would be equal to we'd have 1 12 times your a is 10.7 I'm going to leave the units off for now so 10.7 + 9 plus 10.5 is 30.2 okay so this would be 30.2 you can put meters in there if you want it's up to you and then basically I would divide this by two so you could say s is 15.1 okay so let's say 15.1 again if you want to put meters there you can in the end you know with area it's whatever units you have squared okay so if you want to work with that that's fine so then from here what you're going to do is you're going to plug into this kind of long formula so the area is equal to the square root of you're going to take this s and you're going to multiply it by your s minus your a then your s minus your B okay then your s minus your C okay very easy formula to remember so basically if I plug in here I would say the area is equal to and again I'm going to leave the units off to make this shorter you would end up with meters meters meters meters so meters to the fourth power when you take the square root of the fourth power you get the second power or squared which we know from an area formula that's what you get okay so you're going to end up with meter squared so I'm just going to tell you that in advance okay so basically for S this is going to be 15.1 then times you would do 15.1 minus 10.7 which is 4.4 okay then s which is 15.1 minus B which is 9 that's going to give me 6.1 okay and then s which is 15.1 minus your C which is 10.5 which would give me 4.6 now of course you don't have to set it up this way you can just punch it straight into the calculator I just wanted to give you a little insight into what's happening so I'm just going to crank this out in the calculator and see what I get so I get about let's say 43.2 if I round it to the nearest 10th for the area okay so I'll say the area of that triangle will say is approximately I'll go 43.2 okay let me slide this down don't forget your units you want to put your meters squared because again if you had meters here here here and here okay you would had meters to the 4th power when you take the square root of that you would have got met squared okay so that's where that comes from all right let's look at another example so this is side angle side here so basically with this one it's going to be faster using that area formula with sign so with that one let me just copy this down so C let's say C is 5 m we'll say a is 71° and we'll say B is going to be seven meters let me copy this real quick and let me come down here and let me paste this in here okay so you remember that formula where we said the area so the area is equal to2 times you'll say in this particular case I have B and C so BC and then times the sign of the angle that's included so 71° so what is B what is c one is 7 MERS and one is 5 MERS so let's just throw this in here so 7 m and then 5 MERS and I can't fit this in here so let me drag this down a little bit so 7 m and then five me slide this down right there so I can make that fit meters so basically you'd have what you'd have 35 m 2 over 2 let me write that like that so 35 m^ 2/ 2times the S of 71° let's punch that into a calculator I'll say it's about 16.5 M squared so let's say the area so the area is about 16.5 M squar okay so we already know how to do it that way the way I'm about to show you is going to be much slower but we're going to get a lot of practice the first thing is I would want to find this missing side lowercase a okay so how do I do that so A2 is equal to what's not involved here you have b^2 and then plus your c^ 2 and then minus your 2 * B * C and then times your cosine of the a okay so let's go ahead and say a is equal to the square root of all this stuff so b^2 is what it's seven forget about the units we'll say it's 7^ 2ar + c^ 2 is 5^ 2 - 2 * you're going to have 5 * 7 okay and then let me make that a little bit better because my handwriting is getting pretty sloppy and then times the cosine of this 71° so let's punch this into a calculator so I get about 7.2 okay but how you round here is really going to matter because you're going to end up rounding twice so I'm going to put 7.2 but that's going to really throw the answer off so let me put a is about 7.2 M okay so let's get rid of all this and then let's quickly use our Heron's formula so we would say first and foremost that s is equal to we would sum the sides so you'd have 5 + 7 + 7.2 which is 19.2 / 2 is 9.6 and this is going to be in terms of meters so you can write that and then basically with your area formula so your area it's going to be equal to again the square root of you take this guy right here again forget the units so we're going to put 9.6 times you would do 9.6 minus in this case 7.2 which is 2.4 and then you'd have times you do 9.6 minus your 7 which is going to be 2.6 and then you would do 9.6 minus your 5 which would be 4.6 punch this into the calculator and you'll get your area is a approximately I would say about 16.6 M squared here so you can see there is a difference based on the rounding okay here with the better formula we got 16.5 M squared here because we ended up rounding twice we got 16 .6 m s so again if you get side angle side you want to use that first Formula I'm just showing you can do it this way if you get this on a test sometimes they want you to do that just to get some extra practice in this lesson we want to give an introduction to vectors and specifically here we'll focus on finding the magnitude and also the direction angle all right so when we work with vectors we're basically going to see this as a directed line segment okay so this is going to have what we call a magnitude or a length and then also O A Direction which is given by the arrowead okay so if I look at this directed line segment with an initial point of o in a terminal point of P you can see that it has a magnitude or a length Okay and it also has a direction given by again the Arro head now the way you would notate this would vary by textbook but I would take my initial Point here of O okay and I would list that first that tells me it's the initial point then I would take my terminal point p and I would list that second okay that tells me that's the terminal point so the order here when you're using two letters is going to matter okay and then I want to put a little arrow on top okay so this is the typical way that you will see this notated now you could also use a lowercase letter okay something like let's say V okay something like that and do the same thing right but this one is going to give you the information about hey o is the initial point and P is the terminal point this guy just tells you you have a vector okay now let's erase this for a moment I want to show you a few other ways that you might see this notated so in your book you might see this okay where they just put that guy like that okay you also might see this listed with a bar on top and then lastly you might see them just use a bold face font okay so you might see them just do that now I'm going to stick with this for this tutorial in a few minutes I'm going to start talking about the magnitude or the length of this guy okay and the notation for that is basically where we're going to use these vertical bars so the magnitude of this Vector op I could put in vertical bars to show that or sometimes the textbook will use two okay two vertical bars okay so go with whatever your textbook goes with I'm going to use single bars here because that's what my textbook uses now if we look at these two directed line segments or these two vectors the first thing is if I'm using the two capital letters to name them this guy would be named op okay that Vector op here is going to be your initial point and here's going to be your terminal point right so this guy would be the vector o like this now this guy is not going to be the vector op okay because the initial point is here now and the terminal point is here right these have different directions look at the way the arrow heads are facing okay this one is facing this way this one is facing this way okay so different direction same magnitude but this one's going to be called P so this will be Vector P okay let me get rid of this and make that a little bit better now I can say right here that this Vector op is is not equal to this Vector p and the reason they're not equal is because for two vectors to be equal they have to have the same magnitude or the same length and also they have to have the same direction these do not have the same direction even though they do have the same length so I can say this Vector op is not equal to this Vector P all right so now let's think a little bit about the magnitude and how we find that also something called the direction angle and then the component form okay so let me start with the component form because I think everything kind of builds off that so first and foremost you see on this sketch the way this is drawn we have our initial point at 0 comma 0 or the origin okay so let me label this this is 0 comma 0 then we have our terminal point our terminal point at this a comma B okay so because this guy has an initial point at 0 comma 0 we call this a position Vector okay or we say it's in standard position now when you have a position vector or when it's in standard position you can just list the coordinates for the terminal point that's all you need because you know the initial point is at the origin okay you don't need any more information so the way we list this is called component form okay so here I've named my Vector U so the vector U is equal to you'll see we have these angle brackets here and we're just listing the A and the B all this is is the X and the Y okay from the ordered pair we're just listing it inside of those angle brackets so if I said that I had a vector U okay is equal to let's say I had 3 comma 4 well I know and I didn't make those brackets very well I know that the X location is three and the Y location is four in terms of the point okay the terminal point for that Vector so I would start with the initial point at 0 comma 0 and then I would draw a terminal point at 3 comma 4 okay and I would just draw a directed line segment that's all I'm doing okay so additionally we also have something called let me just erase all this we don't need this anymore we also have something called the magnitude of the vector which we already talked about it's the length of this guy right so if I said what is the length of this guy right here so if I just drew this line right here what is the length of that well again normally if we form a right triangle okay we can do this with the distance formula let me just call this side C for a moment so I'll say C which is the hypotenuse typically okay so C I'll say c^2 is equal to this a 2 plus b^2 so we know if we want C by itself we say C is equal to the principal square root we don't need the negative one of a^ 2 + b^ 2 okay well now instead of C I just have the magnitude of U so I'm just going to replace that okay that's all I'm doing so the magnitude the magnitude of U is just equal to the square root of a 2 + b^2 now in the case where you're not given a position Vector okay you're going to either have to put it into component form or use your original distance formula right remember the distance is equal to you have the square root of you have the quantity xub 2 - x sub 1^ 2ar plus the quantity y y sub 2 - y sub 1^ 2 okay in this particular case your point down here is 0 comma 0 so if you do this it just becomes a this is going to be a right for the X location a minus 0 which is a being squared then plus this would be B minus 0 okay so it would be B just being squared okay so that's all it is so let's get rid of this and we see how to find the magnitude now let's talk about how to find the direction angle because this one's quite confusing okay so basically the direction angle is going to be this positive angle Theta here between the X AIS okay and our position Vector now it turns out that it's pretty easy to find this overall okay all you really need to do is some basic trigonometry in these problems if you're given a and you're given B so if you're given this and you're given this and you're trying to find this angle here what can you do well remember the tangent is y overx okay in this case it would be B over a so the tangent okay of this angle Theta is going to be equal to B a and you could say specifically where a does not equal zero okay but you're going to have to do more work than this because again if you involve the inverse tangent function remember that guy is going to only return something from quadrants one or four okay so you have to basically think about this by getting a reference angle in most cases and then figure out what quadrants are in so you're going to have to do a bit of detective work okay let's go ahead and take a look at a simple problem so I'm given a vector P okay and the component form gives me a horizontal component of -2 and a vertical component of5 okay so if you wanted to sketch this guy you would have one point at the origin okay at 0 comma 0 okay that's your initial point and then you'd have a terminal point at -2 comma 5 let's just look at the graph real quick so you see your initial point is here at 0 comma 0 okay and your terminal point you're going two units to the left and 5 units down so it's at -2 comma 5 okay it's just a directed line segment that's all it is so we think about the magnitude of this guy again that's pretty simple the magnitude of this Vector p is going to be equal to the square root of again you're going to square each one and then sum the results so I'd have -2 squared okay make sure you put in parentheses then plus you'd have -5 squared okay so you want the negative and the five being squared and the negative and the two being squared make sure you use parenthesis okay so what I'm going to do here is I'm going to say that the magnitude of this Vector p is going to be equal to the square root of let's go ahead and say -22 is 4 okay plus if you do -5^ SAR that's 25 so that would be 4 + 25 or 29 okay so this is the magnitude of this Vector P okay nothing more than just the distance formula now if we think about the direction angle this is where things get a little bit more tricky let me get this out of the way okay if I go back I want you to observe a few things and I can just erase this we're trying to find this angle Theta here okay if we think about this if we swing this around we see that this guy is in quadrant three okay so in quadrant three if we think about tangent it's going to be positive right remember this all students take calculus okay but if I use my inverse tangent function I'm not going to get something in quadrant three I'm going to get something in quadrant one so if I go back and I think about this guy remember the tangent of theta is equal to you take your yvalue or your vertical component which is neg five over you take your x value or your horizontal component which is -2 and so basically I could say the tangent of theta is really really equal to five halves right so if you do the inverse tangent on this so you do tangent inverse of this five halves you're going to get something let's say it's approximately I'll go with 68.2 deges okay so 68.2 de and again I'm just rounding so if I know that this guy right here is a result of the inverse tangent function well again if I'm in quadrant three this is not my angle okay so people mess up all the time and say okay well Theta is going to be 68.2 degrees that's wrong because I'm in quadrant 3 so if I go back you can visualize what's going to have to happen here okay I've got to go 180 Dees first okay and then I also need to add on top of that that reference angle to get the rest of the way okay so I need 180° plus the reference angle to get this angle here Theta so if I go back all all I want to do is say Theta is going to be equal to 180° plus 68.2 de and you can put approximately here let's just leave the equals for a moment I'll just go ahead and change it in the end so this is going to be what it's going to be 248 2482 de okay so let me erase this we don't need this anymore so that's our Direction angle it can be a little bit confusing just make sure you understand what quadrants you're going to be in and that when you're working with the inverse tangent function you're getting something in quadrant one or four only okay so let's go ahead and put this as 2482 de okay also I should note that if you get something in Quadrant 4 from the inverse tangent function remember it rotates clockwise to get that okay so that's not going to be your angle either okay so you got to do a lot of detective work when you're doing these okay so we found that the magnitude of our Vector p is aare of 29 and the direction angle is approximately show you something that's a little bit harder so we have this vector v here okay and the horizontal component is-7 the vertical component is -6 so again if I wanted to find the magnitude so the magnitude of this vector v this is equal to the square root of you could say -7 SAR plus - 6^ SAR okay like this so the magnitude of this vector v is equal to let's go ahead and say this is 49 + 36 so that's going to be 85 in here so basically the square root 85 really can't do anything to simplify that so that's just going to be our answer okay let's get rid of this and let's just drag this up here and again finding the direction angle usually takes a little bit longer because you got to plug some stuff in now if I think about this again the tangent of theta is equal to again it's going to be the vertical component -6 over the horizontal component -7 might as well just say this is 67 but again if I do the tangent inverse of this guy right here I'm going to end up with something in quadrant one where is this guy at if I graph it again my initial point is here at 0 comma 0 my terminal point I go 7 units to the left and six units down so it's at -7 comma -6 okay so if I look at this guy I'm going to be in quadrant three right this is the angle Theta that I'm trying to find so I've got to swing all the way around here so remember what we have to do we've got to go 180 degrees and then plus that reference angle to get there so let's go back and let's punch this into the calculator and that's going to give us about 40.6 de okay so again when you think about this I'm going to say that my direction angle Theta I'll just put approximately here I'll go 180° plus I'll go 40.6 de okay so what is this going to give me it's going to give me 220.6 de so let me erase all of this let's just put our answer here so 220.6 de so again the magnitude of this vector v is < TK of 85 the direction angle is approximately 220.6 de all right let's take a look at another example so this one involves a little bit more work so we have this Vector PQ okay so we know because of the way this is listed right from the beginning of the lesson we said that the first letter here p is the initial point right so the coordinates are 2a9 the second letter Q is the terminal point so the coordinates 1a4 okay let's go ahead and sketch this real quick so if we look at this guy we go two units to the right and nine units down so that's right here this is going to be your P or your initial Point again this is 2 comma 9 and then the other one here if we go one unit to the right and four units down we're going to be right there so that's going to be Q okay that's our terminal point so again this is 1 comma -4 so the first thing you want to do if you have a problem like this you want to put it in the component form again this is where the coordinates for the terminal point are given only because the initial point is at 0 comma 0 or the origin okay so you'll see that I already have this graphed right so this is 0 comma 0 or the initial point and then you'll see the terminal point is at -1 comma 5 so where does this come from how did I get that well if you go back here it's a simple little formula all you really want to do let me come over here and say for this Vector PQ okay and sometimes books will rename this with a lowercase letter once it's in component form that's up to you if you want to do that okay I'm going to keep it as it is so I'm going to say this is equal to I'm going to put my little angle brackets there and basically to get this position here you go with the x value from your terminal point okay so I'm going to take this one right here okay and then I'm going to subtract away the x value from your initial point so Min - two okay so 1 minus two gives us negative 1 now you want to go in that order I know sometimes when you work with these formulas if you reverse them they come out fine but in this case you want to go from the terminal point okay you want to start with that and then subtract away from the initial Point okay so the same thing with x's and y's so with this guy I'm going to take this one minus this one okay this is from the terminal this is from the initial so I want to go -4 minus a 9 which is -4 + 9 which equal 5 okay so that's where that terminal point of -1 comma 5 came from you go back you see this is one unit to the left and five units up so this is -1 comma 5 and you can call this the point Q okay now the point p is at the origin okay and because of this a lot of books again we'll just rename this as the vector v or U or whatever they want to do I'm just going to keep it like this now if you want to find the magnitude okay me make another little border here of this Vector PQ okay what I'm going to do is just take the square root of I'll take this first guy -1 squared plus the second guy five being squared okay so the magnitude of PQ okay like this is going to be the square root of - 1 s is 1 plus 5 S is 25 so this can be 26 so it's just the square root of 26 now if you wanted to come over here and use these points again you could plug into the traditional distance formula you could say and I'm just going to label this as the magnitude of this PQ guy okay I'll show you it's the same basically you're going to subtract this guy minus this guy so x sub 2 - x sub 1 so it's going to be 1 minus 2 let me let me write this out so 1 - 2 this is squared then plus you're going to do this guy that minus this guy so -4 minus A9 is+ 9 okay this is going to be squ and we want the square root of this guy okay so notice that we already did this to find these guys right here okay so we're a step ahead right so that's why we do this first and then we don't have to do all this okay so 1 minus 2 is going to be negative 1 so it's 1^ s which you know is one and then4 + 9 is 5 5 S is 25 so this becomes Square < TK of 26 okay just like it is over here okay so you see it's the same magnitude in each case now let's talk a little bit about the direction angle so Theta this is what well again let me actually erase this let me just say that the inverse tangent I think we can kind of skip ahead a bit the inverse tangent of what well I'm going to take the vertical component which is five over the horizontal component which is Nega 1 which really I could just write as neg five okay like this now when you talk about putting a negative as the argument for this inverse tangent function you're going to get something in Quadrant 4 okay so in this particular case if we think about this this horizontal component is negative this vertical component is positive if I think about it where am I going to be I'm going to the left and I'm going up so I'm going to be in quadrant 2 if you want to see this graph done its own you'll see again my terminal point here is that we go one unit to the left and we go five units up and you see I'm in quadrant two here okay so basically when I do my inverse tangent function I'm going to end up with something in Quadrant 4 and also I'm going to be rotating clockwise okay so I want to just take the absolute value of that and think about that as my reference angle and then what I want want to do in this particular case I would think about the fact that okay if I had an angle of 180° I want to subtract off and go back by the amount of the reference angle right so I want 180 de minus the reference angle to get this angle Theta here okay this angle Theta here now let's go back so I'm going to get about -78 let's say 69 so let me kind of slide this down because I don't think it's going to fit so I'll put it's approximately - 78.6 n degrees now again what I want to do is I want to think about this in terms of absolute value so in quadrant 2 I want an angle with a reference angle of 78.6 n° so all I'm going to do get rid of the negative sign I'm going to say Theta is approximately 180° minus this 78.6 n° and that's going to be about 10.31 de so let's just erase this and put it about or approximately 10.31 de all right for the last two examples we're basically going to reverse things so now we're going to be given the magnitude of a given vector and we're also going to be given the direction angle and we have to come up with the component form okay so let's think a little bit about this angle Theta and let's think about if I had this guy B and I had this guy a well let's think about some relationships here I know that the sign of this angle Theta is opposite okay over the hypotenuse which in this case would be the magnitude of this Vector U okay so I can say that the sign of this angle Theta is equal to we'll go ahead and say this B here which is the vertical component over the magnitude of this Vector U okay so I'm going to be given the angle Theta and I'm going to be given the magnitude of this Vector U so I should be able to plug in there and figure out what B is okay then also if I think about the cosine so so the cosine of theta what is that well that's now adjacent so the adj let me erase this that's the adjacent over the hypotenuse so this would be a over the magnitude of this U guy okay so we can use these two we'll just basically solve them for a and b okay and then we'll plug in and we'll have our answers so let me just paste this in here and let me drag this over a little bit we can solve this for a and b to come up with a nice little formula to use here so basically if I multiply both sides here by the magnitude of this Vector U you're going to see that b is equal to you would have the magnitude of this Vector U and then times the sign of this direction angle Theta okay then here a is equal to the magnitude of this Vector U times the cosine of theta okay let's get rid of this now we have a little formula to use so I'm given the magnitude of some Vector n okay so I want to find the component form and basically all I'm going to do is think about this and this so I want the a and the B so the a goes here the B goes here okay that's all we're doing so the first one for the a you're going to think about going down here the magnitude in this case of our Vector n is 27 so I'm going to have 27 let's put a equals this so 27 times the cosine of the direction angle Theta in this case is 300° now first off a 300° angle is in Quadrant 4 cosine is positive in quadrants 1 and four okay so we know that this guy's going to be positive now 300° has a reference angle of 60° and cosine of 60° is going to be a half so I can just put times a half here and this is basically 27 over2 okay so 27 over two okay then for B for B it's equal to what so in this case again the magnitude of this Vector n is 27 and times the sign of this guy 300° so again think about the fact that in Quadrant 4 your sign is going to be negative right sign is positive in quadrants one and two only so in quadrant four it's going to be negative okay so I'm going to put negative and then the reference angle is 60° so s of 60° is square < TK of 3 sare < TK of 3 over two okay so all this is you're not going to be able to really simplify this too much you're going to say this is negative let me put this out in front so the negative of 27 * > 3 over 2 let me close my brackets down so this would be your component form for this Vector n let's look at one more of these so again all I'm going to do here let's go ahead and write this vector k like this is equal to I'm going to set up my brackets and again this is a comma B so a is equal to again with the horizontal you're thinking about cosine right so all it is is this guy right here times the cosine of this right here so 32 times the cosine of 120° now this again has a 60° reference angle but now I'm in quadrant 2 2 and cosine is negative in quadrant 2 so this is going to be half right SO2 so this end up being -6 right so if we put this in here as -16 because this will cancel with this and give you 16 okay and obviously this is times negative 1 so that's where the negative comes from okay let's get rid of this let's think about B now so B is equal to you're going to take the same 32 the magnitude of this vector k times now the sign okay the sign of this 120° well I'm in quadrant 2 okay I have that same 60° reference angle but now sign is positive okay so this is going to be times < TK 3 over 2 and so this is going to cancel with this and give me 16 so this is 16 * the squ of 3 okay so 16 * the S < TK of 3 so that's going to be our component form and maybe I should make that three a little bit better okay for this vector k again your horizontal component is NE 16 your vertical component is 16 * the < TK of 3 in this lesson we want to continue to talk about vectors and here specifically we're going to focus on operations with vectors all right so let's start out here by talking about how to add two vectors together so the sum of two or more vectors is also a vector this is referred to as the resultant vector or some people will just call this the resultant okay so basically let's say I had this simple example here very generic and on the left I have this Vector with a horizontal component of a and a vertical component of b and I want to add this to this Vector on the right where there's a horizontal component of c and a vertical component of D so for my resultant vector or just the resultant if you want to say that basically I'm going to do a plus c so a plus c okay then I'll do a comma then I'm going to do B plus d so B+ D okay that's all it is it's very very simple you're going to sum these horizontal components so a plus c that's going to give me the horizontal component for my resultant vector then I'm going to sum these vertical components so B plus d that's going to give me my vertical component for my resultant vector and you can do this with more than two let's say for example I had something like two and then three here okay and then let's say plus we'll do maybe four and then five here and let's say we do plus let's say we do six and then you could even do like negative so negative 9 okay so something like this so all we would do if we wanted to find the sum here you would say that you would have 2 + 4 + 6 so 2 + 4 is 6 6 + 6 is 12 okay so that's the horizontal component for this resultant vector and then for the vertical component you're going to do 3 + 5 +9 so 3 + 5 is 8 8 +9 is going to be negative 1 okay so this resultant Vector the horizontal component is 12 and the vertical component is1 all right let's also talk about how to show this geometrically we can do this with a triangle and we can also do it with a a parallelogram so using the triangle method let's think about this we have this Vector a and this Vector B they're being added together so let's say I started by just sketching Vector a so this is the initial point and this is the terminal point so I've just sketched Vector a then for Vector B I'm going to use my horizontal and vertical components to sketch this guy again the initial points here your terminal point would be here let's say okay and then your resultant vector or basically the vector a plus the vector B is going to be this guy right here so this Vector right here so you'll notice that the initial point for this Vector is going to be the same as the initial point from the vector a then the terminal point from this Vector is going to be the same as the terminal point from this Vector B okay so that's all you're doing so for this example let's go ahead and say we have the vector T okay we're going to add to this our Vector U our Vector U so what is this going to give us so for the vector T you have a horizontal component of six and a vertical component of8 for the vector U you have a horizontal component of three and vertical component of four so in this resultant Vector here to get the horizontal component I'm going to take this six and add it to this three right so just sum the horizontal components so 6 + 3 is 9 okay for the vertical component I'm basically going to sum the vertical components here so8 plus4 is going to be4 okay so this guy basically would have an initial point at the origin okay it would have a terminal point at 94 so let's look at this graphically let's start out by just redefining things so we had this Vector t Okay in component form this was 6 comma 8 so we had a horizontal component of six a vertical component of8 then for the vector U we had a horizontal component of three and a vertical component of four then for this resultant Vector so the vector t plus the vector U we basically what we said 6 + 3 was 9 we said8 + 4 is -4 right so a horizontal component of N9 and a vertical component of4 okay so looking at the graph here you see for Vector T we started with an initial Point here at the origin right it's just in standard position so the terminal point is going to be right here which is at 6 comma ne8 okay so we want six units to the right and we want eight units down so that's as expected this Vector U when you try to sketch that now the initial Point has to be at 68 so you're using the three as the horizontal movement right so I'm just going three units to the right and then I'm using the four as a vertical movement so now I'm going four units up so one 2 3 unit to the right that puts me at nine on the x axis and then four units up 1 2 3 4 that puts me at ne4 on the Y AIS so this point right here would be 9 comm4 okay and this is consistent with what we said right so basically this Vector right here this resultant Vector has the same initial Point okay as this Vector t and has the same terminal point as this Vector U all right now let's think a little bit about this parallelogram method this one's a little bit harder to draw especially if you're not working on a coordinate plane but essentially what you want to do here is have the initial point for the vector a and for the vector B to be the same okay so everything's going to have the same initial point then you'd have your normal terminal point your normal terminal point so these are easy to draw especially if we're on the coordinate plane the next part is a little bit more challenging you're going to complete a parallelogram essentially what you're going to do is you're going to take this side right here and just slide it down okay so you would have it like that okay and then basically this point right here is going to help you set up this diagonal so this is going to be your resultant Vector okay now you could also do it this way okay where you take this side right here and basically copy it and slide it down and if you want to form the full parallelogram you need to do that right so basically with the parallelogram okay in case you're unfamiliar with this you're going to have opposite sides that are going to be equal in length and they're also going to be parallel right so this side and this side right here are the same length and they're parallel and then this side and this side are the same length and they're also parallel now let's take a look at this with an example we have this Vector T which is a horizontal component of ne1 and a vertical component of8 we have this Vector U which is a horizontal component of four and a vertical component of3 so if I did the vector t plus the Vector U again all I need to do is for this guy right here for the horizontal component I'm going to do -1 + 4 so that's just going to give me three okay for the vertical component I'm going to do 8+ -3 which is going to give me five okay so all I want to do is sketch this all right so let's write this out again so the vector T is going to be with a horizontal component of Nega -1 and a vertical component of eight then the vector U is going to have a horizontal component of four and a vertical component of -3 so the resultant Vector when you do this Vector t plus this Vector U this is going to be-1 + 4 which is three and then 8+3 which is 5 okay so you can see this graphically it's a little bit more busy than the triangle drawing but it's still there so if you look at the first guy so the vector T here's the initial point at the origin and then the terminal point right here is going to be at -1a 8 so one units to the left8 units up so this is- 1A 8 then if you look at your Vector U okay your initial point is also at the origin the terminal point here okay would be at this point 4 comma -3 okay so four units to the right three units down and then basically if you wanted to draw this you already know the terminal point of this resultant Vector is here okay which is at 3 comma 5 and you could basically sketch this guy right here okay and this guy right here those sides there again this side right here okay is going to be the same length and parallel to this side right here then the other one's the same thing so this side right here and this side right here same length and then are also parallel okay so when you look at this guy you would then draw this Vector here okay where basically the initial point is the same as the other two vectors and your terminal point is right there okay so it's going to be at that 3 comma 5 okay let's talk about this multiplying a vector by a scalar also very easy so let's say you had some real number K and it was multiplied by this Vector with a horizontal component of a and a vertical component of b well the result of this is going to be a vector with a horizontal component of K * a and a vertical component of K * B okay so the reason they call it a scaler is you can think about this as scaling your vector okay so graphically you can think about it like this let's say we had some Vector U okay so this is U to start if you had half times this Vector U we can see that it's half as long okay if you add two * this Vector U now it's twice as long okay compared to this original Vector U now when you get a negative involved what it's going to do is it's going to give you an opposite direction okay so basically if I had the negative of this Vector U it's going to flip the direction right so it's the same length but the direction is flipped right it's the opposite direction then if I had3 Hales times this Vector U well now it's the opposite direction and basically compared to this original Vector u in terms of its length it's going to have2 of the vector you added on right so it's like I took this Vector right here and added this Vector right here onto it and and then I gave it the opposite direction to form this Vector here which is -3 * the vector U okay let's find the -2 times the vector v and then I'll show you this graphically so we have the vector v horizontal component is3 vertical component is four so if I wanted -2 times this vector v then all I would do is -3 * -2 which is positive 6 and then I would do -2 * 4 which is8 okay so this guy right here this -2 * this vector v is going to be six for the horizontal component and8 for the vertical component now graphically the original vector v again when we look at this is going to have a horizontal component of -3 and a vertical component of four okay so you see it starts at the origin that's the initial point and then the terminal point is right here again at -3 comma 4 okay so 3 unit to the left four units up now if I look at -2 multiplied by that vector v so -2 multiplied by this vector v again I'm scaling this because this guy is twice as long okay that's the two part and then it's going in the opposite direction that's the negative part okay so basically I'm multiplying everything here by -2 so this is six and this is8 and you'll see that the initial point is the same the terminal point is here I went six units to the right eight units down this is 6 comma ne8 okay so you can see this guy right here is twice as long and it's also going in the opposite direction okay so let's combine these two concepts together so so now we have this vector v horizontal component of -2 vertical component of five and then we have this Vector W with a horizontal component of 3 and a vertical component of four we want to find -3 times the vector v plus 2 * the vector W okay so all I'm going to do is set this up and say -3 times this vector v plus 2 * this Vector W is equal to I'm just going to put a neg3 out in front of this so -2 and then five is inside and then plus I'm just going to put a two out in front and then I'm going to put a three and then a four inside okay so that's all I'm doing so what I'm going to do is just first multiply so let me copy this down or I could just put equals to keep this going I'll say basically each part of this is going to get multiplied by -3 so this would be-3 * -2 which is 6 and then -3 * 5 would be -15 okay then plus 2 * 3 is 6 and then 2 * 4 is going to be eight okay so now I just want to find the sum again very very easy I'm going to add the horizontal components so 6 plus 6 is 12 then I'm going to add the vertical components so5 + 8 is -7 okay so in the end this -3 * this vector v plus this 2 times this Vector W gives me this Vector okay with a horizontal component of 12 and a vertical component of7 all right let's wrap up the lesson and look at something that involves a concept from the last lesson so we have the magnet itude of this Vector f is 17 the direction angle Theta is 212° the magnitude of this Vector B is 22 and the direction angle Theta is 10° so what if I wanted to find the vector F minus the vector B what would I do well first I need to find the horizontal and vertical components for each so remember how to do this for the vector F to set this up I would do what I would take my magnitude 17 and for the horizontal component I'm multiplying by the cosine of this direction angle which is 212° and then for the other one it's going to be 17 times the S of 212 degrees okay that's all I'm doing then for B this Vector B what I would do is I would do 22 now the magnitude of this guy times the cosine of in this case 10° okay and then I would do 22 times the sign of this 10° as well okay so you can punch that up and get a proximations and then set up your subtraction or if you want to you can go ahead and do this all in one step so we can basically say 17 * the cosine of 212° minus 22 * the cosine of 10° okay then comma you would have 17 * the sign of 212° minus 22 times the sign of 10° up to you how do you want to do it you get the same answer either way actually the rounding might be a little bit different but let's just go ahead and do it the long way just get some practice so first I'm just going to say the vector f is going to be let me put approximately here because I'm going to approximate this 17 * the cosine of 22° I'll say that's about negative 14 let's just go ahead and say 42 okay this guy 17 * s of 212° is going to be about 9.01 okay so this is my Vector F and again this is an approximation so let's scroll down and get a little bit more room so we can stay on the screen I'll do my Vector B now so this is approximately we'll go ahead and set this up and say 22 * cosine of 10° it's going to be about 21.67% -14 -1 14.42 minus 2.67 comma I'm going to have 9.01 minus 3.82 okay and I put equals it should probably be approximately to be more precise so let me go ahead and put that like that okay to make that better and let's just go ahead and correct this out so I'll say this is approximately and I'll put my little symbol there say about - 36.9 right for 14.42 - 21 .67 then over here 9.01 - 3.82 we'll say this is -1 12.83% so to start at the lesson we're going to talk about something called a unit Vector so a unit Vector is just a vector that has a magnitude of one okay so let's say that we denote our unit Vector with a u okay when you work with the unit Vector you're going to have a special notation some people call this a cap some people call it a hat but basically this tells you you have a unit Vector okay and if you have a unit Vector the magnitude of this unit Vector again let's call it U is always going to be equal to one so for some problems you're going to be asked to find a unit Vector let's call it U that has the same direction as a given non-zero Vector let's call it V and basically to solve this type of problem you're just going to take your vector v and divide it by the magnitude of that vector v okay and that's going to give you your unit Vector U now if you're confused by me saying it it's okay the formula is very very simple so basically this unit Vector U is equal to the vector v okay so the vector v divided by the magnitude of that vector v okay so you can really think about this as multiplying by a scalar one over whatever this is so I'll further break this down and say this is the vector v times one over whatever this result is so the magnitude of this vector v remember in the end this is going to be a number right this is some number okay so if I wanted to practice on this one let's say this vector v okay has a horizontal component of neg5 and a vertical component of -12 how would I find the unit Vector U in the direction of V so basically in the direction of V but now I have a magnitude of one so what I would do is I would first find the magnitude of V so the magnitude of V hopefully you remember that formula you're going to take the square root of I'll have -5 being squared plus -12 being squared so each one of these guys is going to be squared and you're going to find the sum of those two results okay so i know5 s is 25 -22 is 144 so this would be my vector V the magnitude of that guy would be the square root of 25 + 144 is 169 and the square of 169 is 13 right so the magnitude of this vector v is going to be 13 okay so that's easy enough once you have this result you can basically just divide okay divide each one of these components by the magnitude of that vector and you're done right you've set up your unit Vector U okay but if you want to kind of go about this the long way let's just go ahead and use this formula here so you have your vector v I'm going to write it in component form so I'm going to say this is -5 comma -12 and let me make these a little bit wider okay so it's more consistent so like that and I'm going to multiply by one over the magnitude of that vector v so one over in this case it's 13 so you can see this is just a scalar multiple right so I can either multiply by 1 over 13 or I can just divide by 13 doesn't matter right so in the end my unit Vector U okay would be equal to in component form you'd have -5 over 13 comma -2 12 over 13 okay that's all you need to do I want to prove a few things to you one that the magnitude of this Vector U is 1 and also that the direction angle is the same okay so for the magnitude again it's very very easy so the magnitude of this unit Vector U is going to be the square root of so let's go ahead and say this is going to be -5 over 13 being squared plus you're going to have -12 over 13 being squared okay so5 over 13 if you square that 5 S is 25 13 s is 169 so let me write this down here the magnitude of this Vector U is going to be equal to the square root of this would be 25 over 169 then plus -12 squ is 144 and then 13 squ again is 169 okay so if you add these guys you have a common denominator 25 plus 144 is 169 so you get the square root of 169 over 169 this is obviously one and the square root of one is one so you can see the magnitude of this unit Vector U is one okay so that part works out now what about the direction angle okay you're going to see that basically if you think about the direction angle remember the tangent of theta is going to be equal to this -12 okay over this -5 right which in this case you can might as well just say it's 12 fths right but again we have to think about what quadrant we're in because if we do the inverse tangent okay we're going to in this case because this is positive we're going to get something in quadrant one okay but this is in quadrant 3 because we have a negative x value and a negative y value so I go to the left and I go down so I'm in quadrant 3 so what I want to do here say the inverse tangent the inverse tangent of this 12 fths here would be equal to or let's say approximately because we got to punch this into a calculator I get about 67.3 de okay and because we're in quadrant 3 we want to add 180° to this okay to get our measurement for Theta okay so we'll say Theta is approximately I'll say 24738 de okay so this is your direction angle you already know how to find that now if you look at it for this unit Vector U you'll notice you're still in quadrant three right this is negative and this is negative okay so you're still in quadrant 3 and I want you to think about the fact that You' still have tangent of theta is equal to now would be -12 over 13 / -5 over 13 well the negatives are going to cancel right so You' have 12 over 13 times flip this guy 13 over 5 okay and the 13s are going to cancel so I'm back to this 12 fths right so the tangent of theta is equal to 12 fths and again I'm still in quadrant 3 so I go through the same steps and I'm going to end up with a Theta here okay my direction angle which is again approximately V all right let's change our problem up just a little bit so sometimes you'll be asked to find a unit Vector U basically in the opposite direction of some vector v so all you really need to do here is think about the concept of well if I multiplied a given vector by neg1 it's just that Vector that has the opposite direction okay same magnitude but an opposite direction okay so we're going to use that concept here and just multiply our regular answer by ne1 so in other words I would have my unit vector U is going to be equal to we'll have this vector v times what I'm going to do here is put Nega 1 okay instead of positive 1 so negative - 1 over the magnitude of this vector v so let's go ahead and find the magnitude of our vector v so the magnitude of this vector v is going to be equal to the square root of so 4 S is 16 + 12 s is 144 if you add these guys together you get 160 and what I can do is I can Factor this and say this is 16 * 10 okay 16 is a perfect square so this would be 4 * theun of 10 okay so I'll say this is 4 * the < TK of 10 now all I'm going to do here is either divide everything by the negative of this guy right here or you could say multiply by negative one over that guy whatever you want to do I'm just going to make this a little bit quicker and say my unit Vector U that I'm looking for in this case is going to be I'll take the four okay and I'm going to divide it by the negative of four * theare of 10 okay and then let me put my angle brackets in here and then I'll have my 12 and I'm going to divide that by the negative of 4 * theun of 10 okay if I didn't ask you for the opposite direction this would be positive one but since I did it's a negative one there so that's why these are negative here okay let me get some room going here so let me go ahead and set this up as the unit Vector U is equal to put my brackets there so we know we're going to have to rationalize the denominator there so four divided 4 is 1 so let me put a negative out in front let me put a 1 over the > of 10 and I'll just multiply this by thek of 10 over thek of 10 to get that out of the way so this becomes 10 down here okay and up here this is the square root of 10 okay so this is the square root of 10 now it is common to not rationalize the denominator for these examples it's really up to you or your teacher just get with them and see what they want you to do for this one we have 12 / 4 which is three and of course you have the negative so let me go ahead and put three over the < TK of 10 and again I'm going to rationalize the denominator so times the of 10 over theun of 10 okay so we going to multiply by that so in the numerator have 3 * of 10 so 3 * of 10 in the denominator I would have 10 right so basically the way this guy would work is you would have this unit Vector U which again is in the opposite direction of my vector v the horizontal component isga 10 over 10 and the vertical component is3 * 10 over 10 all right so let's continue to talk about these unit vectors and now you're going to see these two extremely useful unit vectors which are defined using using this I and also this J okay so you see on the screen here we have this little graph you have a directed line segment that starts at 0 comma 0 okay and ends or terminates at the point 1 comma 0 so the initial point is 0 comma 0 or the origin the terminal point is 1 comma 0 so we have this unit Vector I okay in component form this is 1 comma 0 like this okay this is not the imaginary unit I so please don't get confused by that then similarly we have this guy that is denoted with a J so we'll say J here with my little hat so this guy it terminates at 0 comma 1 okay and basically we're going to use this guy to rewrite something like this so our vector v in component form we have a comma B okay but using those two vectors I and J basically what you can do again because the vector I in component form is 1 comma 0 okay and the vector J in component form is0 comma 1 you can take your horizontal component and multiply it by this guy so let's say the vector v i could also Define it saying it is this a multiplied by this 1 comma 0 okay then plus and I probably need to move this down a little bit so plus this B here multiplied by this guy 0 comma 1 okay so you know that these are just scalar multiples at this point if I multiply through this is basically a comma 0 okay plus this is 0 comma B okay and then if I add these two together I'm right back to what I start with so I didn't do anything legal now what you can do at this point is just replace in each case this with the unit Vector I and this with the unit Vector J okay that's all you're going to do this is called writing it in the I comma J form or you might hear it as a linear combination of the standard unit vectors I and J okay so I'm going to say the vector v now is going to be written as a times the unit Vector I plus b times unit Vector J okay so if you see something in this form to put it in component form basically again the vector v is what it's just in component form this is your horizontal component right and then comma this is your vertical component that's all it is okay if we look at a little example here let's say I have some Vector p and I want to write it as a linear combination of those two standard unit vectors I and J all I'm going to do I'm going to say this guy P this Vector is -5 this is the horizontal component multiplied by this unit Vector I okay then plus this guy four okay multiplied by that unit Vector J that's all it is again if you go back through this the long way all you're really saying is that this Vector p is equal to you'll have 5times this is what 1 comma 0 then plus four times this is 0 comma 1 and again all I'm doing is if I kind of move this up here because I'm going run our R this Vector p is what now it's -5 okay comma 0 plus and let me move this down a little bit so I can fit everything let me change this up because I'm getting sloppy with my notation there so plus over here this would be 0 comma 4 again if you add these guys together this Vector p is back to what it started with right this is going to be neg 5 plus 0 which is neg5 and then comma 0 plus 4 is four okay so it's just two different ways to write the same thing this way right here you're going to see a lot okay so you need to know how to go back and forth between the two if you wanted to graph something like this the terminal point is going to be at5 comma 4 so if we looked at this on our little coordinate plane here again the horizontal movement here is given by the5 times this unit Vector I okay it's just basically like if I said the horizontal movement here is given by this -5 comma 0 right let me make that a little bit better so basically if I started here and I went five units to the left to end up right there okay that's my horizontal movement given by this5 * the unit Vector I then for the vertical movement if if I'm starting here and if I go up well basically this is given by four times the unit Vector J right four times that 0 comma 1 like this so this is 0 comma four right I'm just moving up four units vertically so when you put those two guys together you end up with this directed line segment with a terminal point at5 comma 4 five units to the left four units up so that's all it is all right let's talk about something called The Dot product now and this is going to lead us into a topic called finding the angle between two vectors okay so we start with the dot product and the dotproduct of two vectors is going to be a real number so this is not going to be a vector this is also called the inner product okay so someone might refer to it that way now the dot product for us is going to be used to determine the angle between two vectors and we'll get to that shortly so when you have the dot product of two vectors let's say u and v so I have u and v those two vectors I'm going to write the vector U then dot the vector v okay so this is the dot product and basically all I'm going to do do is I'm going to multiply the first component here by the first component here so a * C so in other words horizontal component time horizontal component then plus second component here time second component here so vertical component here time vertical component here so B * D it's a very easy pattern to remember once you do a few examples it just becomes second nature okay so let's look at an example so we want to find the dotproduct of this Vector U and this vector v so you're going to do the vector U dot the vector v like this and again if this is confusing for you because again it's as this linear combination of the unit vectors I and J you can always put it in component form so the vector U is again in component form this is my horizontal component this is my vertical component okay do this a few times and then it becomes very easy right so the vector v this is my horizontal component this the whole thing with the negative so let me make sure I do that properly this is your vertical component okay so either way you want to do it but all I'm going to do here is I'm going to say 7 * 3 so 7 * 3 so first * 1 then plus 6 * -4 so second * second so 6 * -4 that's all you're doing okay just match things up and say these guys are being multiplied and then these guys are being multiplied it's all you got to do so once you have this set up 7 * 3 is 21 then basically plus 6 * -4 is -4 so this would be-3 so that's your dotproduct of this Vector U and this vector v okay let's look at another example if you want to pause the video and put these in component form that's up to you I think you can get it like this so the vector U is -2 * the unit Vector I Plus 8 * unit Vector J then this vector v is 4 * the unit Vector IUS 6 * the unit Vector J so the dot product of this Vector U and this vector v so U Dov would be what you're going to have this guy times this guy right so first times first -2 * 4 then plus this guy times this guy so 8 time6 8 * -6 and again if something's negative make sure you take that with it okay so let's put equals here -2 * 4 is8 and then Plus 8 * -6 is -48 and of course this is really easy this is just going to be 56 all right so now that we understand how to find the dotproduct of two vectors we're going to use this now to find the angle between two nonzero vectors so let's suppose we have these two nonzero vectors let's call them u and v okay and we want to find the angle between them and we'll call this angle Theta so we'll say that Theta is greater than or equal to 0° and also less than or equal to 180° or we could say Theta is greater than or equal to Z and less than or equal to Pi if you want to work with radians so we can come up at this little formula here this is derived from the law of cosiness so we have the cosine of this angle Theta is equal to the dotproduct of this Vector U and this vector v over you have the magnitude of your vector U times the magnitude of your vector v okay so that's all it is if you want to find Theta you're going to use your inverse cosine function okay you're just going to be punching this stuff into a calculator so really really easy all right let's look at the first example we have a vector U okay which is -6 * the unit Vector IUS 5 * unit Vector J then we have this vector v which is 4 * unit Vector I Plus 8 * the unit Vector J okay so if I want to basically set this guy up first and foremost if you're still confused by these linear combinations of the unit vectors I and J you can always write this in component form okay to ease your confusion again all I'm going to do to do that is take this guy as a horizontal component okay and then this guy is the vertical component that's all I'm going to do so if you're confused at this point just do that for a little while and then eventually this will click for you so then this guy for the vector v in component form it's going to be four comma then you're going to have eight okay that's all it is so if I want my DOT product okay so the dot product of this Vector U and this vector v this is going to be what -6 * 4 that's -4 then plus5 * 8 that's going to be -4 okay so this gives me -64 then if I want the magnitude of this Vector U okay this is going to be the square root of so I have -6 s that's 36 and then plus5 squ that's 25 so this would be the square root of 61 okay then the magnitude of this vector v let me change my ink color here this is going to be the square root of 4 S is 16 and then 8 squ is 64 so 16 + 64 would be the square < TK of 80 you could simplify that if you wanted to again you're punching this stuff into a calculator so if you're trying to save time I wouldn't even worry about it but since this is a math tutorial I'm going to take the extra step and write this down here let me just put this as a square of 61 let me write this instead of the square root of 80 let me say that 80 is 16 * 5 and 16 is 4 * 4 right so I'm going to say this is 4 * five right because 16 is just a perfect square okay so now we have everything we need let me tighten this down and put- 64 here hopefully have this copied down I'm going to get rid of it because I need the room and I'm basically going to write my formula here so the cosine of theta is equal to again the do product so of this Vector U so the vector U dot the vector v and then this is over the magnitude of this Vector U and then this is multiplied by the magnitude of this vector v okay so what I would do here is just plug in to the calculator and essentially you're going to say thata is equal to the inverse cosine of all this right so your dotproduct of the vectors u and v so this is going to be 64 then over you're going to have the product of these two magnitudes so the magnitude of vector U is square of 61 and then the magnitude of vector v is 4 * of 5 so that would be 4 * the sare < TK of 305 okay and again you don't have to write all this out if you're using a calculator just punch it into and you're done but you might want to consider the fact that you should put parentheses around this guy when you punch it into your calculator if you keep getting the wrong answer that's usually what's going on it just depends on your calculator okay so what I would do is punch this guy into the calculator as the argument for the inverse cosine function and I'll say Theta is approximately 156.35 de okay so that's basically all you're going to do and again if you keep getting the wrong answer with your calculator if it says something like domain error something like that then make sure you wrap this part right here inside of parentheses and that should clear that up for you okay let's take a look at another one we should know the formula at this point let me go ahead and do the dotproduct first so the dotproduct of u and v so U Dov here is going to be what 2 * 3 which is 6 +8 * 5 which is going to be 40 so this is 46 okay and then basically the magnitude of this Vector U is the square root of you have 2^2 which is 4 then plus you have 8^2 which is going to be 64 so 4 + 64 is 68 right so square of 68 again you could simplify that it's 17 * 4 so it's technically 2 * thek of 17 now again if you punch this into a calculator you know who cares but I'm just writing this down for the sake of our tutorial okay now the magnitude of the vector v we're going to have what you have the square OT of 32 is 9 then plus -5^ 2 is 2 5 so this is going to be of 34 now you can't simplify the squ of 34 so let me erase this and I'm going to write this as the squ of 17 okay let me write this like the square of 17 times the square of two like this okay so let's punch into the calculator first let me do this cosine of theta is equal to again you've got your dot product which is the U Dov okay so the vector U dot the vector v over you've got the product of these two magnitudes okay so the magnitude of this Vector U times the magnitude of this vector v okay and then if I plug in to my calculator what I want to do is say Theta is equal to cosine inverse of all of this okay so the dotproduct of u and v is 46 over you're going to have the magnitude of U * the magnitude of V so essentially what you have is 2 okay times 17 * sare of 17 is 17 okay so that's just going to be 34 okay and then times the square root of two now again you don't have to go through all this if you were writing this out on a test you might want to but for the purposes of punching it into a calculator who cares right it's just going to work either way so again if you're having problems with this wrap this in some parenthesis and if I punch this into my calculator I get a Theta that's approximately 16.93 de all right so before we wrap up the lesson let's talk about the concept of orthogonal vectors and this is just a fancy word to say that you have perpendicular vectors okay so they're going to meet at at a right angle so when the dot product for two nonzero vectors again u and v is zero then what's going to happen is the top part of this formula here is zero and because you have nonzero vectors down here the bottom part or the denominator will not be zero so 0 divided by something that's not zero is always going to be zero okay so then you could say the cosine of theta is going to be equal to zero and where does this happen again if Theta is greater than or equal to 0 de and less than or equal to 180° well it must be true that Theta is equal to 90° so the angle between these two vectors u and v is going to be 90° meaning these two guys are perpendicular vectors or in the language of vectors we say they're orthogonal vectors okay so all you need to do if you want to check to see if you have these orthogonal vectors is just do the dot product and see if it's zero okay that's all you want to do so to start out this problem I have this Vector U which is going to be four * the unit Vector r I Plus 8 * unit Vector J and then I have this vector v which is going to be 6 * unit Vector IUS 3 * the unit Vector J okay so if I wanted to find out if these two vectors were perpendicular or again if they were orthogonal vectors I could just find the dotproduct and see if it's zero okay so in other words with your formula cosine of theta is equal to you've got your dotproduct okay which is the dot product of u and v so U Dov okay over the magnitude of this Vector U times the magnit of this vector v okay so we know this formula at this point again if this top part ends up being zero and these guys down here aren't zero which they won't be well then the cosine of theta will be equal to zero and Theta is 90° right so these guys are going to be perpendicular so the dot product here okay we do U dot V is equal to 4 * 6 is 24 and then Plus 8 * -3 is -4 so of course this is zero right so again this top part would be zero the bottom part wouldn't be zero you can go through and find the magnitudes multiplied together and you'll see see that you're still getting zero right in the end so 0 divided by something that's not zero you just get zero right so the cosine of theta is going to be zero so theta equals 90° and of course we have perpendicular vectors if you want to see this your vector U okay which we have right here the terminal point is at 4 comma 8 right because again this Vector U was set up as 4 I okay that unit Vector plus we had 8 J that unit Vector okay and then for the vector v the terminal point was at 6 comma -3 so here's the vector v again you're going to have this 6 * unit Vector I plus you're going to have this -3 or might as well just write minus 3 times the unit Vector J okay so this is how we set this up so this guy has a terminal point at 6 comm3 this guy has a terminal point at 4 comma 8 you can see this graphically that they're going to meet at a 90° angle and therefore by definition these guys are perpendicular or you could say they're orthogonal vectors all right as a last example what if you had something like this so we have our Vector U and it's equal to we have 7 * unit Vector I + 7 * unit Vector J then we have this vector v and it has B okaye some unknown time unit Vector I minus 8 * the unit Vector J how can I find a value for B such that the two vectors u and v would be orthogonal or you could say perpendicular vectors well all you'd want to do is just set up your little dot product okay so the dot product of u and v is going to be what 7 * B so 7 * B and then plus you're going to have 7 * 8 which is 56 so you really need this guy right here you can just substitute this with a value here you really want this to be zero let me just do this down here so you really want this to be zero so that those guys can be orthogonal vectors right so I'm just going to solve this equation for B and to do that I'm going to add let's just go ah and write this as56 or minus 56 so I'm just going to add 56 to both sides of the equation okay so this would be gone let me just write this as 7 B is equal to 56 obviously this is very simple divide both sides by seven you'll find that b is equal to 8 okay very very easy to do these and again if you plugged in an eight there well then the dotproduct from your vector U and your vector v so u. V would be what 7 * 8 would be 56 and then plus 7 * 8 would be 56 so the dotproduct would be zero and in this case if B is 8 well you would have perpendicular vectors or and orthogonal vectors in this lesson we want a review of complex numbers all right so I want to start out by talking about the imaginary unit I again so we say I is equal to the square < TK of1 and this is used to allow us to simplify when we have the square root of a negative number so for example if I had the square root of let's say -4 using the product rule for Radicals I could write this as the < TK of -1 * the < TK of 4 okay I know that theun of 4 is 2 so let me just go ahead and put the < TK of1 * 2 like this and then I can replace the square of1 okay since those guys match up with this I here okay so this can go right there so this is I * 2 or just write 2 I okay so that's a way to simplify the square root of a negative number now also let me erase all this we can square both sides and come up with another definition we could say that I2 is equal to1 and we're going to use this all the time okay okay now using this imaginary unit we can form what's known as a complex number so the complex number in standard form is going to look like this you're going to have a which is your real part plus b which is going to be the imaginary part times I which is the imaginary unit okay so let me label this real quick this a is called the real the real part okay and then the B here is called the imaginary the imaginary part okay now some books will call BI the imaginary part just get with your book or your teacher in terms of what they want to do if you get this for vocabulary now in terms of complex numbers we don't often think about it this way but if you just had the number let's say three well three is a complex number okay three is a complex number because I got always say 3 + 0 I it's kind of like when you think about three as being a natural number right it's a whole number and then it's an integer and then it's a rational number and the way you show it's a rational number you go okay well 3 is equal to 3 over one right those are the same so this is by definition a rational number so it's the same kind of trick here where we just say okay well three is really equal to 3 plus 0 I because this is just zero and 3 plus Z is still three okay so this tells me that any real number that I'm working with is also a complex number so the real numbers are a subset okay of the complex numbers and if you have a complex number okay like let's say 4 + 2 r this is called a non-real complex number that's how you can differentiate between the two also you can have something that's called a pure imaginary number let's say you just had the 2i part okay so this you could write as 0 plus 2 I which is just 2 I this is also a non-real complex number but a lot of books will call this a pure imaginary number okay let's go ahead and look at some examples we're just going to fly through these again this is just stuff we've already seen if you want to simplify the square of like - 147 you would use your product rule for Radicals I would just put this1 here times you would try to factor this well 1+ 4 is 5 5 + 7 is going to be 12 I know that 12 is divisible by three so 147 will be divisible by three and if I divide 147 by 3 I'm going to get 49 so I can put a three here and I can get a square root of 49 here okay and this guy is a perfect square so the square root of 49 is 7even so what I can do here is say that this is equal to < TK of1 * < TK 3 okay then time 7 now this guy right here I can always replace with I so really I'll end up with just saying that I have 7 I okay times the square otk of three so this would be our simplified answer for this Square t of 147 again 7 I * the sare < TK of 3 all right another similar problem we have the square root of - 180 so for this guy if I think about 180 I would Factor this so + 8 is 9 so I know it's divisible by 9 so let me go ahead and put nine there and this would be 20 right so 9 is 3 * 3 those are prime 20 is 4 * 5 4 is 2 * 2 so what I'd want to do since N is a perfect square and so it's four multiply those together you have 36 so basically you'd want to write this as sare < TK of-1 * < TK of 36 * < TK of 5 okay so that's going to allow you to simplify and basically sare of 36 is 6 and square root of ne1 is I so I'm just going to put that we have I times 6 or I should really rearrange this and put 6 I and then just times the sare RO of 5 okay so simplifying these types of problems very very simple this is our answer again the square root of 1880 this is going to simplify to 6 I * theun of 5 all right so Additionally you will see complex numbers when you work with equations such as a quadratic equation we know that if the discriminant that b^2 minus 4 a c part if that's less than zero you're going to get complex Solutions okay so we're going to see that here so I'm just going to solve this I'm going to use the quadratic formula if you want to complete the square you can do that on your own so I have 2x^2 + 10x + 5 = 8X I'm going to subtract 8X away from each side to put this in standard form so this is going to cancel over here and be zero so I'm going to have 2x^2 + 2x + 5 = 0 and basically I'm going to record the values for a b and c okay so what's a it's the coefficient for X squ so this guy right here here so that's a two and then what's B it's going to be the coefficient for x to the first power so that's also a two and then what's C that's going to be your constant term so in this case that's five okay and then you're going to plug in so x equals remember this is negative B plus or minus the square root of B ^2 minus 4 a c all over 2 a okay this is something I've memorized throughout the years if you don't have this committed to memory it's okay something you can really look up quickly okay so I'm just going to scroll down a little bit and I'm going to plug in so X is equal to the negative of 2 plus or minus the square root of you're going to have 2^ 2 - 4 * 2 * 5 and all I'm doing is I'm just plugging in okay this is over 2 * 2 so I plugged in for B A and C and that's where we got this from now let's just go ahead and solve this real quick it's just a matter of punching this into your calculator so X is equal to -2 plus or minus the sare < TK of you have 2^ 2 which is 4 minus 4 * 2 is is 8 8 * 5 is 40 go ahead and do 4 - 40 that's going to be - 36 and you can see because this part right here ends up being negative you're going to have these complex Solutions okay then this is over the four because 2 * 2 is four and then the square Ro of 36 you can say that is 6 I right squ of 36 is 6 and then square of negative 1 is I let's come down here a little bit and let's go ahead and say that X is equal to -2 plus or minus I'll put 6 I over four now we can factor out a two okay and make this a little bit better so let's go ahead and say x is equal to let me pull this down here a little bit I'll factor out my two I'll have - 1 plus or minus 3 I okay then over four I'm going to cancel this with this and put a two here so we'll say x is equal to you have -1 plus orus 3 I over your two now if you want to write this in standard form for the complex number then you can put this we go ahead and do solution set notation I'll put one2 okay like this and then I'll go plus or minus we'll do three halves and then times I like this okay so that's your solution set notation maybe I can make those brackets a little bit better let me try to make this one a little bit better like that okay and of course this is two different solutions right you have - one2 plus 3es * I and then you have2 minus three Hales time I all right let's start working on some operations with complex numbers now so the first thing I'm going to do here is a simple multiplication problem I'm doing this to draw your attention to a common mistake so we know with the product rule for Radicals something like Square Ro of a Time Square Ro of B this equals the square root of a this only works when they're both not negative okay so if you try to do this here you're going to end up with the wrong answer so what you have to do when you have negatives involved like this you have to convert them over okay simplify first and then go through and multiply so what I would do here is I would say this is I * the > of 3 then times for this one let me go ahead and write this out so we'll have two * I'm going to put theun of 1 * the < TK of 4 * theun of two okay just so you can see what I'm doing squ of four is two so 2 * 2 is 4 okay and then basically you would have square root of1 which is I so this becomes 4 I * thek of two okay so from here now if you multiply you see that you have I * 4 I which is going to be what that's going to be 4 I 2 okay of 3 * of 2 we can write as aunk 6 now what is i s by definition remember earlier we said this is 1 so you can say you have 4 * -1 * > of 6 which is just going to be the negative of 4 * the > of 6 now you'll notice if you didn't follow this formula okay and go ahead and convert this first you would end up with a wrong answer right you would end up with a pos4 * the of 6 which is not the correct answer okay so you want to make sure you convert these over simplify them first before you multiply okay let's look at some addition of complex numbers really really easy topic to add two complex numbers so for example if I have a plus bi I and then I add this to let's say c plus di okay this is what add the real parts so a plus c okay and then plus you're going to add the imaginary parts so we'll say B plus d okay and this is multiplied by I that's all it is if you were subtracting okay so if this was a minus then this would become a minus and this would become a minus and that's basically it now when you subtract things away remember you can do addition of the opposite so you could always go plus and then minus and then minus like this okay so if you want to do that you can do that either way your answer is going to be the same okay so let's go ahead and get rid of this and let's work on this problem here so we have 4 + 2 I plus we have -7 + 7 I okay so what do we do we add the real part so 4 plus7 let me show this so 4 plus7 okay and let me put a parenthesis there and then plus you add the imaginary part so 2 + 7 so 2 + 7 okay then times I out here and let's go ahead and simplify this so 4 +7 is -3 and then plus 2 + 7 is 9 and then times the imaginary unit I so this is your answer here okay and it's in standard form so -3 + n i all right for the next problem let's do a subtraction so you can do this as addition of the opposite if you want so I'm just going to go ahead and say this is 7 + 2 I okay I'm going to say plus and I'm going to change the sign of everything inside it's just like if I put a plus here and a negative one outside and distributed this to each term okay that's all I'm doing okay I think it makes a little bit simpler so I'm going to change this to positive five and I'm going to change this to positive I'm going to write a one and then times I remember if something's hanging out by itself like X you can put a one in front as the coefficient okay so now I can just add so this will be 7 + 5 okay and then plus you're going to have 2 + 1 okay and then time I so this gives me 12 + 3 I as a result and again if you want to do this with subtraction you can it's going to look a little bit different so what you can also do is you can say that you have 7 minus in this case this is a negative five so be very careful okay and then you have plus you're going to have two minus this is a negative one so again be careful okay and then times I here 7 -5 is 12 and then plus 2 - 1 is three and then time I so same answer either way I think it's more straightforward if you do addition of the opposite let's look at some multiplication again with these type of problems you can generally use foil okay and sometimes you'll get longer ones just depending on the situation but just pretend like you're multiplying polinomial okay so I'm going to use foil here so I'm going to do the first terms -4 * 2 is going to be8 and then my outer ne4 * 4 I would be - 16 I my inner I would do -3 I * 2 that's - 6 I and then my last I would do -3 I * 4 I which would be-2 I squar okay so i^ s is negative 1 so let's go ahead and replace that so I'm going to have8 and then I'll go ahead and move this down and say minus 12 * 1 well Nega * negative is positive so I might as well just say this is + 12 okay and then I have this would be Min -6 i - 6 I so I can say that's - 22 I so8 + 12 is 4 and then we'll say minus my 22 I okay so that's going to be my answer there and I boxed that off a little bit too closely so 4 - 22 I all right so we should know at this point about conjugates if you have complex conjugates the formula is a little bit different so if we have something like a plus b time a minus B we know this is a^ 2 minus b^ 2 okay as an example let's say something like x - 2 okay then I have X let's say say plus 2 I know this is x^2 which is the first guy minus 2^ 2 which is the second guy which is four okay I don't need to foil this out that's the quicker way to do it but if I'm dealing with complex conjugates the sign is going to change it's not going to be minus it's going to be plus so for example if I have a + bi I * a minus bi what happens is a * a is a 2 the outer and the inner are going to cancel and then the last you have your minus b^ 2 * I 2 well this i^ 2 here is1 and that's what's changing the sign so you end up with a 2 + b^2 okay so if I look at this pattern here I have 4 + 3 I * 4 - 3 I I can just say okay the first guy squared four squar is 16 plus this three guy here the second guy just the just the B part just the imaginary part so the three guy squared which is n so that's 25 okay this is always going to be just a number so in this case it's just going to be 25 okay again if you wanted to go through this the long way we can just for the sake of completeness we would do first * first so 4 * 4 is 16 and then the outer would be -12 I the inner would be positive 12 I so you see those guys are still going to cancel then for the last you have 3 I * 3 I so that's Min - 9 i s and again the I squ here is going to change the sign and I cired that a little bit too widely but basically this is what this is 16 you might as well say minus your 9 * -1 which is 16 basically plus 9 which again is 25 okay let's talk about dividing complex numbers now so we know that if the imaginary unit I is in the denominator since I technically is the square root of ne1 it represents a radical and it needs to be rationalized right so we want to rationalize the denominator when we see this situation so if I have something like negative 3 - 4 I over 5 + 4 I what can I do remember we just talked about multiplying by the complex conjugate If I multiply this by 5 - 4 I over 5 - 4 I what's going to happen is in the denominator here I'm going to have a real number now okay we don't care what happens in the numerator we're trying to get the radicals out of the denominator so in the denominator here again I can just take this guy five would be squared that's 25 then plus this four would be squared that's 16 so this ends up being 41 okay so that's the denominator now so we're good there in the numerator you go through and just do four right so-3 * 5 is5 the outside would be -3 * -4 I which would be plus we'll go ahead and say 12 I the inside would be -4 I * 5 which would be 20i let's just go and combine like terms this would be a negative 8i so a negative 8 I so I'll put Min - 8 I there and then for the last you have4 I * 4 I that's going to be plus 16 i^ 2 i^2 is NE 1 so let's get rid of this and just change the sign so basically have -15 minus 16 so let's say this is -31 so the top here is -31 and then I'll put Min - 8 I okay now of course if you want to write this in standard form you want to break this up and say that you have3 1 over 41 let me put equals here so this is the real part and then we'll put you put plus negative or you just put minus I'll put 8 over 41 like this okay let me make that one a little bit better and then I'll pull my I out here okay so this is your a or your real part and then this is your B okay or your imaginary part okay then you have your imaginary unit I and of course you could put plus negative there if you want to that's up to you if you put minus there it's really the same thing so that's why I'm putting that okay let's talk about simplifying powers of I which is a really easy concept once you realize you go through this little cycle so I to the first power is just I i^ 2 is equal to1 I cubed is i^ 2 * I right so this is ne-1 * I which is negative I and then I to the 4th power is I 2 let me make that better so i^ 2 * I sared which is basically what -1 * 1 which is 1 Now using this pattern it's really easy to rewrite things okay in terms of one of these guys right here and then you're just going to simplify okay you can put that as your answer so this involves the rules of exponents so what I'm going to do here is I'm going to look at this right here so this number is 151 okay so thinking about what number going down would be divisible by four so I just take 151 punch this into my calculator divide by four gives me one where you're going to get 37.75 but I'm going to write this as 37 with a remainder of three okay so basically the three divided by four is where you're getting the0 75 from 37 * 4 is 148 so I can use that to my advantage I can say well this is i to the power of 148 time I to the^ of 3 so this is completely legal right I basically have just used my rules of exponents we have I and we have I so the same base we would add exponents and we' be right back to where we started now what I can do since this number right here is divisible by four I can use a little trick and I can say power to power rule I to the 4th power raised to the 37th power because again 148 divid 37 is 4 so 4 * 37 will be back to 148 right this is power to power rule then times I cubed okay now why does this work I to the fourth power we see over here is one so this guy right here is one raised to the 37th power 1 to the 37 power one to any power is always one right so I can basically just get rid of this and say this is a big fat one now times I cubed okay so I've dealt with that and now I can just go to my little reference here and say well what's I cubed it's negative I okay so that's basically what you want to do when you have a problem like this when this guy is posi let's look at another example so we have I to the^ of 38 so there's a really easy way to do this but let me do it the slow way first I'm going to write this as 1 over I to ^ 38 okay and then of course I can just do 38 / 4 and see where that puts me it's going to be 9 with a remainder of 2 or 9.5 so really I could write this I could say this is equal to 1 over I I'm going to go ahead and put I to 4th power raised to the 9th power okay then I'll put times 1 over we'll go ahead and say I raised to the power of two okay that remainder right there now we know that this guy right here is one and one to the 9th power is one so I can just keep going like this and say this is what this is 1 * 1/ i^ 2 I s is 1 so I can say this is 1 * 1 over1 so this just ends up being 1 okay so really easy to simplify this but this is kind of a long way to do it the shorter way if you're working with a negative exponent like this well now what I'm doing is I'm thinking about the smallest number that's divisible by four okay that's going to be larger than this number and I'm thinking about this in terms of its absolute value so if you think about just the number 38 okay going up okay what's the closest number that's going to be divisible by four well it's not 39 it's going to have to be 40 okay so in other words I would want to multiply this by I would want to say I to the^ of 38 time I to the^ 40 that's all I need to do because what's going to happen here is that I to the^ of 40 is just one I can multiply anything by one it's unchanged so this is going to end up telling me that I have what I to the^ of -38 + 40 okay which is I to the^ of 2 or i^ 2 which by definition is1 so for me this is a little bit quicker if I get this type of problem but you can do it the other way if you'd like all right so the last thing I want to do is review a few things about the complex plane so let's say we wanted to graph something like 5 - 7 I what we do is set up a little complex plane so instead of having X and Y we now have the imaginary okay which is vertical and the real which is horizontal okay so before this is X and this is y now this is imaginary this is real okay so if I have something like 5 - 7 I well again this is the real part we talked about this earlier and then this right here the seven is the imaginary the imaginary okay part so all I need to do is move five units horizontally okay five is positive so I'm moving to the right so I'm going to go one two three four five okay so right there and then I want to move seven units down vertically okay so I want to move seven units on the imaginary axis going down because it's negative s so one two three four five six seven and it's going to put me right there okay so this is going to be my 5 - 7 I plotted on the complex plane okay so the last thing would be the absolute value of a complex number so if we have something like the absolute value of 2 + 6 I again remember the way you would do this this is going to be the distance between this number and zero or the origin on the complex plane okay so your distance formula normally is what it's the square root of you have x sub 2 - x sub 1^ 2 plus you have y sub 2 - y 1ty squ well one of the points is at the origin right so it's 0 comma 0 so this is 0o and this is zero so this is 0 and this is zero so really I could simplify this and say the distance between this guy and the origin is the square root of drop the X sub2 and just say x^2 plus drop the Y sub2 and say y^ 2 okay now in the terminology of complex numbers instead of saying X squ I'm gonna say a squ instead of saying y^ squ I'm gonna say b^ squ okay so what I'm going to do is say that the absolute value of some complex number let's say A Plus bi is equal to the square root of a 2 plus b^2 so this is the square root of 2^ 2ar okay 2^ 2ar and make that better 2^ 2ar plus this guy would be 6^ squar so this is the square OT of 4 + 36 so it's the square root of 40 okay which is > 4 * > of 10 which in the end let me get rid of this so I have some write this and the end is going to be 2 * of 10 so this equals 2 * of 10 okay if you want to see this graphically you could plot this guy this 2 plus 6 I on the complex plane so two units to the right on the real axis six units up on the imaginary axis so that gives us this 2 plus 6i so that's this guy right here and then basically what I'm doing is just forming a right triangle right so this guy right here okay is going to be my hypotenuse that's what I'm looking for is the length of that guy so this is the absolute value of the two plus 6 I the absolute value of that complex number and again if you look at this distance from here to here let me Circle this in a different color so let me use this color here so from here to here well this is six units right so this is six units let me put a six there and then if you look at the distance from there to there it's two units so again with the Pythagorean theorem it's 2^2 + 6^ 2 is going to be equal to which in this case would be the absolute value of 2 + 6 I okay we might as well just say that square right the a squ plus b squ = c^2 well if I take the square root of each side well then I'm back to my formula okay which gives me the square root of 4 + 36 which is a sare TK of 40 okay and again we just simplify that so we know that in the end the absolute value of this 2 + 6 I is going to be 2 * the of 10 okay so you can see that graphically or just from using your formula in this lesson we want to learn about adding complex numbers graphically also we'll learn about converting between rectang angular and polar forms okay so our first topic of adding complex numbers graphically is very very easy to do as long as you understand how to add vectors you're basically good to go so we return to our complex plane which remember is also called the Argan diagram and basically we replace the X notation here for the horizontal axis with real we replace the Y notation for the vertical axis with imaginary so if you have a complex number like a plus bi okay where again the a is known as the real part okay and the B is known as the imaginary the imaginary okay part well basically I could use this to plot a little point on this complex plane okay I'm going to go a units horizontally okay and I'm going to go B units vertically okay so if I had something like let's say 3 + 7 I well the coordinates if you're thinking about this as working on your regular XY coordinate plane where you have some value here comma some value here well it's 3 comma 7 right I'm just moving three units on the horizontal axis now it's called the real axis and I'm moving up seven units okay on the vertical axis now called the imaginary axis so three units to the right seven units up would put me right there okay so I would go ahead and label this as the complex number 3 + 7 I okay if you wanted to do another one let's say we do something like -5 let's say minus 4 I okay so the coordinates there would be -5 comma -4 so I'm just going five units to the left on the real axis and then four units down on the imaginary axis so right about there let's go ahead and say this is -5 - 4 I for the purposes of making sure that you understand how to do this you should pull out a sheet of paper and just make a rough sketch of an Argan diagram we're going to use it throughout the lesson so you might want to have that handy and go ahead and pause the video and try to just graph this guy think about where it would be on your complex PL all right so hopefully you gave that a shot this -7 + 6 I again the real part's -7 so I'm moving seven units to the left the imaginary part is six so I'm moving six units up okay so seven units to the left six units up which would put us exactly right here okay so this would be your -7 + 6 I 7 units to the left and six units up all right now additionally As you move higher in math we start to think about graphing a complex number using vectors okay and specifically here we're going to use a position vector or a vector in standard position okay so this just means that the initial point is at the origin if you think about this in terms of a complex number your origin is going to be at 0 + 0 I or you could say 0 comma 0 if you want the coordinates okay so if the initial Point's at the origin and the terminal point would be at this 8 + 5 I so again it's eight units to the right and 5 units up let's see this position Vector graphically so this would look something like this you're initial point is at the origin so right there your terminal point is at 8 + 5 I so again you're just going 8 units to the right and then five units up and you have a directed line segment or vector okay so this is how we're going to show this 8 + 5i when we're asked to graph it as a vector all right so now let's use the fact that we can graph a complex number as a position Vector to show the sum of two complex numbers graphically okay so we have this 6 + 5 I plus 9 + 3 I let's just add this together real quick just so we know what we're supposed to get and then we'll go to the coordinate plane so we know we would add the real parts so we do six + -9 okay that's your first step then plus we would add the imaginary parts so 5 + 3 okay and this is multiplied by the imaginary unit I so what I want to do here is do 6 +9 that's -3 and then plus you have 5 + 3 which is 8 and then times the imaginary unit I so -3 + 8 I is your resultant Vector okay so let's look at this graphically and again you can do this with the parallelogram method or the triangle method whatever you want to do but essentially you'll see here that we have the origin here okay which is our initial point for basically everything okay so the 6 + 5 I let's say you sketch that first so here's the terminal point then you have this -9 + 3 I okay so if we sketch that guy that's the terminal point and then again if you want to do the parallelogram method this guy gets shifted down to make this side here okay to make this side here this guy gets shifted down to make this side here okay and this point right here is going to be your terminal point for the resultant Vector you see you get the -3 + 8 I again if you want to use the triangle method you can do that as well basically you're going to shift this guy down here so you would have that Vector going like that okay and you would have your triangle looking like that okay so whatever you want to do you can just show this graphically I prefer this method because this is what's shown in most textbooks and basically you're keeping in line with the fact that this is a position vector and this is a position vector and then this is your resultant Vector all right so what we want to do now is move into writing a complex number in What's called the trigonometric form also known as the polar form okay so typically we see our complex numbers written as a plus b i where again the a is the real part and the B is the imaginary part or some books will say bi is the imaginary part so what we're going to do is just switch this over to the x + Yi where X is now going to be the real part and Y is going to be the imaginary part now in terms of notation you're going to sometimes see that the I is Switched so you might see x + I * y this is done for clarity sometimes you'll have a radical involved so you just have to be very very careful when you have I written like this to understand what the imaginary part is okay that I is always separate from that you have to remember that and I'll show you that specifically here with some examples now in terms of getting this guy from what we call rectangular form which is the form that you normally see it in the a plus b or the X Plus Yi into this Polar form you need to consider a few things so we've graphed this guy as a vector a position Vector to be specific so the initial point is at the origin 0 comma 0 the terminal point is at the complex number x + Yi or you could say the coordinates here are X comma y okay on the complex plane so if you think about a few things first off the length or the magnitude of this Vector is labeled as R we've seen this a bunch of times where we think about okay well the length of this guy is the absolute value of the complex number okay going back we said the absolute value of a plus bi I is the square root of a^ 2 + B2 again this is just the Pythagorean theorem right if I go through here and I think about how to get the length of this r or the magnitude of this Vector well this guy here would be R 2 is equal to the length of this which is X so X2 plus the length of this which is y so y^2 a 2 + b squal c s just going back to the Pythagorean theorem if I take the square root of each side I get that R is equal to the square root of x^2 + y^2 okay so this is something we're going to need today when we want to find our R value we just use this little simple formula again if you can't remember it it's just the absolute value of the complex number that's all it is so let's go ahead and cut this away and let's just paste this on this little sheet here we'll just go back and forth so we also need to know about a few other relationships so we think about this angle Theta here this direction angle for our position Vector the cosine of theta is what so the coine of theta is what remember it's adjacent which is X over Hy hypotenuse which is r and then we can also say that the S of theta is what it's opposite which is y over hypotenuse which is R okay let's grab these and cut these away and paste these in here again we're going to be using this in a second and basically what I want to do is I want to solve these for X and Y okay so to solve this for X I would multiply both sides by R and I would have X is equal to R * the cosine of theta to solve this for y I do the same thing I'm going to multiply both sides by R so I would have R * the S of okay so we're going to use this in a little while let's go back up there's one other thing we need so we recall when we work with vectors we figure out the direction angle Theta using the tangent right so we say that the tangent of theta is equal to Y overx where X is not zero okay so we're going to use this as well but again you're remember from working with vectors you have to be really really careful with the inverse tangent function okay with the inverse tangent function you're going to get something in quadrant one or four only one if your argument is positive and four rotating clockwise if your argument is negative so let's come in here and let's paste this in and I guess we can put that about right there okay so the polar form let me just start with a rectangular form so we have x + Yi if you want to write yourself a little note and say okay well this is a plus b i just to keep track of this again X is the real part just like a y is the imaginary part just like B okay that's all it is just a little notational change so this is the standard form or the rectangular rectang UL I can never spell that form Okay then if you want the polar form we're going to do a little bit of work so let me do this on another sheet and I'll actually come back so we have the x + Yi and I'm just going to be plugging in for X and Y if I go back you see that X is R * cosine of theta and Y is R * s of theta I'm just going to plug in so R * cosine of theta R * cosine of theta will be plugged in for x and R * s of theta will be plugged in for y so R * s of theta okay so what I'm going to do here say this is equal to I'll have R * the cosine of theta then I'm going to put plus I'm going to put R * s of theta and I'm going to wrap this in parentheses just for clarity this is going to be multiplied by the imaginary unit I okay so again this is X this whole thing right here and then this is y this guy right here and this multiplying the imaginary unit I okay so I'm going to go a step further and just Factor the r out now so let me put equals and I'm going to come down here with this let me drag a little arrow down here if I factored out the r where would I be I would have R times inside of parentheses I would have my cosine of this Theta and then plus again I pulled out the R I would have my S of theta but what I'm going to do is I'm going to put the I out in front so I times the S of theta okay and you have to be very very careful when I is written this way because sometimes it's going to get mixed up and again I'll show you some examples where you need to be careful of this okay so this is your Polar form that's all it is some books will abbreviate this and they will put they will put R time c i s it's an abbreviation so the C then you have the I and then the S okay so that's how you can remember R * c i s and then times the angle Theta okay so if you want to write it as r s Theta or write it like this this is more popular this is your Polar form so let me write this up here we have R times you have your cosine of theta plus I times your s of theta like this this is your Polar form or again some books are going to call this the trigonometric form and the short short hand for this is R * CIS of theta all right let's just look at some examples of converting back and forth between the two this is a very very easy process but you do need to work a few examples so you get used to things okay you'll also see a lot of these come up over and over again because they're using values on the unit circle so what I have here is 2 multiplied by the cosine of 150° plus I * the S of 150 de so this is in the polar form right you have some value R here which in this case is two and it's multiplied by cosine of some angle Theta in this case is 150° plus I * your s of some angle Theta okay so all I would do here basically is figure out what is the cosine of 150° what is the S of 150° basically replace things and then use my distributive property to distribute the two into everything and you're going to have it in the format of x + Yi okay just like you want so first and foremost what is the cosine of 150° remember this has a 30° reference angle and we're in quadrant 2 cosine is negative okay and the 30° reference angle cosine of 30° is of 3 over2 so I can say this is the negative of the > 3 over2 okay then the S of 150° same 30° reference angle but again I'm in quadrant 2 so sign is positive this is going to be 1/2 okay and positive2 at that so all I need to do now is just go through and basically replace things let me get rid of this I think we don't need this anymore I'll say that I have what I'm just going to put equals here I'll say two times here I'll put the negative of the < TK of 3 over2 and then plus I'm going to put the I here okay times I'll put my 1/2 okay be very careful with the I again it gets switched around a lot sometimes it's here sometimes it's there just be careful with so let's just erase this we don't need it anymore and now I'm going to distribute the two to each term inside the parenthesis so I would have 2 * Nega of the 32 and then plus I'll do 12 * 2 and then you can do time I okay so basically we see that this would cancel with this and this would cancel with this and in my rectangular form I have the negative of the square of three basically plus one I or could just say plus I okay so this is my rectangular form or my standard form and again this right here is the polar form or some people call this the trigonometric form all right let's look at another one so we have the polar form here we have six time you have the cosine of 135° plus I * the S of 135° so what is the cosine of 135 degrees so again you're in quadrant 2 so this is negative what is the sign of 135° again you're in quadrant 2 so this is positive remember 135 degrees you have a 45 degree reference angle so in each case you're going to have squ of two over two right this guy is just going to be negative because cosine again is negative in quadrant 2 and this guy would be positive okay so basically I could replace this and say I have 6 times this is going to be the negative of the > 2 over2 and then plus I * you're going to have the < TK of two over two okay so if I distribute the six into each term I would have 6 * the negative of the > two over two plus I * let me go ahead and switch the order here so this makes more sense we'd have the > 2 over 2 * 6 and time I and I'll just go ahead and cancel this this cancels with this and gives me a three this cancels with this and gives me a three so basically I'm going to have the negative of 3 * thek of two and then plus you'll have 3 * the square two and then times I now again A lot of times for clarity you're going to move this around let me drag this down here so it's a little bit better so you can see things I erase that and put that back in instead of having the I here most books will do three I like this then times the square of two this is just to make it completely clear that the I is not under the square root symbol okay so the rectangular form here or again the standard form is -3 * 2 + 3 I * of2 and then again your Polar form or the trigonometric form is going to be 6 times this quantity cosine of 135° plus I * s of 135° all right let's look at something more tedious now so we're going to convert from our rectangular form to the polar form I'm going to give you three easy steps for this the first thing that I always do is I figure out what quadrant my complex number is in you can do that by just thinking about the real part and the imaginary part so in other words if you were to graph this again on the complex plane it's a comma B so if this is positive and this is positive then you're in quadrant one okay if this is negative and this is positive I'm going left and then up so I'm in quadrant two okay if they're both negative I'm in quadrant three right I'm going left and then down and then lastly if this is positive and this is negative well I'm going to be in quadrant four right because I'm going to the right and I'm going down so just think about your real part in your imaginary part and ask yourself the question if I was to graph this as a vector okay where would I be am I in quadrant one two 3 or four okay so that's the first thing you want to do now if I look at this guy right here I have 2 - 2 I * 3 again this I is here for clarity okay if you want to you can write this as 2 minus 2 * of 3 * I so it's Crystal Clear that this part right here is the real part and this part right here you can go ahead and take the negative with it is the imaginary part right so a plus b i okay so we know that the real part is two so that's positive the imaginary part is -2 * 3 that's negative so I'm going to the right okay and then I'm going down that means I'm going to be in quadrant four right because again this is positive and this is negative and you're thinking about this in terms of the complex plane Okay so now I know that Theta is in Quadrant 4 we need that for later on when we're getting our angle Theta so if we want to get this trigonometric form of the Polar form it's R * cosine of theta plus I * the sign of theta okay that's all it is so I need to figure out what R is and I need to figure out what Theta is and I'm done so how do I figure out what R is R again is just the absolute value of the complex number okay we've been doing this forever it's the square root of the real part Square so 2^ 2 plus the imaginary part again if you're writing it like this it's Crystal Clear -2 * 3 so -2 * of 3 okay this guy being squared some students make a mistake when it's written like this again they put I in there for some reason again just go ahead and write it like this so it's Crystal Clear that this is the imaginary part you want the a s plus the b s or the X S Plus the y s whatever the notation is okay the real part squared plus the imaginary part squared okay that's what you want so we have R is equal to the square root of this is four and then basically you have -2 SAR which is 4 and you have square root of 3 SAR which is three so 4 * 3 is 12 so i' put + 12 here okay so R is equal to theare < TK of 16 so R is equal to 4 okay so the first part of that okay the first part of that is done so R is equal to four and this looks like I made a radical here I was just making a border so let me kind of clarify that and put this maybe like this or something that would be better I think and maybe I can just do it upside down I'm just trying to make a border that's not a radical so R equals 4 okay so now I need to find Theta so what is that going to be again you use the fact that the tangent of theta is y okay or the imaginary part over X or the real part okay so the imaginary part you have your -2 * > 3 over your real part which is two okay so you can go ahead and cancel this with this and this is the negative of the of three okay now what I would do okay is not punch this into the calculator as the tangent inverse of negative square of 3 because you're going to get something in Quadrant 4 rotating clockwise now your teacher may allow that as an answer but typically you want something between zero and 2 pi where zero is included and the 2 pi is not okay so what I'm going to do is I'm going to say that I want the tangent inverse of the square < TK of 3 and this is going to give me 60° okay this is going to be my reference angle and I'm going to be in Quadrant 4 okay remember we figured out we're in Quadrant 4 for this reason so in Quadrant 4 if I had an angle with a reference angle of 60° to find that I would do 360° minus 60° which gives me 300° or in terms of radians it would be 5 piun over 3 so now that we've done all this work okay we are ready to write this guy in polar form and I guess you get rid of this we don't really need it anymore I'm just going to write this and say that basically this guy and I'll just even erase this would be the r which is four times you would have cosine of your angle so you can use either one you want you can do it in radians or degrees so 300° then plus again I'm going to put I first and then times your sign okay of that same angle 300 degrees if you want to write this with radians all it is I'll put equals again so four times you're going to do cosine of 5 piun over 3 okay and then plus I * the S of 5i over 3 so always go with what your teacher says if they want radians or degrees just ask them okay so we basically have this done and also if you wanted to you can use that little abbreviation let me just show you that real quick you could do this is 4 * CIS so again the C the I and the S that's where they deviations coming from and then times let's just go a and say 5 piun over 3 all right let's look at another example so we have -2 * 3- 2 I so again the first thing you want to do is figure out what quadrant is Theta going to be in again you can speed this up at this point again here's the real part it's going to be negative here's the imaginary part it's going to be negative as well in other words if you were to graph this you would have -2 * sare of 3 this would be your location on the real axis and then you would have -2 that' be location on the imaginary axxis so in other words this is negative and this is negative so I know I'm going to the left and down so I'm going to be in quadrant 3 and I put equals here I should just put is in quadrant 3 okay so we can keep that straight now what I want to do is quickly find my value for R I'm just going to make that a little bit more clean so my quadrant three I'll just put for my angle Theta so for R again it's the absolute value of this complex number however ever you remember it remember it's the square root of the real part squared plus the imaginary part squared however they're giving that to you okay so I'm going to take the square root of I'll take -2 * theun of 3 being squared and plus I'll have my -2 being squared okay so from here I know that basically -2 2 is 4 sare of 3 S is 3 so 4 * 3 would be 12 and again we already know this so this is going to end up being four right because this is 12 + 4 that's going to be 16 16 is 4 right but to go through the steps this would be four again 12 + 4 is 16 of 16 is going to be four so R is four we know that so let's find Theta this is the one that takes a little bit longer so we know that tangent of theta again is YX or you could say B over a if you think about this in terms of the traditional way we think about complex numbers it's going to be the imaginary part over the real part so the imaginary part is -2 so -2 and the real part is -2 * of 3 3 okay so this is going to cancel and I'll have 1 over theun 3 if I rationalize the denominator there so times theare of 3 over theare 3 this is the sare of 3 over three okay so basically if I punched up the inverse tangent of 3 over 3 I'm going to get something in quadrant 1 remember I'm looking for something in quadrant 3 okay so let's go ahead and set this up I want to go inverse tangent okay inverse tangent of squ Ro of 3 over 3 like this so you hit that up on your calculator you're going to get 30° okay and again in quadrant 3 if I have an angle with a reference angle of 30° to find that I would do 180° plus 30° okay which would give me 210° so Theta is going to be 210° so now that we have all the required information I can just put equals I'm going to take my r value which is four and I'm going to multiply by again you have your cosine of this angle 210° plus I * the S of this angle which is 210° so that's basically it that's your Polar form now if you want to write this with radians or write it as an abbreviation you can keep putting equals so 210° in terms of radians would be 7 pi over 6 so you could say this is four times you'd have the cosine of 7 pi/ 6 plus I * the S of 7 piun / 6 and again one more time time this C this I and this s you can use it to abbreviate so you could say this is four and you go c i or S and then you'll say your angle so again you can do degrees or radians whatever we'll do 7 pi over 6 so another way to write it this is the Shand all right let's look at one more example so we have the negative of the of 31 over2 plus the of 93 over2 * I again just be careful sometimes they'll put the I up here because you're working with a radical just be able to separate the imaginary part from the real part okay you need to know what those are and you need to not get I mixed into that okay so what I'm going to do here again is think about what quadrant am I going to be in okay this is negative this is positive so I'm going to the left and I'm going up right so I'm in quadrant to so Theta is in quadrant 2 okay very important to understand where I am now once I understand that I'm in quadrant 2 again when I go to find Theta okay I'm going to basically get a reference angle and then use that quadrant 2 well let's find R first because that's a little bit easier so R equals again the square root of take your real part being squared so NE of < TK of 31 over two we're going to square that plus take this guy the imaginary part so the square < TK of 93 over two this guy squared so R is equal to what if I Square this I'm going to get 31 over 4 so the square root of you'll have 31 over4 plus this guy would be 93 over 4 okay if you add add those together you're going to get 124 over 4 so 124 over 4 if you make that division you're going to get 31 okay you can't simplify that 31 is a prime number so basically R is just the square of 31 okay so R is equal to the square OT of 31 okay let's work with Theta now so the tangent of theta is what it's the imaginary part or Y or B however you want to think about it so squ < TK of 93 over2 over the real part which is X or basically a however you want to think about that I'm just going to flip this guy because when you divide fractions you multiply by the reciprocal so I'll put the negative of 2 over the 31 that's going to allow me to cancel this with this I know that from the quotient rule for Radicals I can divide here so 93id 31 is 3 so I can basically say this is the negative okay the negative of the square < TK of three now we know from the first first example this guy is going to have a reference angle of 60° if I did the inverse tangent ofun 3 I'm going to get 60° so because I'm in quadrant 2 what I want to do is I want to take 180° and I want to subtract the reference angle of 60 degrees that's going to get me to 120° so Theta is 120 deg here okay so Theta Theta is20 degrees or in terms of radians you could say this is 2 pi over 3 okay so basically I'm ready to set this up I can go ahead and say this equals and let me just drag this out of the way put this down here I'll say my r value which is the square of 31 times let's just go ahead and use degrees first so I'll say cosine of 120° plus I times the S of 120° okay and again you can keep going right so if you want to put this in terms of radians I'll put equals I'll do < TK 31 * the cosine of 2 piun over 3 and then plus I * we'll do the S of 2 piun over 3 okay like this and last if you want to do your abbreviation take your r value which is square of 31 and then times that c i s or the SS again the C the I and the S so CIS that's how you can remember that and then times your 120° in this lesson we want to talk about the product and quotient theorems all right so in the last lesson we talked about how to write a complex number in what we call the trigonometric form most people call this the polar form and here what we want to do is learn how to quickly multiply or divide two numbers if they're in polar form using something called the product or also the quotient theorems okay so I'm going to start off with the product theorem and basically we have these two generic complex numbers written in polar form right remember when you think about the R value if it's as a vector if you think about it that way this is the magnitude or the length of the vector or you could also think about it as the ABS abolute value of the complex number okay so that's the R value in each case when you think about Theta okay this is going to be if you think about this as a vector it's the direction angle okay or if you think about it here a lot of books will also call this the argument okay so you hear that terminology so here we have this R sub 1 times the quantity cosine of theta sub 1 plus I * s of theta sub 1 okay then times we have this other complex number so R sub 2 multiplied by the quantity cosine of theta sub 2 plus I * the s Theta sub2 okay so all I need to do to derive my little product theorem okay is just use some foil so I'm going to start out by multiplying R sub1 * R sub2 so R sub 1 * R sub 2 then I'm just going to set up some brackets here and I'm going to put these two inside so I'm going to go the cosine of theta sub 1 plus I * the S of theta sub 1 and then I'm going to go cosine of theta sub 2 and then plus I * the S of Theta sub 2 okay so all we're going to be doing is foil nothing fancy here so I'm going to put equals R sub 1 * R sub 2 and then let me open up these brackets and if we're doing foil we're going to do the first term so this guy times this guy so this would be my cosine of theta sub 1 time the cosine of theta sub 2 then I want to do my outer so this one times this one let me go ahead and put plus I have an i there let me put that out in front and then I'll just do my cine of theta sub 1 * my S of theta sub 2 okay I'm going to put this down here because it's going to run off so I'm going to put plus and I'm going to erase this and this and now we're doing the inside right so it's basically going to be let me change the color this one times this one so you would have I * the S of theta sub 1 times the cosine of theta sub 2 okay and then lastly we have the L right so this guy times this guy I * I is i^ 2 now you can put plus i^ 2 but remember this is 1 by definition so we're going to change this in a moment then you have your s of theta sub 1 time your s of theta sub 2 okay let me close the brackets there because this is being multiplied by all of those now as I just said i^2 is1 so we can erase this and just put a negative sign here okay and now what I want to do is I want to group everything where the terms that have an I are next to each other and the terms that don't have an ie are next to each other so I want this one and this one to be next to each other and then let me change my highlighter color here I want this one and this one to be next to each other okay so let me go ahead and do that real quick I'm going to put equals R sub1 * R sub 2 and then inside the brackets here I'll go cosine of theta sub 1 * cosine of theta sub 2 and then I'm going to do minus I'm going to go s of theta sub 1 * s of theta sub 2 okay and then we'll go ahead and say plus let me scroll down and get a little bit of room here I'll say I time the cosine of theta sub 1 * the S of theta sub 2 and then I'm going to put plus I have my I time the S of theta sub 1 time the cosine of theta sub 2 okay so let me close the brackets there now the reason we group these guys together is because to further simplify this or to get into the form you're going to see in your book you're going to use an identity so if we go back to our little handout and we go to the sum and difference identity I want to call your attention to this first one so we have the cosine of some angle a plus some angle B okay in our case it's Theta sub 1 and Theta sub 2 okay so the cosine of a Time the cosine of B minus the S of a Time the S of B okay so what we have is this guy on the right and we want to transform it to what it shows on the left okay that's perfectly legal so what we're going to do is say the cosine of theta sub one times the cosine of theta sub 2 so this is like having a and b okay minus the S of theta sub 1 * the S of theta sub 2 well we could write that as the cosine of we'll have the Theta sub 1 plus the Theta sub 2 okay so I'm just using that identity to make this more compact so let me put my bracket there let me put my R sub1 * my R sub 2 let me scroll down just a little bit to get some room here okay and basically what I want to do now is for this second part where I have let me highlight this this guy plus this guy I'm first going to factor the I out okay so I'm going to put plus let me do this over here let me put plus and put an i there and inside what I would have is the cosine of theta sub 1 okay times the S of theta sub 2 and then you'd have plus you'd have your s of theta sub one times your cosine I probably can't fit this so let me move this down here let me move this down here so times the cosine let me make this a little bit better so the S of theta sub 1 * the cosine of theta sub two so then let me close the bracket off here so that's officially done okay and basically you see once I factored this I out that here we can go back and we see that we have S of a plus b is the S of a * the cosine of B plus the cosine of a * the S of B well we have the same thing okay it's the S of theta sub 1 time the cosine of theta sub 2 plus the cosine of theta sub 1 * the S of theta sub 2 now it's not in that order okay it's reversed here but you could rewrite it right because this is addition so I could say the S of theta sub 1 * the cosine of theta sub 2 plus the cosine of theta sub 1 time the S of theta sub 2 okay it wouldn't matter so let's go ahead and put equals I'm going to rewrite that one as well and I'll say that I have my R sub 1 * my R sub 2 inside of brackets I'll have cosine of theta sub 1 plus Theta sub 2 so then I'll say plus I'll put my I here and I'll do my S of I'll say my Theta sub 1 plus my Theta sub 2 okay let me close the brackets there so this result here is what you're going to see when you look at your product theorem in your book okay so basically you multiply the two absolute values together that's all you need to do that's going to give you your R outside and then inside when you're working with this guy you have the cosine of you're adding the two arguments together so now it's the Theta sub 1 plus the Theta sub 2 plus your I times sign of again you're adding your two arguments together so your Theta sub 1 plus your Theta sub 2 so most books will just say your multiplying the absolute values and adding the arguments if you want a simple way to remember that now another way you might see this again in compact form we might see this R sub1 times the c i s or the S right that comes from this the cosine the i in the S okay it's an abbreviation let me just erase that so basically you have this R sub1 times this SZ and I'll put Theta sub one if it's multiplied by this R sub 2 * this SZ okay and then Theta sub 2 this is going to be be what again multiply these two guys together these two absolute values so this is R sub 1 * R sub 2 and then basically you would have your CIS and then you're adding these two arguments together so your Theta sub one plus your Theta sub 2 okay so this is the shorthand probably a way easier way to remember it all right so to get things started let's go ahead and take a look at a very simple example we have these two complex numbers in polar form and we just want to find the product so we have three times this quantity we have the cosine of 150° plus I * the S of 150° then we're multiplying by we have 3 * this quantity cosine of 90° plus I times the S of 90° so again the very very easy way to do this is just to multiply the absolute values together so the absolute value of those complex numbers so the 3 * the three that's going to give me nine okay and then I'm going to set this up I'm going to go the cosine of I'm going to add the arguments so I'm going to add 150° plus 90° okay that's all I want to do so the cosine of 150° plus 90° okay and then plus my I times I'm going to do my sign of it's the same thing just adding the arguments so 150° plus 90° 150° plus 90° okay so basically you're just going to simplify this and I'm going to have 9 times again this quantity now it's going to be the cosine of 150 degrees Plus 90 degrees is 240 degrees so I'll say this is the 240° plus I * the S of we'll say 240° and again you can also write this using your abbreviation so you could say that this is the nine times the c i s or this is WR the C and then the I and then the S and then go ahead and take that argument the 240° okay that angle there so this is another way to write it and then additionally they might ask you to give it back to them in rectangular form so let me just erase this and this and this okay we'll do that real quick so again for rectangular form you need to figure out what is this and what is this and then you're going to distribute the nine to each term inside the parenthesis so for a 240° angle we have a 60° reference angle so the cosine of 60° is going to be a half and the S of 60° is going to be the square root of 3 over2 okay but because I'm in quadrant 3 cosine and S are both negative so so I can say that this is -2 and then for this one I'll say it's 32 so I might as well just say minus the 32 and I'll just put my I behind here like this okay so from here I'm just going to distribute the nine to each term inside the parentheses so I would have 9 Hales like this and then minus the 9 gets distributed so 9 * < TK 3/ 2 and then times the imaginary unit I so of course if they ask you for Polar form it's going to be this one okay and then if they ask you for the rectangular form you need to go through and figure this out and then you get this one all right so let's look at a slightly more tedious example so what we're going to do here is we're going to have our complex numbers given to us in rectangular form we're going to convert them into Polar form just to get some practice and then we're going to use the product theorem to find the product of those two complex numbers okay so what I'm going to do here let me put equals I'm going to convert this one into my polar form first so to that my r value would be what remember that's the absolute value of the complex number so you're going to take the real part which is netive of 2 * s of two and you're going to square that you're going to add that two you're going to have your imaginary part which is also the negative of 2 * two negative of two * square of two okay being squared okay so basically if I think about -2 being squared that's four the square root of two squar is two so in each case you have 4 * 2 which is 8 so you're going to end up with 8 + 8 which is 16 okay so you basically have R is equal to the > 16 which is 4 okay so I can say that R equals 4 here now for the argument Theta or you could say the direction angle if you think about this as a vector the tangent of this Theta is the imaginary part over the real part right the B over the a or you could say the Y over the X however you think about this so I'm going to take the -2 * S 2 -2 * 2 over -2 * 2 well this is basically going to be one right so the tangent OFA equals 1 okay if we use our inverse tangent function so tangent inverse of one what is this going to give me well if I'm working with degrees I'm going to get 45 degrees okay but this is not our answer right because remember we have to consider what quadrant we're in this is just going to give me the reference angle so in this particular case this is negative and this is negative right so if I was to graph this complex number on the complex plane I would go to the left and I would go down right so I'm in quadrant 3 so in quadrant 3 if I had a reference angle 45° I would want 180° plus 45° which would be 225° so Theta here is 225° okay so now that gives us the information to write this guy out I can say that my r value is four and then my argument here Theta is 225 degrees so i' have cosine of 225° plus I * the S of 22 25° okay so the first one's done now we just need to do the second one so again let me put times here for this one the R value is the square root of take the real part so negative of the square of three this is squared then plus take the imaginary part you can write a phantom one in there this is really negative 1 so I'll put negative one here and this will be squared okay so basically the negative of the square Ro of three being squared is 3 and - 1 s is 1 so you're going to have 3 + 1 which is four so r is the sare < TK of 4 which is two okay so let's put two here and then again for the direction angle Theta I think about the fact that the tangent of theta is the imaginary part which is NE 1 over the real part which is the negative of theun of three now let me go ahead and get rid of the negatives so I'm just going to erase that from here and I'm going to rationalize the denominator soqu 3 over 3 multiply by that and this gives me a tangent of theta which is going to be theun of 3 over 3 okay so let me erase this and actually make this a little bit more compact so the square root the square root of 3 over3 so again if you punch this into your calculator if you do tangent inverse of < TK 3 over 3 you're going to get 30° okay but again this is going to be a reference angle because this is negative and this is negative so again I'm in quadrant three right if I want to plot that guy I'd have to go to the left and go down so in quadrant 3 if I have a reference angle of 30° the to find that I have 180° plus 30° that's 210° so my r value is a two okay put my parentheses there i' have cosine of I would have Theta is 210° here okay so 210° plus I times the S of 210 degrees okay and again you can use radians if you want it's just a matter of what your teacher is working with I'm just going to work with degrees here for now okay so now to find the product again multiply the absolute values so 4 * 2 is 8 and then I'm going to set this up I'm going to go cosine of I'm going to add the arguments so 225° plus 210° so 225° plus 210° okay plus I times the S of again I'm just going to add the arguments 225 de plus 210° so 225° plus 210° okay let me close that down so it we'll say this equals 8 times okay this quantity here is going to be the cosine of 225 deg Plus 210° would be 435 degrees so I'll say this is 435 degrees and then plus I times sine of again that 435° now for some teachers this answer is fine for others they want you to give an angle that is between 0 and 360° or 0 and 2 pi if you're working with radians where in each case the zero is going to be included and your 2 pi or 360° is not going to be included okay so basically you would subtract since we're working with the 360° from each one of these to get your co-terminal angle and in this case it's going to give you 75° okay so I can say this is equal to 8 * the quantity you have the cosine of 75° plus I * the S of 75° okay so just get with your teacher in terms of if you need to do that most books will go to this some books will leave it like this and say it's fine so it's really up to your teacher all right now let's move on and talk about the quotient theorem so again I'm going to go through where this comes from I know a lot of you find this quite boring but it's good to be able to see where things come from it also gives you a lot of practice with manipulating things so let's say we had these two complex numbers in polar form and now we want to do the division okay so we have this R sub1 * the quantity cosine of theta sub 1 plus I * s of theta sub 1 over R sub 2 * the quantity cosine of theta sub 2 plus I * s of theta sub2 okay let me set this up over here because we're not going to have room to do what we need to do so this is R sub one time the quantity cosine of theta sub 1 plus I * we have sin of theta sub 1 okay that's your numerator your denominator here what you're dividing by your R sub 2 time the cosine of now Theta sub 2 plus I * sin of theta sub 2 okay so the idea here is that you would multiply the numerator and denominator by the complex conjugate of the denominator remember if you had something like a plus b i the complex conjugate would be a minus bi so here you can go through and multiply okay the r sub 2 into each term but you're going end up factoring it back out so all you really want to think about is the inside here the terms are going to be the same okay it's just that the sign is going to be different so what I'm going to do is I'm going to multiply by I'm going to have this R sub 2 and then times the quantity you're going to have cosine of theta sub 2 okay so that's the same this is going to be minus down instead of plus and then the term is the same so you have I * s of theta 2 again the r sub 2 here would just be factored out if it was distributed in so you can think about it as just being factored out there okay so that's why it's out there okay so I'm just going to write this in the denominator here so R sub 2 and then cosine of theta sub 2 minus I * sin of theta sub 2 okay like this now this is quite lengthy so I'm going to do the numerator and denominator separately the numerator is going to be very similar to what we did earlier it's just that we're going to have a minus involved in some different places so we're going to end up using some different identities all right so let's go ahead and attack the numerator and denominator separately let me copy this and I'm going to go to a fresh sheet here and let me just paste this in okay and my multiplication symbols out of whack there so if we go through here again we start this in a similar way to the last one so R sub 1 * R sub 2 okay and then inside of here you would have this cosine of theta sub 1 plus I * the s of theta sub 1 okay and this would be cosine of theta sub 2 and then minus your I * your s of theta sub 2 okay so again all we're going to be doing is some simple foil so I'm going to go equals R sub1 * R sub2 like this and I'm just going to foil things out so first times first so again cosine of theta sub 1 and then times cosine of theta sub 2 all right so the next thing we would do is the outer so I would do this one times this one so I'm going to put negative I'm going to put my I out in front and then the cosine of theta sub one and then times the S of theta sub 2 okay then I'm going to go let me erase this then I'm going to go the inside so this one times this one so I'll go plus I'll have my I and then the S of theta sub 1 times the cosine of theta sub 2 okay so far so good now I just need my last so let me go ahead and highlight this one's already done and then this one so I know this is negative I have I * I which is i^ 2 and then I have the S of theta sub 1 and then times the S of theta sub 2 okay let me close the bracket there all right so remember i^2 is1 so this is like if we had minus a Nega 1 which is basically like having plus one okay so you can change this you can change this to a big fat plus okay so from here now what I want to do is I want to get my terms with I together and the terms that don't have I together so now words like we did before this will be with this okay and let me color this differently this will be with this okay so we're going to arrange it in that way so let me go ahead and put equals here let me slide down just a little bit and get some room going I'll have my R sub1 * my R sub 2 I'm going to open up the brackets so the cosine of theta sub 1 time the cosine of theta sub 2 okay then I'm going to put this next so plus the S of theta sub 1 times the S of theta sub 2 okay and then I'm going to put you have a minus here and a plus here it's actually to our advantage to rearrange that okay you can put it like this but in the end you can rearrange it because you're going to need an identity so I'm just going to put this as plus I'm going to use this one first so I'm going to say I * the S of theta sub 1 time the cosine of theta sub 2 and then I'm going to put minus the I * the cosine of theta sub 1 time the S of theta sub 2 now you saw that last time we we ended up factoring the I out so let's go ahead and do that again so I'm going to wrap this in parenthesis and I'm going to get rid of this okay and I'm going to wrap this in parenthesis as well okay so now what I'm going to do is I'm going to go back and I'm going to think about an identity that would help me out here so I have cosine of some angle let's call it Theta sub 1 time cosine of some other angle Theta sub 2 then plus we have S of some angle Theta sub 1 * s of some angle Theta sub 2 if we go back you see that cosine of a minus B follows that pattern right cosine of some angle a time cosine of some angle B then plus s of some angle a * s of some angle B so I can use that to rewrite this I'm going to go R sub 1 * R sub 2 and then inside of brackets here I'll have cosine of I'll go ahead and say now this is the Theta sub1 minus the Theta sub 2 okay so that part is done now when we look at this one s of theta sub 1 * cosine of theta sub2 minus cosine of theta sub 1 * s of theta sub 2 if we go back that's this guy right here right WR s of a minus B this is s of a * cosine of B minus cosine of a * s of B that's why I had to rearrange things because it would have been this way the minus first and the plus you can always rearrange it but it might not be obvious so that's why I did that so let's go back and now let's write this as plus don't forget the I and then times your s of this is going to be your Theta sub one minus your Theta sub 2 okay so go ahead and close this down so this is your numerator if you want to copy this let me go ahead and copy this and I'm back on this screen now so let me c down a little bit I'm going to put equals here and let me go ahead and paste that in and I already had an equal so like I'm doing it twice so let me go ahead and move this over just a little bit so I'm going to say this equals this and actually let me put this a little bit further down and then I'm going to put this over we now need to work on the denominator okay so let me go ahead and actually I should probably have equals here to make this complete and let me grab this Den minator now let me copy it let's go to another sheet and let me paste this in now okay and we don't need this and I need a times here okay so for this guy what you need to realize is that you could first multiply your R sub2 * your R sub 2 so that's going to give me R sub 2 I'm going to wrap this in parentheses for clarity so this is going to be squared and then we're going to have the cosine of theta sub 2 plus I * the S of theta sub 2 okay then times the cosine of theta sub 2 now minus the I * the S of theta sub 2 so notice that you have these complex conjugates here okay because you have the same term here and here okay and then you have the same term here and here and this looks really weird it looks like a face so let me make that like that and then basically your signs are different okay so what I would do here is just use my formula so I would go ahead and say this is R sub 2 being squared times essentially I would do the first guy squared so let me put this inside of parentheses I would say cosine squar Theta sub 2 okay and then minus your second guy being squared so if I squared I I would have I squared if I squared s of theta sub 2 I would have sin^2 Theta sub 2 okay like this now what are we going to do from here just think about this for a second minus i^ 2 remember this is1 so this is basically like minus a NE 1 which is basically plus now hopefully you see this right away you have cosine 2 Theta sub 2 plus sin 2 Theta sub 2 what is that that's one if you go back you see for the Pythagorean identities sin squ theta plus cosine Square th equals 1 right so here we have cosine squ of theta sub 2 whatever that angle is plus sin square of theta sub 2 same angle whatever it is okay and so this is going to be one so I can say this is R sub 2 being squared * 1 which just equals R sub 2 being squ okay so if I go back up this is going to be my denominator so let me bring this down here and I'm just going to put this over R sub 2 and this is squared okay so remember this is multiplication here this is being multiplied you could wrap this in parentheses to make that clear this is being multiplied by this so I can cancel a common factor of this guy with this guy so essentially what I have here okay is going to be what you're going to see for your quoti theorem in your book your R sub1 over your R sub2 okay so you're dividing the absolute values and then times this quantity here you have your cosine of you're subtracting the arguments okay so I'm doing my Theta sub 1 minus my Theta sub 2 okay then plus I times the S of again I'm subtracting the arguments so Theta sub 1 minus my Theta sub 2 and again if you want to write this in a more compact form well if you had R sub one times your SZ of theta sub 1 / R sub 2 * your s of theta sub 2 okay what this would be equal to is your R sub one over your R sub 2 times your SZ of your Theta sub 1 minus your Theta sub 2 okay so let me go and wrap that in parenthesis and that's basically the shorthand I think this is the easiest way to remember things and let's just go ahead and look at an example all right so what we'll do here we'll have this 2 * > 7 * the qu we have the cosine of 7 piun / 6 plus I * the S of 7 piun / 6 and this will be over you have your 3 * the quantity of your cosine of 11 pi over 6 plus I * s of 11 piun over 6 so of course these two complex numbers are in polar form so we can immediately use this quotient theorem okay but to get a little bit of extra practice let's write our answer in polar form first and let's also go into rectangular form okay so let's go ahead and say this is equal to I'm going to divide the absolute values so I'm going to say 2 * 7 over 3 okay this is going to be times the quantity what you're going to do here is you're going to have your cosine of you're going to subtract these arguments so I'm going to do 7 pi over 6 minus 11 pi/ 6 and I'll take care of this in a moment I'll just leave it like this for now and then plus I times your s of again you're doing the same thing so 7 piun / 6- 11 piun over 6 okay so just like that all right so let me close this down so that's proper let's put equals here let me scroll down and get a little bit of room going so let's go ahead and say we have 2 * < TK 7 over 3 okay let make that three a little bit better and then inside of here I'll have the cosine of if I do 7 Pi minus 11 Pi that would be -4 PI right you already have a common denominator so let's say this is -4 pi over 6 and of course you can reduce the fraction each is divisible by two so this would be be two and this would be 3 so this is -2 piun over 3 okay then I'll do plus I * s of the same thing right it'll be your negative your -2 pi over 3 let me go aad and write this down here like this and let me go ahead and close the bracket there okay so again some teachers will accept this some of them want you to find the co-terminal angle so what I would do here is I could say -2 piun over 3 + 2 pi right or if I want a common denominator this would be multiply this by 3 over3 so this would be 6 piun over 3 okay so 6 piun over 3 so your co-terminal angle here would be 4 pi over 3 so you could really write this and you could say that this is 2 * 7 over 3 and then inside the brackets I'll say this is the cosine of 4 pi over 3 okay plus I * the S of 4 pi over 3 okay so for your rectangular form what we'll do is we'll have 2 * 7 over 3 we need to think about the cosine the coine let me make a border here the cosine of 4 pi over 3 okay so this is going to be 240° if you convert it into degrees so cosine of 240° if that's easier for you again it's going to have a 60° reference angle or pi over 3 in terms of radians so if you think about cosine of 60° again it's a half right and this angle here 240° or 4 pi over 3 that's going to be quadrant 3 so we know cosine is negative there so this would first be the negative of a half so let me put some parentheses here and say this is netive - one2 let me put plus my I so I don't forget it if I think about the sign of 4 pi over 3 again 60° reference angle we know that it's going to be in quadrant 3 sign is negative so let me go ahead and make that minus and basically pi over 3 the sign of that is going to be sare of 3 over2 right so it's the negative of the square of 3 over2 you put the I out in front like this or typically with this situation I'll put it here so we'll close that down now and all I need to do is just distribute this inside to each term right so basically have 2 * 7 over 3 * -2 and then minus you'll have 2 * 7 over 3 * < TK of 3 / 2 * I okay so let's come down here and just simplify and we'll have our rectangular form so the twos are going to cancel here you'll have the negative of theare of 7 over3 and then minus the twos are going to cancel here 7 * 3 is the < TK of 21 and then this is over three and of course times I so here is your rectangular form if you're asked to give that answer all right for the last problem let's go ahead and take a look at one that's a little bit tedious so we have this 1 - I * 3 this is over you have the negative of the of 14 over2 minus you have the 14 over2 * I so you'll notice that this division here involves these two complex numbers in rectangular form so what we're going to do is we're going to put these into Polar form we're going to use our quotient theorem and we'll just leave our answer in polar form okay so let's start out by converting each one of these over so I have 1 - I * 3 just going to do this on a separate page so 1 - I * 3 so the real part is one the imaginary part is the negative of the sarek of 3 okay so let's go ahead and say my R is going to be the square < TK of 1^ 2 which is 1 plus plus the negative of theun of 3 SAR which is three okay so obviously this is four and the square otk of four is going to be two okay so let's put this as two okay so now I need to find my Theta right so my argument for this guy and so the tangent the tangent of theta is going to be equal to the negative of theare < TK of 3 over 1 okay might as well just say it's the negative of the square of 3 so let me go ahead and do the inverse tangent function fun okay with the square otk of 3 forget about the negative let's punch that into the calculator so what you're going to get there is 60° or pi over three if you want to work with radians let's work with radians since we work with degrees for the multiplication examples so I'm going to say this is pi over three or you can just put 60° if you like working with degrees okay so let's put that there now this is going to be my reference angle let's think about this guy the real part here is positive so I'm going to the right and then the imary part here is negative so I'm going down right so I'm going to the right and I'm going down that means I'm going to be in quadrant four so what I'd want to do is I'd want to subtract 2 pi minus this pi over 3 okay to figure out what my Theta is going to be okay you could write this as 6 pi over 3 so 6 pi over 3 and this is going to give me 5 pi over 3 okay again if you wanted to do this with degrees you do 360 degrees minus 60 degrees you would get 300 degrees so however you want to do that so let's go ahead and write this since we have all the required information I'm just going to say Theta here Theta is 5 pi over 3 okay so in polar form we know that this is 2 multiplied by the coine of this 5 pi over 3 and then plus I times the S of our 5 pi over 3 okay again if you want to put 300 degrees in there feel free okay so I'm G to get rid of this let's go ahead and look at the other one so we have this negative of the of 14 over2 minus the of 14 over2 * I so let's go back so we have the negative of the Square < t of 14/2 okay then minus so let me put this down here so it's a little bit more consistent we have this square root of 14 / 2 * I so my R is going to be equal to the square root of I'm going to take this guy so it's negative in each case just forget about it we're going to be squaring it let's go ahead and say it's the sare < TK of 14 over 2^ 2ar well if I square the square root of 14 I get 14 if I Square two I get four so this is going to end up giving me 14 over four okay so 14 over 4 which is going to be 7 over 2 so you would have 7 over2 okay plus 7 over2 okay so this would end up giving me 14 over 2 which is 7 okay so this will be the square Ro T of seven okay so that's all that is so that's my R value let me drag this up here so it's out of the way so now let's figure out what is the Theta okay so again if you look at this the real part and the imaginary part in each case is the same thing right so if I did tangent of theta is equal to if you did the negative of the of 14/2 then over the negative of the of 14 over two that's just going to end up being one okay so if I did the tangent the tangent inverse of this guy of one I get 45° or in terms of radians it's going to be pi over 4 okay so you can put that or or you can put 45° whatever you want to do so for this guy this is negative and this is negative so I'm in quadrant three right I'm going to the left and I'm going down so in quadrant three I would want to take pi and I want to add pi over 4 to get that angle so Pi + pi over 4 or again you could do 180° plus 45 degrees whatever you want to do okay so for this one I'm going to multiply this by 4 over 4 so this will end up being 4 pi plus pi which is 5 pi and then over the common denominator of four okay so this is 5 pi over 4 or 225 degrees if you're working with degrees so let's put 5 pi over 4 like this let's get rid of all this so in polar form this guy is going to be we have our Square < TK of 7 multiplied by we're going to have our cosine of 5 pi over 4 plus I * the S of 5 piun over 4 okay so this part it's done let me copy this let's come back up to our problem paste this in real quick and I'm just going to get rid of this and what I'm going to do is I'm going to put equals like this and I'll just drag this over and I missed something so let me get all of this drag this over here let me put my little division bar and I did that terribly so let me try that one more time and let me just scooch this down just a little bit so it lines up perfectly all right there we go so now what I want to do is just use my quotient theorem okay very easy at this point so 2 / theare 7 so 2 / 7 of course you can rationalize the denominator there I can multiply this by theun 7 over thek 7 so this is 2 * the of 7 over 7 so 2 * 2 * the < TK of 7 over 7 so that's that part that's done and then basically what we want to do times this quantity here you have the cosine of you're going to have 5 pi over 3 minus 5 pi over 4 and I'm making this terribly so let me kind of edit this a little bit so this is a five and this is a four and of course we need a common denominator let's just stop and get this so we need to go three and then three we need to go four and then 4 so this would be 20 pi over 12 this would be 20 pi over 12 minus it would be 15 pi over 12 so 15 pi over 12 okay so this would give me what it would give me 5 pi over 12 so the cosine here of 5 piun 5 piun / 12 then I'm going to have Plus plus I have my I times the sign of again it's the same thing so the 5 piun over 3 minus the 5 piun over 4 so that gives us the 5 pi over 12 okay so this is going to be our result in polar form again if you wanted to figure this out in rectangular form just evaluate the cosine of 5 pi over 12 evaluate the S of 5 Pi over2 get those values and then just use your distributive property here to set that guy up in this lesson we want to talk about powers and roots of complex numbers all right so in the last lesson we talked about multiplying two complex numbers together when they're in polar form using something called the product theorem so what we're going to do here is just go a step further and we'll talk about these powers and roots of complex numbers so what I'm going to do is I'm just going to start out with a simple complex number something generic in polar form so I'm just going to write R times the quantity cosine of theta plus I * the s of theta okay so remember when we talk about R that's the absolute value of the complex number when we talk about Theta it's referred to as an argument okay if you think about the complex number as a vector it's going to be the direction angle okay so let's say that I wanted to raise this to let's say the second power right so I'm squaring this and basically what I would do here I would just expand this right so this would be R times the quantity of the cosine of theta plus I * the S of theta okay I'm going to multiply by itself so R * the quantity the cosine of theta plus I * the S of thet so you can see here that you have R * R remember you're multiplying the absolute values together so this gives me R 2 notice that this guy right here this exponent here on R matches this exponent here okay so this is going to be important then times the quantity you have cosine of remember you add the arguments so theta plus Theta is 2 thet then plus I * the S of theta plus Theta you add the arguments this is 2 Theta so I want you to notice some things again this two here this exponent matches this exponent here and then this guy here inside the parentheses it's multiplying the argument so the general rule is whatever this exponent is and this can be any real number for today we're just going to work with positive integers but whatever this guy is it's going to basically be your exponent on R and then it's going to be multiplied by your argument inside the parenthesis okay now I can do another one with you just to show you this so let's say we did something like R * the quantity cosine of theta plus I * the S of thet okay let's say now we Cube this guy okay so we know that this would be equal to basically this guy times this guy times another one okay this guy times this guy would give me this guy so I can go ahead and write R 2 * the quantity cosine of 2 theta plus I * the S of 2 Theta okay then multiplied by one more of these so R * the quantity cosine of theta plus I * the S of theta okay so what is this going to give me again R 2 * R is going to be R cubed so R cubed again notice this exponent here matches this exponent here okay so basically after that inside the parentheses you're going to have the cosine of 2 thet plus Theta is going to be 3 Theta okay plus I times the S of 2 theta plus Theta is 3 Theta so again this exponent here of three is going to be your exponent here on R and then inside the parentheses is going to be multiplying your argument okay and this should make sense because if I'm raising this to the third power or I'm cubing this then basically I have three of these so I have one two and then a third one right so it makes sense that I have one two three Rs so it's going to be R cubed right R * R * R and then you have this one this one and another one so if you're adding the arguments you have theta plus theta plus Theta which is 3 Theta okay so that's what's going to work if you raise this to the fourth power you see this same thing fifth power sixth power whatever you want to do any real number okay so formally this is our theorem here you can write this down for your notes basically if you have some complex number in polar form so let's say R * the quantity cosine of theta plus I * the S of theta and this is raised to some real number n okay again we're going to use just positive integers today then basically you can say it's R to the^ of n times the quantity you have cosine of n * theta plus I * s of n * Theta okay so this n is going to be multiplying your argument and it's also going to be your exponent on r or the absolute value of that complex number again for the Shand remember this is the C the I the S that's where this abbreviation comes from so we have R to the^ of N and then you have this CIS the S some people say and then we have n * Theta all right let's jump in and look at an example so we're going to start with something very very easy so we have the negative of 5 * 3 over 2 + 5 I and this whole thing is being squared now obviously this is in rectangular form and since this is just being squared you could expand this and just perform the operation that way but what we're going to do is use our little theorem so we're going to first write this in polar form then we'll use the theorem okay so I'll say this is equal to I'm going to put my R * the quantity cosine of theta plus I * the S of theta okay and this is going to be squared okay so I'm going to fill out my R and my Theta so to get those remember your R okay which is the absolute value of the complex number is the square root of you're gonna take your real part which is the negative of five * square of three over two and you're going to square this and then plus you're going to take your imaginary part which is 5 over two okay and you're going to square this so let me scroll down and get some room going and we'll come back up in a minute basically if I have -5 * 3 over 2^ 2ar the 5 S is 25 the sare < TK of 3 S is 3 2^2 is 4 so this is 75 over 4 so let's just replace this and put 70 5 over 4 and that is a terrible five so let's fix that and then the 5 half squar is just 254s so let's put plus 254s now 75 + 25 is 100 and 100 divided 4 is 25 okay and obviously 25 is a perfect square square of 25 is five okay so let's say this is five now let me come up here just a little bit so let me scroll up just a little bit and I'm going to erase this R and I'm going to put a five in there for the actual value and now I need to figure out what Theta is so lots of ways to think about this remember your tangent of theta if you're thinking about this as x + y i well it's going to be y overx okay if you're think about this as a plus b i then it's B over a it's the imaginary part over the real part okay you have to be careful about this because when you use your inverse tangent function you're only going to get something in quadrant one if this guy's positive and something in Quadrant 4 rotating clockwise if this guy's negative okay so you got to be real careful here so let let's come up and think a little bit about this so the Y part is five halves okay so that's five halves then I'm dividing by a fraction so I'm going to multiply by the reciprocal the X part or the real part is going to be the negative of 5 * of3 over two I'm going to flip that so I'm going to put 2 over 5 * 3 okay I'm dividing fractions that's why I did that okay so what I would see here is that this would cancel with this and this would cancel with this so I have the negative of one over the < TK 3 rationalize the denominator this is > 3 overun 3 and so this gives me the negative of you have the square 3 over three okay so let's think about this for moment if I type this into my calculator okay as the inverse tangent of this guy I'm going to get something in quadrant four and I'm also going to get a negative angle okay so it's going to rotate clockwise that is not what I want okay so what I'm going to actually do is I'm going to do the inverse tangent of I'm just going to make this positive so > 3 over 3 this is going to give me my reference angle okay so if I punch this into the calculator I get pi/ 6 okay if you want to work with radians or you get 30° if you want to work with degrees now this is not my answer because I'm not in quadrant one okay let's get rid of this we don't even need this anymore okay let's think about where we are if you were to graph this complex number on the complex plane then basically you think about the fact that this is negative so the real part's negative and this is positive right so this is positive so from the origin I would go to the left and I would go up that puts my complex number in quadrant two okay so in quadrant 2 I want an angle with a reference angle of pi/ 6 or 30° okay so basically you can say Theta is equal to you could do PI minus your pi over 6 okay and I can do 6 pi over 6 to get a common denominator so this is 5 pi over 6 which is 150° right if you wanted to do this with degrees you do 180° minus 30° so again it's 150° so let's do 5 pi over 6 so 5 Pi 5 pi over 6 so that's the tedious part so let's just erase this and this and go 5 pi over 6 of course you can put 150 degrees in there if you feel more comfortable with that that's up to you okay so now that we have this done finding our result here is really easy right using our little theorem this R value here is raised to this power so five is squared five squared is 25 that's easy then times the quantity the cosine of this gets multiplied by this exponent so we'll have let me do this in a different color 2 * 5 pi over 6 okay then plus let me return to this other color so plus your I times the S of again it's going to be 2 * 5 piun over 6 okay that's all it is it's very very simple most of the work is just putting it in polar form so once we have this we can think about okay well this would cancel with this and give me a three this would cancel this and give me a three so this is going to be 25 * the quantity you have your cosine of 5 piun over 3 okay plus I * the S of 5 piun over 3 okay and 5 pi over 3 is 300° if you want to write it like that again up to you now this is the answer in polar form if you want to put it in rectangular form again it's all over the place in terms of what your teacher would ask for let just get a little quick practice with that so the cosine of 5 piun over 3 again the reference angle there is pi over 3 okay or 60° so the cosine of pi over3 is going to be a half right and cosine in Quadrant 4 is going to be positive so this equals 12 now the S of 5 pi over 3 again pi over 3 if you think about the sign of pi over 3 well that's going to be > 3 over2 but we're in Quadrant 4 so sign is negative so this is the Nega of the of 32 okay so let's go ahead and replace these I'll go ahead and say this is equal to 25 * I'll say this is2 so this is going to be2 and then plus you have I multiplied by this is going to be the negative of thek of 3 over2 probably better to kind of switch this around here and then maybe scooch this down a little bit for formatting let's close this down Okay so let's go ahead and distribute this 25 in so this would be 2 5/ 2 and then basically I have plus negative here might as well just write minus it's a little bit easier I'll say minus 25 * 3 over two now you can put your I out here and then sometimes people put the I there it's really up to you it's very clear here that it's not under the square root symbol so I would just leave it like this so this would be your answer in rectangular form your 25 over2 - 25 * 32 and then time I all right let's look at another example so this one I'm going to give it to you in polar form make it a little bit quicker so we have 2 multip by the quantity cosine of 7 piun 4 + I * the S of 7 piun 4 and this whole thing is raised to the fifth power okay so if it's in polar form already it's super easy right this r value here this two is going to be raised to the fifth power so you have 2 to the 5ifth power we know that's 32 but we'll just simplify that in a moment then inside the parentheses you'll have your cosine of again this exponent here of five is going to be multiplied by the argument so 5 * 7 piun over 4 okay maybe make these parentheses a little bit bigger then we'll go plus I * the S of 5 multiplied by the 7 pi over 4 let me close the parenthesis there so we'll say this is equal to let's go ahead and say this is 32 right 2 to the 5th power then times the quantity 5 * 7 is 35 so this would be the cosine of we'll say let me make that s a little bit better so we'll say this is 35 piun / 4 and then we'll say plus I so plus I * the S of 35 pi over 4 now again some teachers don't care you can leave this as your answer if they say we want it in polar form that's really up to you if you need to put it in rectangular form it's kind of hard to do that with 35 Pi over4 I mean I know we can use calculators but if you wanted to think about this off the unit circle for myself I always like to put it between zero and two Pi with zero included in 2 pi knot okay so if you think about 35 over 4 forget about the pi for a second 35 over 4 that's really close to 36 over 4 36 over 4 is 9 okay so that tells me if I think about 2 pi 4 Pi 6 Pi 8 Pi okay I would want to subtract 8 Pi off to get myself a co-terminal angle so if I did 35 piun over4 minus 8 Pi of course I could write this as 32 piun over 4 32 pi over 4 this is 3 Pi 3 pi over 4 okay so it looks a little bit easier to work with so I'm just going to change this again some teachers require it some don't care so 32 * the quantity cosine of 3 pi over 4 let me make that a little bit better okay then plus I times the S of 3 pi over 4 okay so that's done so if you want your answer in polar form again this one if your teacher doesn't care this one if they do okay if you want your answer in rectangular form well then again I've got to go through and figure out what is cosine of 3 pi over 4 and what is s of 3 pi over 4 so if you're not used to working with radians if you haven't memorized your unit circle this is 135 degrees so you know this is in quadrant 2 so sign is positive okay and this guy is negative and remember with 135 degrees you have a 45° reference angle or in terms of radians it's pi over 4 okay so for both of those it's > of 2 over2 if you think about pi over 4 right but this one's negative and this one's positive so this would be the negative of the of 2 over2 and this would be posi > 22 okay so we would say that you have equals I'll go 32 * the quantity I have the negative of theun of two over two and then plus you'll have your s < TK of two over 2 * I okay again the I gets changed around based on what book you're looking at what teacher you're working with sometimes they put it out in front there it's really up to you so what I'm going to do here is just say that 32 is multiplied by this 32 ID 2 is 16 so this would be -16 * < of2 okay plus 32 * this the 32 ID 2 is 16 so I'm going to say 16 * > 2 and then time I now again this is all over the place in terms of how the formatting goes typically with this you'll see 16 I and then times theun of two again you don't want to think that it's underneath this square root symbol so that's why this is done okay so this would be your answer in polar form right here okay and then rectangular form we say -16 * > 2 + 16 I * < TK of 2 all right so now let's talk about the root Ro of complex numbers so I'm going to use this first example here to derive a little formula we can use if you don't want to watch it because it's a bit tedious you can just skip ahead to where we're doing the examples so first and foremost every nonzero complex number is going to have exactly and distinct complex nth Roots so you might read this in your book and think what in the world does that mean well what it means is that every given non-zero complex number is going to have exactly two square roots 3 cube roots 44th Roots 5 fth roots Ro so on and so forth so what we can do is use this example here to come up with a little formula okay let's say that I asked you to find the three complex cube roots of this number okay what does that mean well let's say I had let's just go back to regular numbers let's say I had just the number eight okay so the number eight what if I wanted to find the cube root of that well I'm looking for a number that when cubed gives me this number okay so we know this is two right so two cubed gives me eight so I'm going to do the same thing I'm going to be looking for a number that when cubed produces this number here okay so all I'm going to do let me just rewrite this so I'll just say that 3 * the quantity cosine of let's say 45° plus the I * the S of 45° is going to be equal to again I'm just going to take a generic complex number in polar form so I'm going to say R * the quantity cosine of I'm going to use Alpha here instead of theta so plus I'll do I times the S of again I'm going to use this Alpha okay so now I'm just going to raise this to the third power there is nothing different between this and when I just showed you with eight being equal to 2 cubed okay I'm just taking this number raising it to the third power so this number right here is going to be a cube root okay of this number because raising to the third power produces this just like two is a cube root of eight because 2 to the 3 power is eight okay it's all it is so from this point on what I'm going to do is just simplify things and come up with a little formula okay so I'm just going to say 3 * the quantity cosine of 45° plus I * s of 45 degrees okay is equal to if I use my theorem from earlier R would be cubed so R cubed then times the quantity cosine of I'm multiplying the three by the argument so 3 * my Alpha there plus I * the S of 3 * Alpha okay so in order for this to be true you've got to think about the parts of this equation so this would have to be equal to this okay and then this would have to be equal to this and then this would have to be equal to this okay so in other words R cubed R cubed would have to be equal to three which tells me if I take the cube root of each side okay the cube root of each side that R is going to be equal to the cube root of three okay we'll see this in our little formula in a moment for right now this is what we have okay so let me bring this up here so it's out of the way and then the next piece would be that the cosine the cosine of 45° is equal to the cosine of 3 * Alpha okay and then also the sign the S of 45° is equal to the S of 3 Alpha okay so let's think about this for a moment we know about co-terminal angles okay and we know that let's say the cosine of 45° has the same value as the cosine of if I added 360° to that so if I said the cosine of 45° this would be equal to the cosine of 45° and you can pause the video and punch it into your calculator you should know that those are equal it's also equal to the cosine of - 315° okay if you wanted to do it that way okay so knowing this fact what we could do is set 3 * Alpha so 3 * Alpha equal to this 45° okay plus I'm going to do 360° times some integer K okay now we don't have to are negatives I'm going to show you how Solutions just keep repeating themselves I'm just going to work with 0o 1 and two okay and I'll show you that when you get to three four five six things repeat also if you go negative 1 Nega -2 neg3 you're just getting repeat Solutions okay so what I would do here is I would simplify this a little bit and I would divide both sides by three okay so both sides by three so I'd have that Alpha is equal to 45° divid 3 is 15° okay then plus 360° divid 3 is 120° and then times K okay so to get the values for Alpha you're going to see in our little formula you plug in if I'm trying to find the cube root then I'm going to plug in zero one and two okay so I start with zero then I go to one then I go to two if I was trying to find the fourth root I would go 0 1 2 3 so it's always one less than that number okay so if it's four you're going to three if it's five you're going to four if it's six you're going to five you know so on and so forth so let's go ahead and go through and say our first Alpha would be 15° plus 120° * 0 so 120° * 0 so this is basically going to be 15° then another Alpha would be 15° + 120° * 1 so this is going to be 135° so 135° then our other Alpha is going to be 15° plus you're going to do 120° * 2 and this is going to give me 255 de okay and then again we don't need this but if you did go to three so Alpha is equal to 15° plus 120° * 3 well this is 360° + 15 is 375° 375 degrees and 15° are co-terminal angles okay so if I was to say the cosine of 15° or the cosine of 375° it's the same okay so you don't need to go three and four and five and six you're just going to get repeat Solutions Okay so don't need that you also don't need to do the negatives because you're just going to get repeat Solutions so let me erase this so coming back now let me show you that these Solutions work okay so let me box this off remember originally we were looking for the three cube roots so the three numbers that I could plug in here and Cube them and get back to this number so let me show you with the first solution for Alpha so we have our R value which is the cube root okay of three and then it's multiplied by the quantity cosine of alpha so one of those would be 15° plus I * the S of 15° okay let me get rid of this one then the next one would be the cube root of 3 time the cosine of it would be 135° plus I * the sign of 135° okay let's get rid of this one and then your last root here so the last cube root would be the cubot of 3 * the quantity the cosine of 255° plus I * the s of 255° okay so let's get rid of this now if you were to cube each one of these you're going to get back to this right here and we can just do it mentally we don't need to actually go through it if we Cube this number here again following that theorem from earlier this guy right here would be cubed so cubing the cube root of three gives me three so it gives me back that number then if I multiply 3 * 15 Dees I get 45° If I multiply 3 * 15° I get 45° so you can see that cubing this first first one right here gives us this back now I'll let you do this one and this one on your own just Cube them you're going to see that the angle you get in each case is going to be co-terminal with this angle here that's 45° in this case it would be 45° and in this case it would be 765 de okay so let's go from our example and make a nice little formula so if n is any positive integer and R is a positive real number and Theta is in degrees then the non zero complex number let's say R * the quantity cosine of theta plus I * the S of theta is going to have exactly indistinct nth Roots given by the following so we're going to say that we have the nth root okay of this R you saw earlier we were looking for cube roots so we had the cube root which was n n was three of in our case it was three so it was the cube root of three okay and basically we're going to multiply this by the quantity you'll have the cosine of alpha plus I * the S of alpha okay and basically what Alpha is going to be I'll do degrees first and then I'll do radians so Alpha is going to be Theta okay so this guy right here divided by your n okay so you saw earlier what we had was 45° divided by 3 which gave us 15 degrees okay then plus you're going to do 360° times k k is just some integer and I'll talk more about that in a second divided by n so in our example you saw 360° / 3 okay which gave us 120° and then multiplied by K where basically K is going to be equal to 0 comma 1 comma 2 comma 3 we'll put the three dots here to show the pattern continues and then we'll go n minus one okay so in our case our n was three so we went 0 one2 then three would give us a repeat solution so with four so with five so with six you know so on and so forth now if you're working with radians then you would adjust this formula and basically instead of 360° you would just put 2 pi in there okay in terms of radians nothing else is going to change so just depending on if you're working with degrees or radians you would either want to use 360° for degrees or 2 pi for radians all right let's go ahead and look at an example so we want to find all nth Roots okay and we're going to write our answer in polar form so we see that n is set equal to four so we want to find the four fourth roots okay so I'm going to start by writing this guy right here this -3 * 3 - 3 I in our Polar form okay and so let's go ahead and say that the R value would be what the square root of we take the negative of 3 * > 3 we're going to be squaring that plus you're going to take your -3 okay we're going to be squaring that so the r is going to be equal to we'll say the square root of -3 2 is 9qu of 3 S is 3 9 * 3 is 27 so this is 27 + -3 being squared is 9 27 + 9 is 36 so r equal the square < TK of 36 so R equals 6 Okay so we've got that part down let me go ahead and say that in polar form this -3 * > 3 - 3 I would be equal to i' put a six there times you're going to have your cosine okay your cosine of theta I'll replace this in a second plus your I * your s of theta okay okay so for Theta again remember that the tangent of theta is equal to take the imaginary part so -3 over your real part which is -3 * < TK of 3 and so with this we see the negatives are going to cancel the threes are going to cancel so you have 1 over < TK of 3 again rationalizes the denominator so sare of 3 overun of 3 so this is going to be what Square < TK 3 over 3 so if I did my inverse tangent so my inverse tangent here of of 3 3 over 3 let me make that three a little bit better if we punch this into our calculator you're going to get 30° or pi over 6 if you want to work with radians let's just work with degrees so I'm going to say this is 30° so this is my reference angle so let me erase this and let's think about where this complex number would be if we graphed it so this is negative and so is this right so I have a negative real part and a negative imaginary part so from the origin I would go to the left and then I would go down okay so that means I'm going to be in quadrant three so my Thea here would be 180° plus the reference angle of 30° which is 210° okay so I can just erase this and this and I'll go ahead and put 210° there so 210° once you have this in polar form it's just a matter of plugging into your formula you want to do your nth root of R so in our case n is four and R is six so we want the fourth root of six so we want the fourth root the fourth root of six and you're going to have four of these but I'm just going to start out with the first one okay inside of here what are we going to have we're going to have the cosine of again we call this Alpha plus I * the S of alpha okay and Alpha since we're working with degrees Alpha is going to be equal to you have your Theta divided by n in this case the N is four plug that in a second plus you have your 360° / n * K where we say that K is equal to in this case because it's four it's going to be 0 comma 1 comma 2 and three okay so you just go up to one less than whatever n is that's all it is so in our case let's just plug this in so n is going to be four so let's erase this and put four in each case Okay and then Theta is coming from here so it's 210° okay so let's put 210 Dees here so if I divide 210 degre by 4 it's going to give me 52.5 de so 52.5 de and and then 360° divided 4 is going to give me 90° okay so 90° so basically to set this up the first one would be again if this guy was Zero plug that in this is gone so you just get 52.5 degrees okay so you can put that in there so 52.5 and 52.5 now you can pause the video and you can raise this to the fourth power and you'll see that you get back to this guy right here okay and then of course you can convert that back into rectangular form if you wanted to to get back to that original form we started with okay but let's go ahead and keep going and basically let me drag this down just a little bit and I'm going to slide my screen just a little bit down okay and basically now I'm going to do the fourth root of 6 time the quantity I'll do the cosine of and then I'll leave that blank for now and then I times the sign of and let me leave that blank for right now so the next one would be this guy would be one now okay so I'm going to plug in a one so basically 90° * 1 is 90° and 90° plus 52.5 is going to be 142.5 de and I forgot my degree symbol here and here so let me put that in so 142.5 de let me make that decimal point a little bit better so from here all I'm going to do is add on 90° okay two more times so I can just get rid of this okay once you understand the pattern and basically I can just copy this and so 142.5 + 90 is and this would be again 232 232.103 22.5 and 322.34 fourth roots for that number I'm leaving these in polar form all right let's take a look at another example I think this is going to be pretty easy for us at this point but I want to throw something in here sometimes your book will ask you to graph the roots on the Argan diag or the complex plane so let's go ahead and do this so we have8 + 8 I * 3 again n is going to be four so I'm looking for the four fourth roots now again this is given to us in rectangular form so I need to convert it into Polar form first okay so the R value times again the quantity cosine of theta plus I * the S of theta so let's do this quickly so R is going to be what the square < TK of 82 is 64 and then if you have 8 * of being squared it's 64 * 3 which is 192 okay so you're adding those two together and that's going to give me 256 if I took the square root of that I would get 16 okay so let's just erase this and put 16 here then for Theta it's going to be where the real part is negative so I'm going to the left the imaginary part is positive so then I'm going up so from the origin if I go to the left and I go up I'm going to be in quadrant two okay so let's say that Theta will be equal to 180° minus my reference angle and the reference angle here let's go 8 time < TK 3 and I'm going to divide that by forget about the negative let's just do eight this is going to cancel with this so this is square otk of three okay so I'm going to say the tangent inverse the tangent inverse of this guy right here is going to give me 60° okay so if I subtract 60° away from 180 de okay because this is my reference angle in quadrant 2 well it's going to be 120° okay so that's my Theta let's get rid of this let's put in 120° and 120 degrees okay so that's pretty quick all right so from here I need to get the four4 roots so in each case I'm going to take the fourth root of 16 right because again it's the nth root of R so n is four R is 16 so the fourth root of 16 is two right because 2 to the 4th power is 16 so I'm going to put two two two and two okay let's go ahead and set this up in each way so cosine cosine cosine and then cosine okay so let's going to go here here here and here remember when you think about the formula if you call this Alpha it's equal to again you take Theta okay which in this case is 120° and you divide it by n so you divide it by four so in this case that would be 30° okay then plus you have your 360° times your integer K divided by your n so 360° divided by in this case it would be 4 would be 90° okay so this would be 90° * K so for the first instance this would be 30° so let's go ahead and put 30° here I'll fill out the rest in a moment okay in the next instance it's going to be plus 90° so this would be 120° then plus 90° so this would be 210° then plus 90° so this would be 300° so let's get rid of this and let's put our plus plus plus plus I I I and I and then the sign the sign the sign and then the S of 30° close that down and this will be 120° close that down and then 210° close that down and 300° okay so these are our four fourth roots okay for this number in polar form okay so how would we go about graphing these complex Roots well basically each of these can be represented as a position Vector in the complex plane so notice that your magnitude in each case is two okay so two two two and two so what you could do to represent that is you could have a little circle with the center at the origin and where the radius is two okay so each one of these as a vector is going to have an initial point at the origin and a terminal point that's going to touch that Circle that you drew okay and then in terms of the direction angle for the vector just get it from here so the first one will be 30° the second will be 120° the third one will be 210° and the fourth one will be 300° okay so these guys are going to be spaced 90° apart okay so equally spaced 90° apart so if we go to our little coordinate plane and I know that most people put imaginary here so you can put that if you want so you can put imaginary and then for this you could put real but again if you write this as X Plus Yi then X could still be your real axis and y could be your imaginary axis it's really up to you or your teacher okay so I'm just putting it this way so when we think about this again the first solution was 2 * the quantity of the cosine of 30° plus I * the S of 30° we abbreviated it here but you can see that I have a position Vector the initial point is at the origin the terminal point is going to touch this little circle that I drew okay with a radius of Two And A Center at the origin okay so you see that this is a 30° angle okay and then if we go another 90° here so this is another 90 ° you can see that little 90° symbol there basically I'm going to get to my next solution so the angle here is 120° again if I started here 30° plus 90° gives me 120° okay for the whole thing and then basically you can see that it touches again our little circle if we keep going around this is another 90° and then this is another 90° okay so it keeps putting you to another solution so this angle here if I started here and went all the way to here this is 210 degrees if I started here and went all the way around to here this is going to be 300° okay so that's basically how you can sketch the graph if you're given this as an assignment again just take your magnitude okay in each case ours was two and draw yourself a little circle with the center of the origin and the radius of that number okay in our case it's two then what you want to do is have position vectors so they start at the origin and their terminal point is going to be on that Circle and your direction angle each case is given by the angle so here it's 30° here it's 120° here it's 210° and here it's 300° all right so let's wrap up the lesson and look at another example here we're going to be solving a simple equation using the roots of a complex number okay let's say we saw something like x cubus 27 and this was equal to zero the first thing we know that since this polinomial here is of degree three okay we have this exponent of three here and not higher well we know we're going to have exactly three complex solutions for this equation okay that's the fundamental theorem of algebra so how could we use the roots of a complex number to solve an equation like this well the first thing is I could add 27 to both sides okay and I could set this up and say really this is X cubed is equal to 27 okay let's pause and think about this for a minute I am looking for a value that I can plug in here and Cube it and get 27 so in other words what number cubed gives me 27 so if I found the three complex cube roots of 27 7 I could plug them in for x and I would have a true statement okay so you might be thinking well 27 is a real number remember all real numbers are complex numbers the real number system is a subset of the complex number system I can write 27 as 27 plus 0 I okay so with this format here I could also write this in polar form this is really easy to do if you think about the R value it's just going to be 27 okay because you think about this as okay well R is equal to the < TK of 27 2 + 0 S 0 S you can just get rid of that okay thek of 27 s is just 27 so our r value is 27 then in terms of this inside part it's easy as well so you'd have the cosine of again theta plus I * the S of theta so you don't need to do any inverse tangent function here or anything like that just think about this complex number if you graphed it on the complex plane you would simply start at the origin and go 27 units to the right okay so there's no movement at all vertically right you're on the x- Axis or the real axis if you want to think about it that way so really this is going to be a0 degree angle right I haven't moved off the real axis so I'll just go ahead and put this as a0 degree angle okay and that's all it is so this is me writing the number 27 first in rectangular form and then in polar form now in order to find my three complex cube roots I'm going to start with my simple little formula so I have the nth root of R so in this case n is three right because I'm looking for a cube root and R is 27 well the cube root of 27 is three right 3 * 3 is 9 9 * 3 is 27 okay so I can set this up I know I have three of them so I'll go three I'll go three and I'll go three okay let me just write all this out so we have cosine cosine and cosine and we're going to use Alpha we're going to use Alpha and Alpha and I'll replace this in a moment so plus plus and plus I I and I and then S S and S okay and I'm going to use Alpha again Alpha again and Alpha again okay so again these are all going to change based on that input for K so to find Alpha let's go ahead and set this up over here put a little border here to find Alpha the idea is that it's Theta whatever this is in this case 0 degrees divided by n which in this case is three okay so 0 degrees divided by 3 is just 0 degrees so this is 0 degrees you can put it there or leave it off it doesn't matter it's not going to change anything I'm just going to leave it for the sake of completeness so then plus you have 360° / n in this case that's three so this is 120° okay and then times your K now K is going to be 0 1 and two basically your pattern is what it's 0 comma 1 comma 2 comma 3 comma dot dot dot n minus one let me put a common there to make that complete and basically you have three so 3 - 1 is two so in this case we would only go up to two Okay so that's how that works so i' start with a zero there and of course 0 de plus 0 de is just 0 degre so the first guy here is just going to be 0 Dees so I can put that in here and here and that should make sense if I cubed this 3 cubed would be 27 and then you would basically have 3 * 0 de which is still 0 deg and 3 * 0 deg which is still 0 de so in polar form I would be right back to this here 27 * the quantity cosine of 0° plus I * the S of 0° which again in rectangular form is this and written just as a real number is this okay so then the other one I'm just going to add 120° each time time if I plugged in a one there this would just be 120° so 120° 120° add another 120° if I plugged in a two there this would be 240° so 240° and 240° okay so these three numbers here in polar form these three complex numbers in polar form if you Cube them they would each give you 27 now typically if you were in an algebra class you would not give this as your Solutions right you would want to convert this back into rectangular form okay so how would we do this the cosine of 0° is going to be 1 and the S of 0 degre is going to be zero so basically you would have 3 * 1 which is three so I would say x is equal to 3 okay that would be my first solution for this one the cosine of 120° and the S of 120 degrees let's think about this there's a 60° reference angle here here here and here right 120 de 240 degre you have a 60° reference angle so the cosign of 60° is going to be2 and the sign of 60° is < TK 3/ 2 okay so it's just a matter of getting the signs right remember sign is going to be positive in quadrants 1 and 2 and 120 degrees is in quadrant 2 cosine is going to be positive in quadrants 1 and four okay so basically cosine is negative in each case because this is quadrant two this is quadrant three and sign is positive here and negative here okay so to get my Solutions let me write this out I'm going just going to scroll down a little bit so we have some room for this one I would have what three times this would be the quantity -2 plus this guy is going to be < TK 3 / 2 okay * I so if I go ahead and distribute the three into each term I would have -3 Hales and then plus my 3 * < TK 3 over 2 * I okay so that's another solution then the other one let me just go ahead and keep this as negative and I'm going to change this to negative now so basically you would have -3 Hales again and then minus the 3 minus the 3 * of 3 over 2 and then time I okay so I would write this in a more compact form I would just do plus or minus here and get rid of this that way you can more cleanly write your three solutions so if you were in an algebra class and they gave you this equation you would want to give this as a solution now in this particular case this is probably not the fastest method to do this but it's one such method that you can get a solution now let's think about this graphically if I wanted to show these Solutions again I'm going to do this in polar form okay so on the Argan diagram or the complex plane again the magnitude is three in each case so that's the length of your vector so if you're making a circle okay with a radius of three that is centered at the origin each one of these vectors is going to touch that Circle okay that's going to be your terminal point that lies there okay then your direction angle for your vector is going to be 0 degrees then the other one's going to be 120° then the other one's going to be 240° okay all of them are position vectors so they start or their initial point is at the origin so graphically this is a rough sketch of this guy again I have a little circle that's drawn and the radius is three and the center is at the origin okay then you have your three position vectors notice that they're all going to have an initial point at the origin the terminal point in each case is on that Circle again where the radius is three and the center is at the origin so your first angle is 0 degre okay it's given by this three SS 0 degre again this is the abbreviation the second angle is 120° okay so we come from here we rotate 120° and now we have that Vector there okay and the terminal points there and then the last one if I rotate around here this is going to be a 240° angle okay so that's how you would sketch the graph of this guy you'll notice that these guys are equally spaced 120 degrees apart right 120 degrees this is 120 degrees and this is 120 degrees as well so this is how you can graph your Solutions if you're asked to do them again figure out the magnitude or the length of the vectors you're going to sketch yourself a little circle okay with a radius of that number in this case it's three and a center at the or so all of your vectors are going to be position vectors initial point at the origin terminal points going to be on that Circle and you just get your direction angle from your roots so we had 0° 120° and then 240° in this lesson we want to talk about polar coordinates all right so up to this point we've exclusively used something called the rectangular coordinate system or some people refer to this as the cartisian coordinate system after the inventor and we've worked with this guy to plot these points or you could say ordered pairs and also to graph equations so I want you to recall that the rectangular coordinate system is going to feature the x- axis which is horizontal and the Y AIS which is vertical okay the meeting point between the two is going to be right here and this is referred to as the origin okay so this is your origin origin okay and when we specify this using an ordered pair remember this is the x value comma the yvalue so for the the origin it's going to be 0 comma 0 okay because the x value is zero and the Y value is zero now if I think about another Point let's just do a few so let's say we do something like 6 comma 7 let's say we do -6 comma 7 let's say we do -6 comma -7 and let's say we do 6 comma -7 okay so these are directed distances okay from the origin so in other words the first guy here okay or the first component as they say the x value tells you how to move horizontally okay if it's positive I'm moving to the right if it's negative I'm moving to the left so for this guy 6 comma 7 I'm moving to the right by six units okay so let me just circle this temporarily and then the second guy there the y-coordinate is telling me how much to move vertically right so if it's positive I'm going up if it's negative I'm going down so let me go ahead and from the origin go up by seven units okay so you basically want the meeting point between the two so you can go six units to the right and seven units up or you can go seven units up and six units to the right okay either way you're going to end up at this point right here this is going to be your 6 comma 7 and I know most of you know how to do this already but basically we just want to recap this before we get into the polar coordinate system now for the next one we have -6 comma 7 so that means I'm going 6 units to the left right because it's ne6 and then I'm going seven units up because it's positive s okay so six units to the left seven units up so this is my -6 comma 7 okay the other one is-6 comma 7 so both are negative so it means I'm going to the left okay because it's a -6 and I'm going down because it's A7 so 6 units to the left and seven units down that's going to put me right there okay so this would be -6 comma -7 and then lastly we have 6 comma -7 so this is positive for X and Y is negative so it means I'm going to the right and I'm going down right so six units to the right and seven units down that's going to put me right there okay so this is 6 comma -7 all right now let's think a little bit about something called the polar coordinate system and to start this out I'm just going to show you how we get the polar coordinates so let's say I had something like the point where we have 5 comma 6 so this is as an ordered pair again this is X comma y so starting at the origin I know that to get to this point I could go five units to the right and six units up so that puts me right there or alternatively I can start the origin and go six units up and five units to the right so basically that gives me the directed distances from the origin we know that but we also want to be able to write this with polar coordinates which is going to be an R value comma a Theta value okay so where does this come from well essentially what we can do is we can create a little position Vector okay so the initial point is at the origin the terminal point is going to be at that point here okay with our 5 comma 6 so what I could say here is that with rectangular coordinates you have X comma Y in this case it's 5 comma 6 so hey start at the origin go five units to the right and six unit up with polar coordinates it's going to be R comma Theta okay where R is the distance from here to here okay if you're thinking about this as a vector remember the length of the vector is called the magnitude of the vector so in this case that's going to be your R right so this is the square Ro of 61 and then the Theta here is the directed angle so in other words how much do I need to rotate counterclockwise so starting from the positive xaxis if I'm facing this way I need to rotate by 5019 de to get to this position Vector okay so let's say this is 50.1 n° this is an approximation and I'll talk about how to get this in a moment but essentially I want you to see that it's just two ways to get to the same point right I could say hey start at the origin go five units to the right and six units up or I can say hey start at the origin okay facing the positive x-axis rotate 5019 Dees and then walk forward by the square root of 61 units okay so two ways to kind of do the same thing now if we want to talk a little bit about where this r value came from and where the Theta value came from see here how I made a little right triangle okay and essentially we see that the length of this vertical leg here is given by the yvalue of six right so this is basically from 0 to six so this is given by six the length of this horizontal leg here is given by the x value of five okay so really easy to work with this you just use the Pythagorean theorem so if you want your distance from here to here or the magnitude of the position Vector the distance squared would be equal to you would have 5^ SAR and then plus 6 squar so D your distance between these two points would be equal to the square root of 5^ s is 25 plus 6 s is 36 if you sum those two together you're going to get 61 okay so the distance between these two points or your r value is a square of 61 okay so that's where that came from in terms of theta okay calculating that basically for most textbooks they're going to use the inverse tangent function remember the tangent of some angle Theta is y overx okay so in this case your yv value is six your x value is five so let's go ahead and say that the tangent of theta is 6 over5 okay and then because we're in quadrant one here we can go ahead and just strictly use the inverse tangent function right so I can just say the tangent inverse of 6 fths like this is going to give me and this is an approximation so we'll say approximately 50.1 n° okay now if you're in another quadrant you have to be really careful because remember the inverse tangent function is only going to give you something in quadrant one okay if this argument is positive or in quadrant four rotating clockwise if this argument is negative so be really really careful we're going to work a lot of examples today and I'll show you how to deal with that all right now what I want to do is just take you through the basics of the polar coordinate system so with the polar coordinate system everything is going to be based on a fixed point which is called the pole or the origin okay and essentially most books are going to name this with a capital letter O so this is our poll okay to get things started from there you're going to create an initial array which is called the polar axis so that's going to be this guy right here okay so that's your initial Ray called The Polar axis and generally speaking this is drawn in the direction of the positive x axis now once we have this set up we can find a point p in the plane using the coordinates R comma Theta okay so we have our pole here okay that's set up we've constructed the polar axis so now if I give you R comma Theta I can locate any point P okay so R is the distance from the pole o okay to that point P okay so basically if you think about this as the vector op what is the magnitude of that Vector okay that's going to be R and then if I think about my directed angle Theta well this is just telling me how much do I need to rotate counterclockwise from The Polar axis to get to this Vector op all right so what I want to do now is talk a little bit about how to convert between the polar coordinates and the rectangular coordinates so what we did here is we just took the drawing from the last section okay and we just put it on top of basically a coordinate plane that we're used to right so the rectangular coordinate plane you'll see that we have the pole that's going to be fixed at the origin of the cartisian coordinate plane and basically the polar axis is drawn to coincide with a positive xais okay so once we do this we can obtain a few insights here okay so we have this point p in our plane which can be assigned the rectangular coordinates X comma y or the polar coordinates R comma Theta so let's say I gave you R comma Theta and I said hey I want you to give me X comma Y how could you do that well if you consider that the coine OFA is equal to what remember I formed a little right triangle here so essentially I can say that it's going to be the adjacent which is going to be X here okay over the hypotenuse which is going to be R here okay so X over R and then if I think about the S of theta it's the opposite which is y over the hypotenuse which is R so the S of theta is y over R and I didn't make that equal sign very good okay in each case I can multiply both sides by R I can solve for x or y okay okay so I can say that X is equal to R * cosine of theta so X is equal to R * cosine of theta okay that's all you need to you to find X and then for y it's going to be R * s of theta okay so y y is going to be R * the sign of theta okay so pretty easy to go from Polar coordinates to rectangular coordinates now let's erase this and let's consider the other one which is a little bit more involved but basically if I start with my rectangular coordinates my X comma Y and I want to go to my polar coordinates R comma Theta let's think about that so essentially what I'm doing here is I'm again returning to the fact that for r i want the distance from here to here okay so if I'm given the X comma y again I'm just using the Pythagorean theorem R 2 is going to be equal to x^2 + y^2 so R is going to be equal to the square root of x^2 + y^2 okay that's all it is so very easy to remember this now the tricky one is to get Theta right because you have all these different scenarios you have to consider if you're in quadrant one you can just use your inverse tangent function right because the tangent of theta is what it's going to be y overx we know this already but if I'm not in quadrant one then basically I need to get a reference angle okay using the inverse tangent function and then from there I need to take the appropriate action based on what quadrant I'm in right so if I'm in quadrant two three or four okay so let's go ahead and think about a simple little example and so what I have here is the negative of 3 * 22 and then comma the negative of 3 * 2 over 2 okay so these are rectangular coordinates I want to go to Polar coordinates and then I'm going to show you how to plot this on the Polar grid okay so essentially this is again my x value and this is my y value okay so what did I just say essentially if I want to find the R value the R value is going to be the square root of x^2 + y^2 okay so that's the first thing I would do so all I need to do let's go the square root of I'll put R out here is equal to I'll have the negative of 3 * of two over two and this is going to be squared and then plus I've got to do this again so the negative of 3 * two over two this is going to be squared let me make this a little bit longer so it's proper and so if I think about this I know that negative in each case is going to be gone right Square negative you get a positive 3^ s is not the < TK of 2^2 is 2 and then 2^2 is 4 okay so essentially what I'm going to end up with is canceling this with this and getting a two okay so you'll end up with nine halves so I'll have nine halves here and nine halves here so basically I would have R is equal to square < TK of 9es + 9es you have a common denominator there so basically 9 + 9 is 18 you would have 18 ID 2 which is going to be 9 okay so R is equal to the of 9 which we know is three okay so let me erase this from over here let me put this as R is equal to 3 now when we think about the directed angle Theta okay this is where it gets a little bit tricky first off consider that your x coordinate is negative and your y-coordinate is negative so if you think about this on your rectangular coordinate plane if your x value is negative and your y value is negative you're going to the left and you're going down so that means that Theta is in quadrant number three okay so let's let's keep that in mind so let's do our inverse tangent function if you have the same thing over itself okay and it's not zero you're going to get one so basically I want the tangent inverse of 1 type that into your calculator and you're going to get 45° so if I'm in quadrant 3 what I want to do is I want to take 180° and I want to add my reference angle there of 45° and this is going to give me 225 degrees okay so this is 225 degrees so let's erase this so I'll just put that these are polar so the polar coordinates okay like this would be you would have three for your r value and then your Theta value be 225° so to plot this guy this is going to be 3 comma 225° on our polar grid basically what we do to construct this grid remember you have your Pole or your origin okay let me make that visible and essentially this is going to be the center for everything okay then you're going to have your polar axis which again is drawn in the direction of the positive xaxis so you have some notches here so this is one this is two three four you know so on and so forth and basically what these are going to give you you're going to make some circles and the first one is going to have a radius of one right so basically up here this is one this would be negative one if you think about as the regular cartisian coordinate system and this would be negative one right but the distance from the center or the pole to any one of those points would be one right so that Circle would have a radius of one then you would have a circle with a radius of two then three then four then five you know so on and so forth then you also have these angles here which are an increments of 15° or if you're working with radians pi over 12 radians okay so essentially if I want this 3 comma 225° I'm finding the meeting point of a circle with a radius of three okay and an angle of 225° so I can come out to this circle with a radius of three and then I can swing around okay to this angle that's 225° so that's going to put me right here okay so this would be the point on this polar grid where we have three as the radius and 225° as the directed angle okay now there's obviously other ways that you could name this point I'm going to get to this later on okay you realize that you could swing around again 360° or you could actually rotate clockwise 360° or multiples of that okay so we'll talk about that later on all right let's take a look at another example so here we're given the rectangular coordinates we have the Nega of 2 * 3 comma 2 so again this is your x value and this is your y value so to get the r value what are we going to do remember this is equal to the square root of you take the x value squared so the negative of 2 * squ 3 okay being squared and then plus you're going to go ahead and take the yv value squared so 2^ squar let's just go and write four okay so if I Square -2 I get four if I square the square root of three I get three okay so this would be 4 * 3 which is 12 so let me go ahead and put a 12 in there and 12 + 4 is obviously 16 and the square root of 16 is going to be four okay so we have the R value again that's pretty easy to get when you consider Theta okay when you consider Theta you have to think about what quadrants are in so if my x value is negative and my y value is positive that means I'm going going to the left and I'm going up right so I'm going to be in quadrant 2 so Theta is in quadrant 2 okay so always figure that out first so then what I want to do is I want to think about okay well the tangent of theta is equal to Y which is 2 over X which is -2 * 3 let's go ahead and cancel this with this and basically I have one over you would have the negative of the square of 3 let's put the negative out in front and we'll put the square of three over here and then I'm going to rationalize the denominator so > 3 over > 3 so this becomes the negative of the > 3 over 3 now I'm not going to use the negative because I want something in quadrant one when I use my inverse tangent function so let me erase all of this okay and basically what I'm going to do is I'm going to say I want the inverse tangent of let me go ahead and make this positive so I'll say the square of 3 over three if I punch that into my calculator I'm going to get 30° okay so the way this works if I'm in quadr 2 okay and I want an angle with a reference angle of 30° I want to take 180° and subtract 30° away from that so basically my Theta okay my Theta would be equal to 180° minus 30° which is 150° okay so 150° and I forgot to put my zero there so let me put that in so the polar coordinates are polar coordinates let me write this out the polar coordinates would be what it's your r value comma your Theta value so 4 comma 150° so to plot this so again the point would be 4 comma 150° as Polar coordinates and again all I'm going to do is I'm going to go out to a circle with a radius of four okay and what I'm going to do from there is just swing around okay till I get to 150 degrees let's go ahead and plot that point right there okay so this is going to be four comma 150 degrees again plotting this on the Polar grid all right let's take a look at an example where we have polar coordinates and we want to convert to rectangular coordinates so first off we have 6 comma 75° let's just plot this real quick so again what I want to do I want to go out to a circle with a radius of six okay so right here and then I'm going to swing around to an angle that's 75° so that's going to put me right there okay so it's going to be my 6 comma 75 all right so let's consider how to transform this into rectangular coordinates now so remember if we want to write this as X comma y the relationship we talked about earlier is that X is equal to R * cosine of theta and that Y is equal to R * the S of theta so these are two things you need to memorize you're going to use it basically all the time now once we think about this we have our R value of six so you might as well just put this as six in each case Okay so six and then six and then what is the cosine of 75° and what is the S of 75° so remember this is not on your unit circle okay so what you can do is use identities okay we talked about this earlier in the course to get an exact value okay if you punch it into your calculator you're get an approximation so you could write 75° as 45° plus 30° okay and you could use your cosine sum Identity or your s sum identity depending on what you're working with to get the exact value okay I'm just going to give this to you because we don't want to spend time working on something we already did so essentially the cosine of 75° I'm going to put some parentheses here and I'll put it as the square > of 6 minus thek of two and this is going to be over four okay and then the sign of 75 degrees again I'm going to put Let me kind of slide this down a little bit so it's not in the way I'm going to put this as we'll have the square root of 6 plus the < TK of two and this is going to be over four okay so this is getting exact values if you want to punch in your calculate and approximate it you can do that too it's up to you so let's go ahead and cancel this with this okay so this would basically be a three and this would be a two okay and then from there I would distribute this in so I would have 3 * of 6 so 3 * of 6 minus 3 * of Two And this is over two okay so this is going to be your x coordinate then here this cancels with this again this is a three and this is a two and then you're going to distribute that three into each okay so let me put this as a two this would be a three and this would be a three let me slide this down so this would be 3 * 2 okay so if you want to write this as rectangular coordinates you can go ahead and say that it would be 3 * 6 - 3 * 2 over 2 and then comma let me erase this is all going to not fit so let me slide this down a little bit so it all fits on the screen and then over here I'd have 3 * 6 plus 3 * 2 and then over two so kind of messy kind of nasty again your teacher might allow to approximate it so it might be a little bit quicker for you do it that way all right let's look at one more example where we're going from The Polar coordinates to the rectangular coordinates so here we have -4 comma 120° so the R value okay is going to be -4 and the Theta is going to be 120° so let's pause for a minute when you get an R value that's negative and you're asked to plot this guy can be a little bit confusing okay when you get a Theta value that's negative that's pretty simple we know that a negative angle just means I'm rotating clockwise right so let's deal with this r value that's negative first let's go to the polar grid and let's first graph pos4 comma 120° and then we'll think about what -4 comma 120° would mean okay so if I step out to a circle here with a radius of force let me go out here let me swing around okay and I'm just going to draw a little terminal Ray I'll do that the best that I can I'm not a good draw R but basically that Point's going to be right there okay so you'll notice that that point is on that terminal right okay so this is going to be the 4 comma 120 deg okay we know how to find that at this point okay now if I wanted to find -4 comma 120° all it means is that I'm basically going to reflect this point across this pole okay so here's the Pole right here and if I reflect it across the pole I'm basically going to go in the opposite direction so I'm going to go this way 1 2 3 four units so that's going to be right there okay you can think about making a terminal right here and think about having the angle let's say I had the angle I just start here and go all the way around that would basically be a 300° angle okay so -4 comma 120° would represent the same point as 4 comma 300° and we're going to talk more about these different forms in a moment okay but for right now you could see a few different ways that you could go about finding this so let me go to a fresh sheet here where this is already plotted so the first thing I'm going to label this as -4 comma I'm going to do 120 degrees okay I could have found this by going to 120 degrees like this and then basically just instead of going forwards I would just go backwards by four units okay that's probably the easiest explanation so I would just say okay I'm here so I'm going to go one I'm going to go two I'm going to go three I'm going to go four so out to a circle that has a radius four right I'm just turning around and going backwards that's what the ne4 is basically telling me okay so that would be one way I could get to my point another way and this might be helpful for you you might want to plot four comma 120° so I got to four here swing out to 120° so I'm going to be right there so this is my 4 comma 120° and notice that if I wanted to swing over here okay I'm basically going to add 18 180° to this angle okay or I could subtract 180° from the angle as well if I wanted to so so I could either say that this is going to be plus 180° which would be 300° or I could subtract away 180° and say this is 60° if I wanted to rotate clockwise okay but essentially those are the ways that you could think about doing this once you graph these a few times it becomes a little bit easier to work with in your mind but just know if you have a negative r value you're basically going to walk backwards okay that's what it means so let's go back and let's think about putting this into rectangular form okay so we know again that our x value is equal to we would have the r value which is -4 times the cosign of 120° and the Y value is -4 times the S of 120° okay so we know that 120° is going to be in quadrant 2 it's got a 60° reference angle and the cosine of 60° is going to be a half okay but cosine is negative okay in quadrant 2 so this would be negative one2 so times -2 so of course this would be pos2 so the x value is pos2 very easy to find the sign of 120° again a 60° reference angle sign is positive in quadrant 2 so the S of 60° is < TK of 3 over2 so this is going to be times theun 3 over 2 I can cancel this with this and I would have the negative of 2 * of 3 so -2 * 3 so then my rectangular coordinates would be 2A -2 * 3 okay so basically I want to go from here into our next point which is talking about Alternative forms for coordinates of a point in polar form so for a given point in the rectangular coordinate plane you're going to have an infinite number of pairs of polar coordinates so sometimes your teacher will say give me you know four examples of other polar coordinates that are the same as this polar coordin I'm giving you sometimes they want a general formula so I'm just going to give you the general formula here and then you can apply that based on the example you get okay so the first one is pretty easy to understand if we have some R comma Theta this is the same as if I said R comma and then your theta plus 360 * n degrees n is just any integer or Theta minus 360 * n degrees now the plus or minus here is given in some books you really don't need it you could just put plus because n is any integer so you can make it positive you can make it negative you can do whatever you want okay but some books will put plus or minus others so just put plus if you go back to let's say this guy right here where we have 4 comma 120° well if you think about it I could add 360° to this and I could say well also the same point could be located if I said four comma this would be 480° right because basically from here I would swing around another 360° and be right back to that same point I could add 360° again and get to 840 de then I could do it again and again and again and the same thing is true if I subtracted 360° away from 120° so some other coordinates would be something like 4 comma -240 de okay so that would all locate the same point okay so now let's look at this second formula this one is a little bit more challenging to understand but remember if you have R and you have the negative of R remember when you think about the negative of R you're basically going to be walking backwards okay so this is what this is related to so again you have this Theta whatever your angle is plus or minus you're going to have this 180° times this quantity so let me distribute this out so I'm going to put plus or minus and I'll go ahead and say you'll have your 180° * 2 N so let's say this is we'll say 360 we'll say n de and then I'll say plus I'll go ahead and say do my 180 degrees like this okay so you could have positive or you could have negative you're going to see that you can actually drop the plus or minus here as well I'll go through some examples and prove that to you most books will just write this with a plus sign but others put plus or minus okay so it's it's right either way so if we go back and we think about alternative ways to locate this point remember this was -4 comma 120° okay so all we're saying is that we could have done the negative of this number so that's four comma and then I could add or subtract 180 degre to this so if I add 180 degrees I would be at 300° okay so in other words I could have gone out to right here and then I could have swung by 180° this way right to get to 4 comma 300° or I could have started there and I could have swung 180° this way this would be clockwise so that would be me subtracting 180 degre so I could have also said this is four comma If I subtract 180 degrees from this I would get -60° now once I've added or subtracted the 180° I can then swing by multiples of 360° right either positive or negative so that's where these formulas are coming from because from here I could say okay well this is 4 comma 300° plus 360° would be 60° okay so that's another way to locate that point so in other words I could have started here and I could have swung around 180° okay and then swung around another 360° to get to that same point okay also what I could have done I could have started at 4 comma -60 and I could subtract away another 360° which would be - 420° again another way to locate the same point so there's a lot of different ways you could have done this so again we put this into this nice little convenient formula okay so basically it relies on the fact that you're first adding 180° and then from there you could add 360° and then times some integer n they could be positive it could be negative could be whatever again you see this plus or minus here but you could drop that and just put a plus and I'll show you that in the example all right so let's just look at an example here so we have the polar coordinates 3 comma 240° and what I'm going to do is ask you for a formula for all the pairs of polar coordinates that describe the same Point again your example might be different they might say find them within the certain range you just have to apply to that okay so what I'm going to do here is first start by saying okay the three would be the same and what I would do here is I would add 360 time n to this right so the formula you saw was theta plus or minus 360 and you would do n de okay like this well Theta is 240° so let me put that in there so 240° and then plus or minus your 360 * n well I don't need the plus or minus because I can get anywhere I want to go by just varying n right n is any integer so it could be positive it could be negative so again you could put plus or minus it's not wrong but really most books will just put plus you don't need the minus so let's go ahead and get rid of that okay so the next one is a little bit more challenging to understand so remember you would put the negative of whatever this number is so in this case it's three so I'm going to go ahead and put3 and then I'm going to put a comma and let me go through and work out the two different scenarios so I'm going to go ahead and say I have 240° okay then I'm going to go plus remember you're going to have your 180 degre and then you're going to have plus your 360 okay n degrees so this is basically where I'm swinging 180 degrees and then from there I can add multiples of 360° I'm also going to do this with the minus sign to show you it comes out the same so let me close this real quick -3 comma I would do 240° and then- 180° and then minus 360 n degrees okay so basically for this one 240° minus 180° would be 60° okay so this would be 60 degrees and then let me just tighten this down a little bit we're going to end up erasing this so it doesn't really matter then this one 240° plus 180° is 420° okay so 4 20° and if you think about it 420° if I subtract off 360° this is going to be co-terminal with 60° so really I could just put this as 60 Dees okay and just slide this down and now we want to think about this do I need the plus or minus here no because I can get wherever I want to go by just plugging in different integers for n right if I want to do negative if I want to do positive I'm allowed to plug in whatever I want there as long as it's an integer so I can go ahead and get rid of this okay and just list it with the plus sign again if you want to put plus or minus it's not wrong but it's just Overkill right so I'm going to list this as my answer so 3 comma 240° plus 360 n° and -3 comma 60° + 360 n° in this lesson we want to find the distance between polar points all right so let's say that I asked you to find the distance between this 4 comma 105° and 3 comma 225° How could we do this we know that if we have the rectangular coordinates right and we're working on the rectangular coordinate plane we can just use our distance formula right so if I have two points let's say x sub one comma y sub 1 and let's say another one is x sub 2 comma y sub 2 again it doesn't matter which point you label is which I can plug into my distance formula let's say d the distance between the two points is the square root of I'm going to say x sub 2 - X sub1 Quan 2qu plus we have this y sub 2 - y sub 1^ 2 Okay so this works for rectangular coordinates but I have polar coordinates okay so what you could do and I don't advise you to do this you could go X is equal to R * cosine of theta okay and then Y is equal to R * the S of theta so you could use that to convert these guys over okay then you could plug in and you'd have your answer but there's an easier way okay we can derive ourselves a little distance formula when we have these polar coordinates so we could plug into this generically change up a little bit of notation and come up with a formula that way I'll do that at the end of the lesson because I think it's a bit boring I want to show you this using thew of cosiness I think it's a little bit easier to follow okay so let me erase this real quick and erase this and let's just start by plotting these two on the Polar grid okay so we'll come down here and first I'll plot this 4 comma 105 degrees so you're G to step out one two three four units okay so we're g to swing around to 105° so this is 4 comma 105° okay then for the other one we have the 3 comma 225° so I'm going to step out 1 2 3 units and again I'm just going to swing around here to 225° and I know this is upside down but you can still read it so this is 3 comma 225 degrees like this okay so we've plotted those two guys so now I want to think about okay if I had a little line segment that was drawn connecting those two points I want the length of that line segment which is the distance between those two okay so the first thing we need is to figure out what is the angle between the two okay so in other words what is this angle right here okay if I was to sketch this what is that well again if you think about the whole part here let's say I start from here and I go all the way out here that's a 225 Dee angle okay but this angle here is only 105° okay right there so if I consider the fact that to go all the way it's 225° if I wanted to take this part out okay I would be excluding this from here and I should probably do that in a different color so if I took this part right here out I would have the angle that I'm looking for okay so basically you're going to do this with subtraction you can take the larger angle here which is 225° and I can subtract away this smaller angle here which is 105° okay and this would give me 120° so that's going to be my angle between these two guys you can see I've already made a little diagram for us with this so again this is just the angle between the two okay so we need this now we can set up a little triangle once we know this we already know that if we had a little line segment drawn from here to here that the length of it would be four right because this guy is lying on a circle with a radius of four this guy right here is lying on a circle with a radius of three okay so this would be of length three so we have a side of three an angle of 120 and a side of four we have the side angle side which allows us to use the law of cosiness to find the missing side so let's just call this a again you can call it whatever you want you can do a b or c just going to call it lowercase a okay so let's come down here and let's remember that let's say you were trying to find this lowercase a being squared so side a this is equal to remember you're going to think about the B and the C that you know so we're going to say b ^ 2 + c^ 2 then minus 2 * you're going to have the B * the C okay okay and then we're going to multiply this by the cosine okay of the angle you know so it's always going to be opposite of this guy so in this case it's going to be the cosine of capital letter A okay in this case that's going to be our 120 degrees okay so now all you have to do in this case is just plug in I'm going to say that a squar so a squ is equal to one of the sides is going to be four the other's three it does not matter how you put it in there and I'll show you that later on but basically just go ahead and say we have 4^ 2 + 3 2 okay and I'll put minus I'll go 2 * I'll go ahead and say four and then time 3 and then times the cosine of 120 degrees okay so once you get it to this point you're just simplifying you can replace a with d okay because you're trying to find the distance between the two doesn't matter it's just a letter so I'll say that basically my d^2 is equal to 4 2 is 16 then plus 3 squ is 9 then minus you're going to have 2 * 4 which is 8 * 3 which is 24 and then multiply by the cosine of 120° again that's going to be in quadrant 2 you have a 60° reference angle so cosine of 60° is a half so cosine of 120° would be - one2 okay so to simplify this 16 + 9 is 25 let's just do it right here we'll say this is 25 and then this would cancel with this the negatives would also cancel so this is going to be plus and basically 24 ID 2 is 12 so I'll say this is 25 + 12 okay which is basically going to give me 30 7 so we have that the distance squared is 37 and if I want the distance I'm just going to take the principal square root of each side and it'll be the square Ro of 37 okay so basically we found that the distance between those two points is the square Ro of 37 now how can we generalize what we just looked at let me write this again we had a s but I'm going to call this d^2 right d for distance is equal to we had basically our two sides that were known so we said B s + c^ 2 let's go ahead and replace this since when we have polar coordinates we're going to have something like R sub1 let's say Theta sub one okay and the r sub1 is the length of a line segment or a side of the triangle and then we also have the r sub 2 comma Theta sub 2 okay so what we're going to do is we're going to plug in instead of the B and the C I'll go ahead and just say this is R sub one I'm going to put this in parenthesis because it gets kind of messy so R sub one that's being squared and this one will be R sub 2 this is being squared it does not matter which point gets labeled as which and I'll explain why that works out in a moment okay so then you're going to put the minus here and then two times again it's these guys remember with the formula it was two times in this case it would be B * C it's basically those two known sides so I would say time R sub 1 and then time R sub 2 and then time cosine of the angle that you know now in this particular case remember how we found this we subtract okay so I'm going to do is I'm going to say that I have Theta sub 2 minus Theta sub 1 the reason it doesn't matter here is that let's say you end up with something that's negative remember with cosine you have this negative angle identity so the cosine of negative Theta okay if it ends up being negative is the same as the cosine of theta okay so it doesn't matter if you produce a negative angle there is the same as if it would have been positive okay so you will get the same result either way none of this stuff matters in terms of the order right you're just doing some simple operations here so the order that you choose the points does not matter very important okay so this would be our formula of course you could take the square root so you could say d is equal to the square root of and I could basically just copy all this and let me go ahead and copy that real fast and let me paste this in and I'll just stick that underneath there so we don't have to write it and let me just make that a little bit longer and I'll just slide this down a little bit so it fits on the screen okay so definitely something you want to copy down down for your notes and let's go ahead and just look at some examples so we have 2 comma 120° and then 4 comma 210° again you can label these however you want comes out the same way so let's say this is R sub one and this is Theta sub one let's say this is R sub 2 and this is Theta sub 2 so the distance between the two let me go ahead and write it out a few times so you get used to it so R sub one being squared plus you have your R sub 2 being squared and then minus 2 * your R sub 1 * your R sub 2 and then multiplied by the cosine of the angle between them so we're going to say Theta sub 2 minus Theta sub 1 okay so let's just go ahead and plug in we know that everywhere we see R sub one we're going to plug in a two so let's put 2^ SAR here and let's put two here okay and then everywhere we see R sub 2 we're going to put four so let's go ahead and put a four there we don't really need parentheses there and let's put a four there and then for Theta sub 2us Theta sub 1 you're going to have 210° minus 120° which would be 90° so let's just go ahead and put cosine of 90° now most of you know that cosine of 90 degrees is zero so you can basically get rid of this right so you can just erase this you don't even need it because you'd be subtracting away zero which doesn't change anything so here we're just going to have the square root of 2^ 2 + 4 2 so 2 squ is obviously four and 4 squ is obviously 16 okay and then if we add those together we would get 20 okay so the square root of 20 remember you can simplify this we could say d is equal to this would factor into 4 * 5 and four is a perfect square right so basically I can say this is 2 * the RO 5 okay so that's going to be my answer for the distance between these two points again if you change up the ordering pause the video you can see this on your own the only difference is you're going to end up with the cosine of 90° and you can say that the cosine of 90° is equal to the cosine of 90° degrees in each case and let me put my degree symbol in there so that's better in each case you're going to get zero okay so remember this identity here let's take a look at another example here we'll work with some radians so let's just go ahead and label this as R sub1 comma Theta sub one let's label this as R sub 2 comma Theta sub 2 and again I'm just going to plug into the formula so the distance is equal to the square root of let's just go ahead and say R sub 1 is 3 so we'll say 3^ 2 plus r sub 2 is 2 so 2^ 2 then minus 2 * you can have R sub 1 which is 3 * R sub 2 which is 2 and then times the cosine of this one's a little bit tricky because you do have to work with fractions here so let me do this off to the side I'm going to have pi over 2 and I'm subtracting away 7 pi over 6 okay I need a common denominator so I'm going to multiply this by 3 over3 so this would be 3 piun over 6 3 piun - 7 piun is -4 piun so this would be4 4 pi and then over that common denominator of 6 and of course you can reduce this fraction this would end up being -2 piun over 3 okay so -2 pi over 3 let's write that in here -2 pi over 3 okay now if you want to remember because of the negative angle identity for cosine this is the same as if I had cosine of 2 pi over 3 so you can just get rid of the negative there that way it makes it a little bit easier if you're working off the unit circle and from here I can just go through and evaluate things 3 S is going to be 9 and 2 2 is going to be 4 and basically you have 2 * 3 which is 6 * 2 which is 12 so this is A2 here 9 + 4 is 13 let's just go ah and do that real quick and essentially the cosine of 2 pi over 3 remember 2 pi over 3 is 120° and we already know the cosine of 120° is negative a half so this is times - one2 I like to wrap this in parentheses because basically you want to do this operation first I know that's pretty obvious for most people but I don't want you to get confused as to what's going on so this is going to cancel with this and give me a six here so basically you would have a -6 right so what you end up with here is minus a -6 which is like adding six so 13 plus 6 would be 19 so basically here the distance between these two points would be the square root of 19 all right what I want to do now is look at two different examples where you have these negative angles and also the negative R values this is a bit trick for students to understand you can basically just take what you have okay as it's given and plug it into the formula and it's still going to work okay so I'm going to use this directly as it's given then I'm going to change it over and I'll show you why it doesn't make a difference so first and foremost we have four comma 30° let's say this is our first Point again it doesn't matter how you label this I'll go ahead and say this is R sub1 comma Theta sub one and then let's say this 4 comma 210 degrees let's say that's the second point so R sub 2 comma Theta sub 2 okay so from here we just plug into the distance formula so D the distance between the two points is the square root of you're going to take this guy and square it so four squared plus you're going to take this guy and square it so four squared and then minus you have your two times this guy times this guy so 4 * 4 okay and then times the cosine of the angle between them okay so what I'm going to go ahead and do is just say it's 210° minus a -30° okay remember minus a negative is Plus positive so 210° minus a -30° okay and let me go ahead and close that down and from here we're just going to simplify okay so I'm going to say the distance between the two is equal to the square root of this would be 16 and this would be 16 16 + 16 is 32 so let's write that and then minus you have 2 * 4 which is going to be 8 and then * 4 again which is going to be 32 okay so you have 32 there and then times you're going to have the co sign of 210° minus a -30° is 240° so let me put 240° there okay so let me just scroll down and get a little bit of room here so we can fit everything on the screen so I'll say the distance is equal to the square root of the cosine of 240° again this is going to be in quadrant 3 now and you're going to have a 60° reference angle so this is going to be - one2 okay so you would have 32 minus you'll have 32 * -2 again if you want to make this clear it's not necessary you can wrap this in parenthesis it's kind of up to you so I'm going to cancel this with this and put a 16 here so basically you would have 32 minus a -16 that's why I like the parentheses there so it makes that clear so basically 32 minus a -16 is 32 + 16 which is 48 okay so this is the square root and let me make this a little bit smaller this is the square root of 48 of course 48 is going to be 16 * 3 and 16 is a perfect square right it's 4 * 4 so you can go ahead and write this as 4 * the < TK of 3 okay so I'm going to say the distance is 4 * the < TK of three now I want you to pause for a minute because some of you who try to plot this if you go back up and you think about these points here 4 comma -30° and 4 comma 210° if you think about the angle between them it's not 240° it's going to be 120° so why in the world did we get the right answer and I'll show you in a moment that if you do it the other way you still get the same thing if we go to a little diagram here you'll see that I've kind of drawn everything out if you start at this point here let me label this so everything's crystal clear this point right here could be thought of as either 4 comma 330° or if you're rotating this way clockwise you could say that it's 4 comma -30° okay and then this point over here you could think about it again as 4 comma 210 10° okay 4 comma 210° like this okay so if we think about the angle between these you can just look here I've already kind of done this for you but essentially if you start at this 330° and you subtract away the 210° you can see this angle between them would be 120° okay the angle on the outside is going to be 360° one full rotation minus this 120° which is 240° okay so that's if I started here and I rotate along the outside there okay so that's 240° okay so we see that the angle should have been 120° so what in the world is going on well it turns out that we have a little identity and it states that the cosign of let's say 360° minus some angle Theta okay whatever it is is going to be equal to the cosine of that angle Theta so this is exactly what we just came across we see that the cosine of let's say Theta is 120° so 360° minus 120° okay would be equal to the cine of 120° this right here would end up being 240° so the cosine of 240° is equal to the cosine of 120° okay so it does not matter if I punch this in as 120° for the cosine or 240° that value is the same okay so when you go through and do your calculation you're going to get the same answer whether you work with it as is or you do a little conversion okay now this little identity here can be quickly proven using the difference identity for cosine okay so if you want to go through that you could basically say that this guy right here let me erase this real quick and we'll go ahead and say that this is what this is going to be the cosine of the first guy so 360° okay times the cine of the second guy so that's Theta then remember with the cosine difference identity the sign is going to be opposite right so I'm going to make this a plus and then I'm going to go ahead and say now that I have the sign of this first guy the 360° and then times the S of theta now I know the S of 360° is the same as the S of 0° and the S of 0 de is going to be zero right so you can basically get rid of this this is going to be zero and the cosine of 360° is the same thing is the cosine of 0 degrees and that's going to be one so you can get rid of this this is a one so basically what you're left with here is 1 * the cosine of theta which is just the cosine of theta okay so that proves that little identity there we can get rid of this and go back to this form and let's go back now and let's see what would happen let me go ahead and erase this we'll keep this kind of here for now let's see what would happen if we converted this over instead of having 4 comma -30° let's add 360° and say that our other point is now going to be and I'll just write this over here we'll say it's 4 comma 330° okay so I'm just going to put this over here and just line this out for a moment and I'll say this is my R sub1 comma Theta sub 1 and maybe we can just erase this for now just so it's clear what we're doing and so in this particular case what would we do now well we're going to say that we still have the same 4^ 2ar + 4^ 2ar and then minus 2 * you have your 4 * 4 okay so that part's all the same and then times the cosine of this part would be different so basically you have your 210° minus your 330° which is 120° again remember because of your negative angle identity with cosine you can just make this positive 120° okay it's going to be the same thing and essentially we already know that the cosine of 120° and the cosine of 240° will be the same right we just saw that identity so we know we would get the same answer so we can go through here and just say okay well this would be 16 + 16 again that's 32 and and then minus you'd have 2 * basically 16 which is 32 and then times the cosine of 120° we know is again -2 okay so you can wrap this in parenthesis same thing we did before this would cancel with this and give me 16 so this is -6 again you get 32 minus A6 which is going to be 48 okay so we end up simplifying this we get the distance between the two points is 4 * the of 3 okay so it doesn't matter if you end up with a negative angle you can work with your formula either way okay let's look at an example with a negative r value so again you have the choice of working with it as is some people tell you to convert it if you plug it into the formula you will get the right answer okay so if you have something like -3 comma 165° and then 4 comma 45° if you wanted to remember the little formula if you have R and negative R what you have to do is add 180 degrees to that okay so that's how I could change this over I could say that this same point could be located by saying I have three comma add 180° to this this would be 15° okay so you can use either one let's go ahead and use the first one and I'll show you it works the same either way so the distance is equal to the square root of let's call this point one so R sub 1 comma Theta sub 1 let's call this point 2 so R sub 2 comma Theta sub 2 and again all I'm going to do is just plug in so I'd have negative3 so -3 be careful there being squared plus you'd have four being squared and then - 2 * --3 okay * 4 and then times the cosine of if you subtract 45 minus a negative 165 that's like 45 + 165 okay let me put my degree symbols in there so that's going to give me 210° okay so this is 210° let me extend this a little bit over all right so let me put here that the distance is equal to the square root of if I Square neg3 I get nine if I Square Four I get 16 if I add 9 and 16 together I'm going to get 25 okay so that's pretty simple here what I'm going to do is I'm going to say -2 * -3 is POS 6 and then POS 6 * 4 would be pos4 so let me put plus 24 times the cosine of 210° now 210° that's going to be in quadrant 3 okay so cosin's negative and it's based on a 30° reference angle okay so the cosine of 30° is < TK 32 and because we're in Quant 3 this would be negative so I'm going to say this is the negative of the < TK of 3/ 2 okay so if you multiply these together what's going to happen is this would cancel with this and give me a 12 okay and it's going to be negative so I might as well just write minus and I'll say 12 times the < TK of 3 okay times theun of three now you can approximate this if you want if you want an exact answer this is going to be it all right so let's work with this point I'll show you it still works so we'll go ahead and say change that r value to positive at 180° to 165° that's 15° Let's Line this all out let's get rid of that and I'll label this as R sub1 comma Theta sub one okay so if you plug in here the sign's going to change and here the sign's going to change okay and this is going to change everything else would be the same right so basically what you'd have here is inside now you do 45° minus 15° okay and that's going to give me 30° so I want you to notice that a moment ago we had the cosine of 210° okay and essentially this had a 30° reference angle but this guy is positive and the other guy is negative but notice that you have a sign change here okay before you had -2 * ne3 now it's -2 * posi3 okay so I'm going to show you a little identity in a moment that proves that this is basically going to work right so if you add 180° to your angle okay it's going to change the cosine of that from negative to POS positive okay or from positive to negative so essentially what I have here is 9 + 16 which is 25 again so that part is the same okay and then here we have 2 * 3 which is 6 * 4 which is 24 so this is minus 24 now you can put plus - 24 if you want okay and we're going to multiply this by the cosine of 30° we know that this is going to be the < TK of 3 over two okay and so from here this cancels with this again it's negative time positive so it's still negative so it's going to be -12 so -12 let me erase that and put -12 * theun of 3 okay so you see you get the same answer either way okay so let's talk about why this works let's get rid of all of this again when I think about something like the cosine of some angle theta plus 180° this is going to be the negative of the cosine of theta so you saw with our example we basically had a Theta that was 30° right if I added 180 degrees to that I got 210° and that's going to be the negative of the cosine of 30° okay again you can prove this with the cosine sum identity so if you wanted to do this you could say that you had what the cosine of the first guy Theta and then basically times the cosine of 180° now you know this is negative 1 so you can just replace that just put a negative out in front and then you're going to go minus right because it's always the opposite sign so it's going to be S of the first guy so Theta times the sign of 180° now the sign of 180° is zero so you can get rid of this and you're just left with the negative of the cosine of theta so it doesn't matter if you get a negative r value if you change that over to a positive then your angle is going to increase by 180 de okay so you're going to change the signs on the middle part but then you're going to compensate for that because your cosine value is going to change signs as well so you're going to come back to the same answer in this lesson we want to talk about the polar form of a line all right so we've spent a lot of time talking about the polar coordinate system and also how to convert rectangular coordinates to Polar coordinates and then take those polar coordinates and plot them on the Polar grid what we're going to do now is take the next step and talk about polar equations okay so we're going to need to know how to convert between the rectangular form and the polar form and we're also going to need to know how to graph these equations on the Polar grid so today we're going to start out with something very simple it's the polar form of a line and basically graphing them and converting them back and forth between the rectangular form and the polar form okay so basically what you need to know is if you're going from your rectangular form to your Polar form it's very very simple you're going to use a relationship that we've talked about many times and that's basically that the sign of theta going back to the original definition is y/ R right you could say opposite over hypotenuse and the cosine OFA is equal to x r the adjacent over the hypotenuse what we're going to do in each case is multiply both sides by R so I'll say this is R * s of theta multiply this by R and that would cancel so you've solve this for y here same thing over here multiply this side by R and this side by R this would cancel over here and what you end up with is a relationship you're going to use to convert this okay so I'm going to take this right here this R * s of theta and I'm going to plug it in for y and then I'm going to take this right here this R * cosine of theta and I'm going to plug it in for X that's all you need to do now depending on your teacher they may have you work with some generic form so this is the standard form of the line you also have the slope intercept form the Y equals MX plus b if you're working with this a lot you may need to convert it over generically okay and then you can just plug in for your a b and c okay or in this case you would just be plugging in for your M and your B okay but it just depends on what you want to do let me just go ahead I'm going to solve this generically and then on the first example what we'll do is we'll just plug it in that way and then I'll also plug it in the other way I'll show you it's the same thing okay so let's just go ahead and plug in R times cosine of theta here and then R times the S of theta here okay so what would we have you would have a multiplied R * the cosine of theta plus you would have your B * your R * your s of theta okay and this just equals c now the idea here is that you're going to factor out the r okay and then you're going to solve for R so if I pulled that guy out okay if I pulled that guy out I would have R times the quantity what's left your a times your cosine of theta so a * cosine of theta plus your B * your s of theta so your B * your s of theta and this is equal to C okay so let's stop for a minute let's scroll down and get a little bit of room going how can we solve this for R remember this whole thing is a quantity and it is multiplying by R okay so what I can do is just divide both sides by this quantity this a * the cosine of theta plus b * the S of theta okay perfectly legal and so what's going to happen is this is going to cancel with this that will be one so you have R by itself and it's equal to and over here let me scooch this down a little bit so this looks a little bit better I'm going to divide this by the same thing so a * the cosine of theta plus b * the S OFA okay so R equals you have your C which is your constant from the ax plus b y c over your a time your cosine of theta plus your B time your s of theta where the a is the coefficient from X and the B is the coefficient for y now you could do a similar process with the yal MX plus b get yourself a general formula for that it's up to you in terms of how many of these problems you're going to have to solve and whether it's worth or not to use a general formula or just plug in each time okay but what I want to say is if you're going to use this formula make sure that your line is in the format of ax plus b yal c it must be in this format to use this formula okay it's very very important because a lot of times you get the slope intercept form or something else given to you okay and people start plugging things in and they get the wrong answer so let's go ahead and just look at a simple example so here we have y = x - 4 okay this is is in slope intercept form so this follows the y = m the slope * x + b the Y intercept okay so we know this if we want to convert this to standard form so we can use that little formula then what we would do is we would subtract X away from each side okay so I would say that I havex + y is equal to-4 now for myself I prefer the high school definition of the standard form of the equation of a line and so I want the coefficient here for X to be positive and to do that I'm just going to multiply everything by Nega 1 so this would be x - Y is equal to 4 okay it just changes the sign of every term when you multiply both sides by negative one okay so let me actually erase this and this okay and at this point basically you can use your formula you know that the coefficient for X is a one the coefficient for y is a negative 1 and your constant is a four so in other words the a value is one the B value is Nega -1 and the C value is pos4 so remember R is equal to you have your C over you have your a * cosine of theta plus your B times your s of theta okay if you do this a few times you will memorize it now what I'm going to do is just plug in so instead of C I have four instead of a I have one so you can just get rid of that and instead of B I have negative one so I can just change this to a minus and that's all it is so R equals 4 over the cosine of theta minus the S of theta okay now you might be thinking how in the world are we going to graph something like this and I'll show you that in a moment but really quickly I just want to show you if you had done this the other way that you would get the same answer right so basically if I plugged in R time cosine of theta there and I plugged in R times the S of theta there I could achieve the same result okay so what I could do is say I have R * s of theta and then what I'm going to do is say I'm going to subtract this right here away from each side so I'll say minus r * cosine of theta and this is going to be equal to -4 okay and from here what I'm going to do is I'm going to factor the r out so I'll say R * the quantity s of theta and then I'll say minus the cosine of theta and this will be equal to-4 okay and then I'm going to divide both sides by this so that R is equal to you'll have -4 over I'm going to go ahead and say I'm going to put cosine of theta first and then plus the S of theta right so the only difference between this and this is the sign and of course what you could do here is you could multiply the numerator and denominator by Nega 1 okay that's perfectly legal because that's a complex form of one so this would become positive this would become positive and this would become negative and essentially what would happen here is your two forms would match up completely right r equal 4 over cosine Theta minus s of theta in each case okay so let's erase all of this and let's talk about graphing now so graphing a polar equation is probably one of the most difficult things to graph and with a line it's a little bit more challenging lines are a lot easier to graph on the rectangular coordinate plane to see this really quickly let's think about y = x - 4 on the rectangular coordinate plane we know that the Y intercept the Y intercept is going to be at 04 and M the slope is equal to 1 right the coefficient for X so essentially I could plot 0 comm4 and then just use my slope of one to get additional points so if I come here I can see that I just drop down to 0 comm4 that's right there and then I can use my slope so up one to the right one up one to the right one you know so on and so forth I can generate this point and this point and this point as many points as you need you only need two points to get a line a lot of times people like three to get a check but you only need two now let's say that I wanted to graph that equation in polar form on the Polar grid What what could we do well you can go back to your original strategy of making a table of values so we would go back just make a little table and say okay I have my r value and my Theta value and I like to plug in values for Theta okay and then get a value for R so what I'm going to do here is I already know that the Y intercept occurs at 0 comm4 that's pretty easy to convert over so this point right here you can use the traditional method to convert it over that's up to you but what I'm going to do is just think about the fact that okay if I'm starting here at the origin I need to travel 1 2 3 four units to get there okay and essentially the angle here is going to be 270° I know the R value is four okay not negative4 but four and the angle is 270° so if I come back I can say that a Theta value of 270° would give me an R value of four and you can check this out if you plugged in a 270° there and also there what would you get well the cosine of 270° is zero okay so essentially you would have four over You' have 0 minus the S of 270° is -1 minus a negative 1 is 1 okay so essentially you would have four there right so you would get four comma 270° as one of your points okay then the other one if you think about your x intercept here well again you can go back to your graph that this guy right right here is 4 comma 0 okay so essentially from the origin I have moved 1 2 3 four units to the right so the R value is going to be four okay but the angle here is going to be 0 degrees right because I'm on the positive X AIS so if I go back that point would be 4 comma 0 degrees okay again you can plug this in plug in a 0 degrees here and a 0 degrees here you would have four over the cosine of 0° is going to be 1 and then minus the S of 0° is going to be zero so this would be 4 over 1 which is four okay so that's another point so 4 comma 0° all right so now that we have these two points we're ready to sketch our graph so again our two points we have the 4 comma 0 degrees so that's going to be right here and the other one is going to be 4 comma 270° so I'm going to be swinging around to right here so that's right there and then basically once you have those two points that gives you a line so you can sketch that guy and you are done okay let's look at another example so here we have Y = -2x - 6 again you can put this in standard form and use your formula you can plug in for X and Y whatever you want to do I'm just going to use the formula so I'm going to add 2x to both sides of the equation and I'll go ahead and say that this is 2x + y is equal to -6 so my value for a is 2 it's the coefficient for X my value for B is one and let me put that invisibly okay that's going to be my coefficient for y and then my value for C my constant is -6 okay so now I'm just going to plug into the formula I'll say that R is equal to you're going to have your C value which is -6 so let me put a negative out in front I'll put the six out up here and then over you have a which is 2 * the cosine of theta and then plus you're going to have your B value which is one so don't write anything times the S of theta okay so that's all it is so this is our equation you can see that using that formula is pretty quick and essentially if I want to graph this again the simplest way for me to do this is to think about this on the rectangular plane if I can get the X and the Y intercept that's going to give me two points that I can really quickly convert over to Polar form and I'm done okay in some cases you won't be able to do that I'll give you an example of that later on so here what I'm going to do is I'm going to say okay well the Y intercept the Y intercept since this is in slope intercept form I know is at 0 comma -6 so think about this if I was at the origin I would be going down 1 2 3 four five six units to get there okay and from the positive x-axis I'm swinging around 270° so essentially my r value is going to be six and my Theta value is going to be 270° and again if you're not comfortable with what I'm doing just take the normal steps to convert this that we talked about over the course of the last two lessons Okay then the other guy here the x intercept the x intercept if we think about that well if I plugged in a zero there I would have basically 0 is equal to -2X - 6 add 2x to both sides so 2x is equal to -6 divide both sides by 2 you get x equal to -3 okay so this would be -3 comma 0 so in polar what would that be well ne3 if I start right here let's say we're going to go one two three units to the left okay don't worry about it being negative it's three units to the left and the angle with the Positive xais we're going to come around 180° okay so this guy is going to be let me kind of scooch this down a little bit this guy is going to be 3 comma 180° okay so that's going to be my two points in polar so if I come over here again I go out to a circle with a radius of three I'm going to swing around to 180° angle so there's that point there then for the other guy I'm going to go out to a circle with a radius of six and I'm going to swing around here to an angle of 270° okay then you can just sketch a line through those two points okay let's now now talk about the harder scenario this is where we're going from The Polar form to the rectangular form so when you work with this you need to understand a few relationships the first one is that the tangent of theta is equal to Y overx just going back to that definition you're going to need this a lot also you're going to need to know that R is equal to the square root of x^2 + y^2 again we've talked about that a lot and we'll see how to use these as we go forward okay for right now I'm just going to use this relationship here okay and so I have that Theta is equal to 2 piun over 3 now because my polar grid is written in degrees I'm going to change this and say Theta is equal to 120 de okay I'm just going to convert that over and essentially when you see something that's Theta is equal to K this is a line that passes through the origin okay we're thinking about it that way now how could I convert this over well to convert it over what I would do is I would would first take the tangent of each side okay so if I take the tangent of each side I would have the tangent of theta is equal to the tangent of 120° okay now pause the video and think for a minute about how you can get this into rectangular form how can I get this into an equation with X and Y now if you gave that a shot first I would say okay what is the tangent of 120 deg well it's the negative of the sare root of three so let's erase this or actually let me keep this in line let me actually keep that step there in case people get confused so the negative of the square of 3 and over here notice that I have the tangent of theta okay this is the trick here you're going to replace that with YX because the tangent of theta is equal to or the same as YX so I'll say YX is equal to the negative of the of 3 so once you realize that it's pretty easy from here all I really need to do is multiply both sides by X so multiply this side by X and multiply this side by X and I'll say Y is equal to I'll say the negative of x * the square < TK of 3 so this is the rectangular form of that equation now if we think about graphing this again if I had something that was Theta is equal to 120° and you stop for a minute and think about that you think about the fact that there is no r value given so it means that no matter what the R value whether it's positive we know we can have negatives that basically I'm going to be on 120° angle so if I think about starting at the pole here the orig if I'm walking forward I'm going to be on this 120° angle here remember how if you have a negative r value you're walking backwards so again I would find 120° and I would walk backwards okay so essentially here R is greater than zero and then down here your R values are less than zero okay so again this part is a little bit confusing again if the R value was less than zero you had 120° angle you would find that angle you would turn around and you would Walk This Way backwards now you could also also draw the same line by saying that Theta is equal to 300° okay so I know people would ask that because essentially if I gave you that one well now the R value here would be positive okay because here I'll be walking forward and then the r values going this way would be negative okay so it's just a way to write the same line okay but there's different conditions there where R is going to be positive in one direction for this one and R is going to be positive in a different direction for this one okay let's look at a harder example so this is one that takes a little bit of thought so here we have R is equal to the cosecant OFA + 45° all right whenever you see cosecant or cotangent or secant go ahead and write it in terms of s or cosine and see if that makes anything easier for you okay remember when we worked with our trigonometric identities that was the General strategy okay so that's what we're going to do here I'm going to say R is equal to we know that 1 over the sign of is equal to the cose of okay that's the relationship so I'm going to say this is 1 over the S of theta + 45° okay now from here I can multiply both sides by this sign of theta plus 45° and that's going to give me that R times the S of this Theta + 45° okay is equal to 1 now again again again we know that we can't just distribute this sign inside right we know that we have a sign sum identity and we need to use that here okay so to do that I'm going to put some brackets around this guy and remember how this identity Works It's The Sign of the first guy so the sign of theta times the cosine of the second guy so the 45 deges then you're going to have plus you're going to go the cosine of the first guy which is Theta times the S of the second guy which is 45° okay okay so go Ahad and close that down and this equals to one okay so from here you see that it's getting a little bit easier because we know that the sign of 45 Dees and the cosine of 45 degrees is 2 over2 so all you have to do is replace that so I'll say R times I'll just use some parentheses here I'll say the square < TK of 2 over 2 and then times the S of theta okay then plus I'll say my Square < TK of 2 / 2times the cosine of theta okay this equals 1 so from here I need to do two things remember we have this relationship where X is equal to R * cosine of theta and Y is equal to R * the S of theta okay so if I distribute this R into each term you notice that I could replace R * s of theta with Y and R * cosine of theta with X okay also I'd want to factor this guy out at the same time I know I'm doing two steps in one but I think you're strong enough to get that okay so what I'm going to end up doing is is I'm going to pull this out so I'm going to say I have the < TK of 2 over two and then I'm going to distribute the r in so I'm going to have R * s of theta and then plus I'll have R time cosine of theta and of course let me make these parentheses a little bit better I'll say this is equal to one now from here again R * s of theta that's just y r * cosine of theta that's just X so what I'm going to do is say have square of 2 over 2 times I'll have basically y + x you could rearrange that as x + y it's really up to you and then I'll say this equals 1 now what I'm going to do is multiply both sides by 2 over < TK of two so times 2 over < TK of two this is going to cancel and this is going to give me I'll just go and write this as x + y is equal to I'll have 2 over < tk2 and of course we always want to rationalize the denominator let me move this over a little bit so I'll multiply this by theun of two over the > of two and what's going to happen is this down here would become two so this would be two and up here this would betimes theun of two well notice that now you can cancel this with this so I'm just left with theun of two okay so here we have x + y equal the < TK of two now let's talk about graphing this so if we have x + y is equal to the < TK of two from here I can figure out that my x intercept would be what if I plugged in a zero for y I would basically say that the x intercept would be < TK of2 comma 0 and then basically my Y intercept my Y intercept would be what it would be 0 right if I plug that in for X comma square otk of two okay so again these are easy to convert over now the square root of two is something you have to approximate most people will say this is about 1.41 okay so I'll go ahead and say that I have about 1.41 comma 0 and then I'll have 0 comma about 1.41 okay now let's say we wanted to convert these over to Polar form for this one I know that if I just start at the origin and I'm just going to the right by 1.41 units my angle is going to be 0 degrees and my r value is going to be 1.41 so this is 1.41 comma 0 degrees okay for this one I'm going to be going up so from the origin I'm going to go up by 1.41 units okay so my R value again is 1.41 and my Theta value here the angle if I think about it this way right that's going to be 90° so this is 90° okay now you can go back and check this let me grab this real quick we'll come back up here and paste this in you can check these real quick if you put in zero degrees there the S of 45° is sare < TK of 2 over2 okay and if you think about it you have 1 divided by that so you're just flipping it and so you're going to end up with two you're going to end up with 2 over the < TK of two again rationalize the denominator okay and you're going to end up with just theare root of two okay so a 0 Dee angle for Theta gives me Square < TK of two for R okay so we know that then the other one you can do this one you plug in 90 degrees there well 90 degre plus 45° gives you 135° you still have a 45 degree reference angle so it's going to be the same thing right so basically I'm going to end up with an R value ofun of two okay now if you wanted one other point you think about your unit circle here here and something that would work pretty nicely you know that the S of 30° is equal to2 and so if I could make this bottom part 1/2 okay somehow some way one divided by 1/2 would be two that would be a nice point to plot okay so what I could do I could think about okay well s of 150 deges 150 degrees has a 30° reference angle s of 150 degrees would also be a half and what I could do is I could add 105 degrees so I could put 105 degrees in there okay so I would have S of 105 degrees plus 45 degrees is 150 degrees okay and the sign of 150 degrees is a half so then your r value there would basically be two right because 1 divided by half is two and the Theta value here don't plug in 150 degrees it's what you're putting in here for Theta okay so Theta is 105 degrees that's what you put in there be very careful a lot of students make that mistake when you're plugging something in for Theta that's what's going in here I am plugging in 105 de I am not plugging in 150° 150° is the result that comes from adding 45° to 105° okay so very important to make that clear now once you have these three points you can go ahead and sketch this guy so let's go ahead and say that this point right here is going to be my point which is an R value of let's say 1.41 approximately and then I'll say the Theta value is 0 degrees okay so that's one point then this point right here is going to be my R value of 1.4 1 okay so you notice how it would be on the same Circle there okay and then basically the angle here is 90° okay so you can see that so let me draw an arrow to that and then also we have this point right here if you go out to a circle with a radius of two and we swing around to this 105 degree angle remember you want 105 degree angle not 150 degree angle but 105 degree angle this guy right here would be a02 so you would have 2 comma 105 degrees okay you're just finding some points and then sketching a line through the points all right let's also talk about horizontal and vertical lines to wrap things up so if you get something like yals B you know this is a horizontal line right so basically no matter what the x value is y always equals that constant B okay and essentially for this I would just say okay I have R * s of theta just plugging that in for y there is equal to B and I would divide both sides by the S of theta okay so both sides by the sign of theta and so if you see something like R is equal to some constant B over the S of theta you know that you have a horizontal line okay that's very clear and I'll show you an example of this in a moment let's also look at the vertical line so you know if you have something like x equals a this is a vertical line no matter what the value is for y x always equals that constant a so here I'm just going to plug in R * cosine of theta and essentially I would have R * cosine of theta is equal to a divide both sides by the cosine of theta so divide by the cosine of theta so you get is equal to a over the cosine of okay so this is your vertical line in polar form all right let's look at two quick examples so we see here we have yals 3 to convert this over again if it's a horizontal line then I know it's going to be R is equal to you're going to have this constant here which is three over the S of theta okay that's all it is and to graph this it's no different than when you graphed it on the rectangular coordinate plane you would basically just go up from the origin three units okay by this number number of units and make a horizontal line if you go here and you look at this guy you can start here I go up to a circle with the radius of three so I'm right there so basically one point of the line and we'll verify this in a second would be three for the radius and then 90° for the angle okay so that guy would be on that line and then essentially you would just draw a horizontal line there some other points you would have six comma you would have 30° okay and then another point you could see that you would have 6 comma 150° notice that 150° has the same reference angle and also since s is positive in quadrants 1 and two it would have the same s value right in each case it's a half so if we go back and we think about does this work well the sign of 90° is 1 so 3 over 1 is three so that works right basically if I plugged in 90° there my R value would be three and then if I did R is equal to 3 over the S of 30° or the sign of 150° well in each case this would be a half and essentially you would have 3 * 2 which is 6 okay so there you would get 6 comma 30° or 6 comma 150° all right so lastly we have X is equal to -3 so following the formula we'll say R is equal to we'll say -3 we'll go ahead and say that's divided by the cosine of theta okay so before we go to the graph again if you think about a point that's going to work think about your coordinate plane if this is 0 comma 0 if I want the x intercept which is going to occur at 1 -23 so this point right here on the rectangular would be -3 comma 0 right so if I'm drawing a vertical line this is where it's going to cross now if you think about this this is going to be what it's going to be an R value of three and then again this angle here is going to be 180° so you can stop and plug that in and see if it works the cosine of 180° is1 right so essentially we would have -3 over -1 which is going to be equal to three okay so that gives me that point right there you could grab other points you know that the cosine of 120° is half and then also the cosine of 240° is negative a half so if you did something like let's say 120° and then also 240° what's going to happen is you're going to be dividing by - one2 which is like -3 times you'll do -2 which is six okay so this would be some points and I know that six and three don't match up but remember we're on the Polar grid so if we look at this guy you can see that from here you're basically going to go out to a circle with a radius of three and then swing around to 180° angle so there's your x intercept as it would be on your rectangular coordinate plane here it's going to be at 3 comma 180° okay from there you can basically make a vertical line but I just want to show you again this point right here would basically be going out to a circle with a radius of six swinging around to 120° angle or going out to a radius of six and Swinging all the way around until you get to 240° so that's that point there in this lesson we want to talk about circles in polar form all right so in this lesson we're going to learn how to convert back and forth between the rectangular form of a circle and the polar form of a circle we'll also look at some strategies to graph a circle on the Polar GD so I'm going to start off with a circle whose Center is at the origin this is the easiest scenario and we're going to think about converting this back and forth now first and foremost when you work with Polar form you have that R okay so remember if you have something like the cosine of theta we say this is X over R if we have the S of theta we say this is y over R and if you multiply both sides by R in each case you would get R * cosine of theta is equal to X and then you would get R * s of theta is equal to Y okay so what I'm going to do here instead of having this r s in rectangular form I'm I'm going to replace this with an A okay so the a is now going to be the radius in rectangular form okay so I'm going to say this is x^2 + Y2 is equal to a s so that's typically the letter chosen in your textbook so what I'm going to do now is think about the fact that okay if I want to convert this to Polar form I can always plug in R * cosine of theta anywhere I see an X and I can plug in R * s of theta anywhere I see a y okay so this is the longer way some of you already know that X 2 + y^2 is equal to R 2 okay we use this when we convert our rectangular points to Polar points so we already know this relationship but let me show you this real quick so if I plug in here I would have R * cosine of theta being squared then plus you would have R * sin of theta being squared equals a 2 okay so because this is multiplication here you can bring this in so you can say this is r^ 2 * cosine 2 thet then Plus r^ 2 * sin^2 th and this is going to be equal to a^ 2 now the idea here if you want to show that basically this side here is basically R squ then you can just Factor this out okay so pull this out in front of a set of parentheses so R 2 here times the quantity what's left what I'm going to have is my cosine Square Theta so cosine squ Theta then plus my sin Square Theta my sin Square Theta this equals a 2 so let me scroll down for a moment get some room going and I want you to think about the fact that this is the Pythagorean identity where you have sin Square theta plus cosine Square Theta is one right and if you reverse that it's the same so this is one so I can go ahead and just cross this out and say it's one and you see that X2 + y^2 is just going to give us R 2 okay so now I have R 2al A2 okay and here comes the confusing part if I solve this for R what I end up with is I end up with r is equal to plus or minus a now this result here is is confusing because when we work with this in the rectangular coordinate plane we don't have a negative radius it doesn't make any sense but if we think about this in polar we can have a negative Rus so it's allowed so basically we're going to see this some books will actually write just rals a and I'll explain this when we get to a problem how you're going to end up with the same graph either way all right let's go ahead and look at a simple example so we have X2 + y^2 = 25 so again this matches your format of x^2 + y^2 is is equal to again the radius here we're going to call it a so a being squared okay and this converts over to you would basically have your R is equal to plus or minus a what you have to consider here is 25 is not a it's a squ right so what I would want to say is to convert this over I could quickly just say this is R is equal to plus or minus 5 right that's all you would really need to do again if you want to do this the long way I want you to consider the fact that we just showed that R2 is equal to x^2 + y^2 okay so basically I can replace that with my R 2 and this equals 25 and again if you solve for R R equals plus or minus 5 so however you want to do this you can do it this way which is a bit longer or you can just use your formula if you have it memorized all right let's talk about graphing this guy now so essentially people get a little bit confused if you have something like R is equal to let's just say five for right now what would I do well there's no Theta involved people are always trying to plot points with Theta and r there is no Theta so that just means no matter what the Theta is no matter what the angle is the r value is always five so starting from the pole okay let's say that Theta was 15 deges I just walk forward by five units right so I'm going to be on this circle with a radius of five Okay then if I was at 30° I'd be on a circle with a radius of five you know so on and so forth you're going to end up basically tracing a circle through those points that are formed to get this graph here okay that's all you're doing now if you consider r equal -5 remember if you had a point let's just say it was 5 comma 45° okay so that's going to be coming out to five a circle with that radius swinging to 45° that's right there okay consider that you could locate the same point by saying you had5 comma 225 degrees right so in other words if I'm facing this way right I swung all the way around to this angle right here okay well what I want to do is go backwards now so I want to go this way by five units so you can trace out the exact same points using that logic so R5 there would be the same Circle okay so that's why a lot of people will just drop this one and say okay this is the graph of R equals 5 but again some teachers want you to put Ral plus or minus 5 like this just because you can have a negative radius with the polar grid all right let's go ahead and take a look at an example where we have an equation of a circle that's centered at the origin and it's given to us in po Polar form we're going to put this in rectangular form we're also going to graph this on the Polar grid so we have that R is equal to plus or- 6 and remember we can say that X2 + y^2 = a^2 can be converted to Ral plus or minus a so now we're going backwards here so we have the R equals plus orus a in this case a is 6 we want to go to this form so I would just say I have x^2 + y^2 is equal to go ahead and square a a is six so that's going to be 36 okay another way you could do this if you don't like using these formulas or maybe they're too tough to memorize you can just Square both sides that would give me that R 2qu is equal to if I have positive 6 or Nega 6 and I Square it I get 36 remember whenever you have R squ you can replace that with x^2 + y^2 okay going over to rectangular and this just equals 36 so either way you're going to get the same answer now graphically again this is very easy to to do so Ral plus or- 6 from the origin or from the pole if you want to think about it that way essentially I'm just walking forward by six units okay so I'm going to a circle with a radius of six in each case no matter what the angle is okay so I would Trace out a series of points okay and again you can trace out the same points using a ne6 for the radius so then you sketch your graph by tracing a circle through those points and you're done all right in most cases we're not going to see a circle whose Center at the origin a lot of times though the circle will pass through the origin and so most of our problems will involve that scenario so I'm going to start off with something generic here you'll see this in most textbooks let's say we had a circle that was not centered at the origin okay and essentially what I'm going to say is that the center of that circle is going to be labeled as P sub Z and then the polar coordinates for that point would be R Sub 0 and then Theta subz so the distance from the origin which we labeled as o to to this point which is the center P subz okay would be R subz so that's the length of this side right here okay so that's given by this R subz and the angle here is going to be Theta subz Okay so we've got that part now the other thing is we can have ourselves a little point on the circle so let's call that P okay and the coordinates there in polar are going to be R comma Theta so the distance from the origin to that point P okay would be R now the angle there this guy right here would be Theta so we have those two things now if I want to figure out what is the length of a which is the radius of this circle okay or the length from this guy right here this point P sub Z to this guy right here some point on the circle P what I can do is I can use my law of cosines right because I have the length of this side here which is R sub Z I'll have the angle here which is going to be given by Theta okay the bigger angle here minus Theta Sub 0 which is the smaller angle here okay and then we'll have this side here which is R okay so I have side angle side so what I can do to get the length of this side a is set up my little law of cosiness here so a^ s is equal to you have your R subz being squared plus your R 2 minus your 2 * R Sub 0 * R * the cosine of theta minus Theta Sub 0 okay now we're not going to actually work with this guy what we're going to do is we're going to think about the first scenario where our Circle passes through the origin okay so now consider the fact that again from the origin if you called that the point O to this guy right here which we said was P sub Z okay the coordinates here now are a comma 0 Dees okay so your Theta is going to be 0 deges and your R sub Z is going to be a right because the distance from the origin to the center is now going to be equal to the radius so so if we go back this allows me to plug in for this R subz I'm going to plug in an A and then I'm going to plug in an A okay so we can simplify this in this case we would have a 2 is equal to you'd have a 2 there plus r^ 2 - 2 a r times the cosine of if I plug in a zero there that's basically gone right so I just have cosine of Thea so from here you can notice that this would cancel and you could move the r sare to the other side so Nega r^ 2 is equal to -2 a r * the cosine of theta I can multiply both sides by1 so essentially I just have r^2 = 2 a r * cosine of theta now here's where everybody gets a little bit lost normally if you have something squared you take the square root of each side in this case we're going to divide both sides by R okay because r s divid r is R by definition and so this is going to cancel here okay and I'll have 2 a * the cosine of theta okay so if you go back to this little diagram here you'll see Ral 2 a * the cosine of thet so that's going to give me a circle that passes through the origin that has a center on the x axis okay and it's all based on the fact that the distance from the origin to the center is given by this a here okay that's the radius of this circle okay now another way that you can derive this if you don't like the geometric definition or going through all that stuff you can think about this in rectangular form this a comma 0° here would be a comma 0 and then it's easy to achieve this result I would say okay well I would say x - a^ 2ar plus you would have y - 0^ 2 is equal to in this case the radius is a so a squ okay let me go ahead and copy this real quick basically what you'd want to do first is expand this so you'd want to say you have x^2 - 2 ax + a 2 plus y - 0 is just y so this is y^2 this equals a 2 you see again you have the same thing on each side so get rid of this and this that's gone okay so now I know if I have x^2 + y^2 again that's R 2 so you can rearrange things if you want to make that crystal clear so X2 + y^2 and then - 2 ax would be equal to zero okay let me put this visibly because some people get confused you cancel that over there it's zero so let me replace this with r 2 and let me add 2 ax to both sides and now now what you do is for X remember if you have X you can always replace this with r * cosine of theta okay so I'm going to put this as R time cosine of theta and again the confusing thing here is that instead of taking the square root of each side I'm going to divide both sides by R that's something you just have to remember okay so what I'm going to do is basically cancel this with this cancel this with this and say let me scroll down a little bit that I end up with r is equal to 2 a Time the cosine of theta okay so if you see something in polar that looks like this you can immediately convert it over and say it's x - A squar okay that quantity squared then plus y^2 is equal to a 2 okay now converting back from this one to this one is a little bit more tricky you have to check to make sure that everything matches up and I'll show you that in an example okay let's take a look at the first example so we have x + 2^ 2 + y^2 = 4 okay so the first thing is if you want to make sure that this matches the x - a^ 2 + y^2 = your a^ 2 again think about rewriting this you have a plus here how could I rewrite that you want to do minus a negative so you just want to match this and see if it works so x minus a -2 qu^ 2 plus y^2 is equal to you basically have -2 being squared does this match yes now if you saw something like let's say this was five would this work out no because if I'm saying this is -2 well -2 s is is not five okay so you have to pay attention to this if you're going to use these shortcuts okay so what I would have here is that a my value for a is equal to -2 not positive2 -2 so when I set this up remember we said that R was equal to 2 multiplied by a multiplied by the cosine of theta okay so here a is -2 so when I convert this over I can just say that R is equal to basically 2 * -2 is4 times the cosine of theta okay once you do this a few times it becomes very very easy to convert these over let me erase all this real quick and I'm just going to show you that you get the same thing if you do this the long way so let me move this up here with these it's more of a Time saer versus when you looked at the ones that were centered at the origin so first I'm going to expand this this would be x^2 + 2 * x * 2 would be 4x + 4 + y^2 = 4 okay again you get this duplicate on each side so you get rid of that and this is zero okay so now you have X + y^2 okay that's going to be replaced with r^ 2 + 4x = 0 okay so from here what I'm going to do is I'm going to replace this with r 2 I'm going to subtract 4X away from each side okay and all I really need to do now is replace x with r * cosine of theta so R * cosine of theta and again this is the key we're going to divide both sides by R okay do not take the square root of each side you want to divide both sides by R so this is going to give me that R is equal to again this cancels -4 * the cosine of thet so however you want to do this you're going to get the same answer okay just make sure that if you're using the shortcut that things match up if this was five or six or 10 or something like that this would not work this would not be a circle that goes to the origin that is centered on the X AIS this is where this formula works now if you wanted to graph this this is a lot more difficult or a lot more challenging I would say versus when you worked with graphing a circle on the rectangular plane the reason is when you work on the rectangular plane you can plot your Center and then from your radius you can just move up by that amount to the right by that amount down by that amount and then left by that amount then you can draw your circle right you're done with this guy you basically have to make some points okay so let me put an R value and then a Theta value here and you just have to use some logic okay so first off I'm always going to try the cosine of zero I know that's one so let me put zero I'll do this in terms of degrees so 0 degrees okay so the cosine of 0 degrees is going to be one4 multiplied by 1 would be4 okay so one point would be -4 comma 0 deg another point I always try 90° because I know cosine of 90° is 0 so that would be4 * 0 which would be 0 so if I put 90° I would get 0 okay so basically this point would be 0 comma 90 degre that's just going to be on the pole okay then other things you could do if you had this as a half right4 * a half would give you -2 if you want something that's positive for the r then you're looking for negative a half okay so you know that cosine of 60° is a half but if you want it to be negative it's got to be in quadrants two or three so you'd want something like 120° okay which is a 60° reference angle in quadrant 2 or 240° which is a 60° reference angle in quadrant 3 so now I would say okay well the cosine of either 120° or 24 4° isga a half -4 * half is 2 okay so this is two and this is two so there's two more points so you have 2 comma 120° and then you have 2 comma 240 degrees okay so let me grab these real quick let me copy them all right so obviously I've already graphed this guy okay but I just want to show you how you can do this so we have4 comma 0 de I would convert this and say this is 4 comma 180° just to make it simpler and and essentially that would be what so I would go out to a circle with a radius of four and then swing around to 180° so right there and then 0 comma 90 Dees doesn't matter the angle if you have a radius of zero you're not going anywhere right so from the pole you're stuck right so it doesn't matter which way I'm looking or turning or orienting myself I'm just going to be on the pole okay then 2 comma 120° or 2 comma 240° so in each case you're going out to a circle with a radius of two and then I'm going to swing around to 120 and then I'm going to swing around to 240° okay so from those four points that's enough to sketch the graph of that Circle if you're using a computer obviously it's going to be perfect but if you're doing it freehand just notate your little points there so you can show the teacher you found some points and then sketch the circle the best way that you can okay let's look at another example so here we're giving this guy in polar form so r equal 6 * cosine of theta now be very very careful when you look at the formula remember it's R is equal to 2 a Time the cosine of theta 6 is not a okay you have to do some extra work here you would say that okay well this guy right here is equal to this guy right here okay so 2 a is equal to six which means if you divide both sides by two a is equal to three okay so once you know that a equals 3 converting this over is pretty quick right so let me write here that a equals three what I would do I would say okay well I know that we have this form of r = 2 a * cosine of and this goes to x - a^ 2 + y^2 is = a squ okay we know that so here I'm just going to plug in everywhere I see an a I'm just going to put a three so this would convert itself over to xus 3ty squar + y^ 2 is equal to 3 being squared which is nine so that's the quick way to do it but again you've got to make sure that you match things up correctly if you want to ensure against a mistake then you can always do this the long way so so what I would do here is I would multiply both sides by R so that gives me R 2 okay which we can replace and it gives me 6 * R * cosine of theta remember whenever I see R * cosine of theta I can replace that with X and whenever I see R 2 I can replace that with x^2 + y^2 okay so I'm going to have x^2 + y^2 is equal to 6X what I'm going to do is move this over to the other side so I'll say I have x^2 and then I'll say minus 6X because I'm subtracting away from each side and then I'm going to leave a space because we're going to need to complete the square then plus y^2 and this equals z okay so let me get rid of all of this just so I have a lot of room to work get rid of this and let's move this up so maybe it's been a while since you completed the square but remember what you want to do is you want to look at this coefficient right here in this case is-6 it's always the one on the variable to the first pile okay so X the first pile so -6 you want to cut in half or multiply by half and then you want to square the result okay so6 / 2 or6 * half is going to be-3 so then -32 is equal to 9 okay so from here I'm going to add nine to both sides to make this legal can't just add it to one and essentially what I'm going to have here is a perfect square trinomial on the left hand side okay I don't need to do anything here because this is just y squ can't do anything with that so with this one I'm going to go okay xus 3 quantity squar is how you factor that then plus y^2 is equal to 9 okay and again it's easier if you use the formula but sometimes if you're not super familiar with how to work with the formula you might as well just go through this to guard against any mistakes okay so let me erase and let's talk about graphing this guy so again I would put some points here so R and then Theta okay so if I'm thinking about some thetas I'm always going to go with 0 degrees cosine of 0 degrees is one so R would be six then if I think about let's go ahead and do 90° okay cosine of 90 deg I know is 0 6 * 0 is zero so R is zero okay so it's going to be at the PO again let me think about the cosine of 60° is going to be a half right so 6 * a half is three okay I'm trying to get some whole number values here so I could do 60 deg okay that would give me three and then also cosine is positive in quadrant four also so I could do 300° that has a 60° reference angle and I'm going to get three here as well okay so this would give me 6 comma 0° this would give me 0 comma 90° this would give me 3 comma 60° and this would give me 3 comma 300° and again if I want 6 comma 0° well basically from the pole I'm just walking to the right six units right so I'm going to be right there and then 0 comma 90 degrees again if the radius is zero I'm not moving off the pole and then 3 comma 60° again I'm going to go out to a circle with a radius of three and then swing around to 60° so that's right there Then 3 comma 300° circle with a radius of three and I'm going to swing all the way around to this 300° right here and again these are enough points here to where you can plot them and then sketch a nice little circle through the points again doesn't have to be perfect just notate your points so your teacher can see that all right so another common occurrence is going to be a problem where the center is on the y- AIS and the circle is going to pass through the origin so in other words from the origin which is a point on the circle to our Center okay is going to be a distance of a so that's going to be the radius okay so the center occurs at a comma in this case could be 90° angle right if I'm thinking about this this is a right angle so it's 90° and essentially if I wanted to come back up here I would change only a few things so let me get rid of this and now this would be 90 de that I'm plugging in here so that's a change okay so first let's go through and do the cosine of we'd have Theta minus 90° remember this is going to be your difference identity for cosine so it's the cosine of the first guy which is Theta times the cosine of the second guy which is 90° let me write that a little bit better so what is the cosine of 90° well we know that's going to be zero so you just get rid of that right but I'm gonna leave it for right now so plus then you're going to have the sign of the first guy which is Theta and then times the sign of the second guy which is 90° okay so we know this is gone we know that the sign of 90 de is one you say that's one so basically this is going end up being s of Thea okay so what I'm going to have here I know that this would be a^ S and this would be a s so get rid of it I'm just going to say that I would have 0 is = to r^ 2 - 2 a r okay instead of cosine of theta minus 90° we know that this is the S of theta so s of theta okay let me move this over so r^ 2 is equal to -2 a r and then times the S of theta let's get rid of this and I'm actually just going to get rid of this right here okay let's move this this up like this and I'm going to multiply both sides by Nega 1 get rid of the negatives and essentially from here I'm just going to divide both sides by R okay like I keep doing so what's going to happen is this is going to cancel with this this is going to cancel with this and you get R is equal to 2 a * the S of theta okay so if R equals 2 a * the cosine of theta it's going to pass through the origin and it's going to be centered on the xaxis if Ral 2 a * the S of theta it's going to pass to the origin it's going to be centered on the Y AIS again if we go back to this guy can erase this if you think about these coordinates here in rectangular this would be what it would be zero that's the X location and then the Y location would be a so now if I think about this in rectangular you would have x minus 0ty S Plus youd have y - a^ 2 is equal to again the radius here is a the distance from the origin or a point on the circle to the center is going to be given by a so I can say this is a squ okay so let me grab this real quick so what I'll do from here I'm going to write this as just X2 because obviously x - 0 is just X then plus I'm going to expand this so this will be y^ 2 - 2 * y * a so let's write - 2 * a y then plus a being squared and this equals a squar so you'll notice that again I could subtract a squar away from each side so basically let's just line this out put a zero here okay and then from here have x^2 + y^2 we know we can replace that with r 2 let's do that in a second I'm going to say this equals to I'm going to move this over here so I'm going to add 2 a y to both sides so I'm going to put 2 a y like this and now all I need to do is take X2 + y^ 2 replace that with r 2 and then the 2 a y I'm going to put 2 a and replace y with r times the S of theta okay and then for the last step I'm just going to divide both sides by R so let me scroll down to get a little bit of room going here and all we're going to do is just cancel one of these with this and this with this and I'll say R is equal to my 2 a multiplied my S of Thea okay let me make that s a little bit more clear all right so for our example we'll look at Ral -4 * the S of theta so again if you see something that looks like R = 2 a times the S of theta this guy is going to pass to the origin and it's going to be centered on the y- axis okay okay so again don't just take4 2 a this part right here let me do this in a different color so this is clear so 2 a is going to be equal to -4 so that means that basically you have 2 a = -4 divide both sides by 2 a = -2 okay so make sure you understand what you're plugging in so if a equal -2 and you want to plug into that formula so we're going to have x^2 plus you'll have your quantity y minus a in this case it's minus a -2 okay and that's going to end up being y + 2 but I'll change that in a moment so this is squared and this equals the radius which is a being squared so you'd have -2 being squared okay so this ends up being x^2 plus you'd have y + 2 okay quantity squar is equal to four okay so that's converting it over to rectangular using our little formula now if you don't want to do this or you're not comfortable with this again it's fine it's no big deal you can always do this the long way okay so for this one I would multiply both sides by R so you get r^ 2 is equal to-4 and then time R * the S of theta if you have R * s of theta that's y okay so just go ahead and replace that with Y replace r s I'm not going to have room for this but you want to replace R 2 with X2 + y^2 so this is x^2 + y^2 and again you're just going to move things over and complete the square that's all you need to do so this is x^2 plus I'm going to go ahead and say this is y^2 and then I'm going to add 4 y to both sides okay and I'm also going to leave a space here here and I'll put equal Z right because I added 4 y to both sides so here I need to cut four and half that' be two and then Square it I get back to four now because I added it over here I've got to add it over here so this equation becomes this so you get rid of all this okay so I'm going to go ahead and say this is x^2 plus if you factor this this going to be y + 2 qu^ s and then this equals 4 okay so this equals 4 so it's not that bad overall if you're really good with going through the steps on this I like to use the formula if I see something thing immediately it's just a lot quicker for me but if you want to guard against errors this is a Surefire way to do it every time so let's think about graphing this now again what I want to do is get some points so I'm going to put R and then Theta so the first one that I'm going to do again I always do0 degrees so I know that the sign of 0 degrees is 0 so4 * 0 is 0 okay and then if I did let's say the sign of 90° well I know that's going to be one okay so if s of 90° is 1 then R would be -4 now if you want to convert that over you could do pos4 and 2070 degrees if you wanted to okay it's a little bit easier to work with now let's think about some other things where is s equal to a half because if this was a half and I multiplied it by4 better yet if it was negative a half and I multiplied it by ne4 I would get positive2 okay so s is 12 at 30 degrees or 150 degrees if you want it to be negative2 then you would want to do 210° okay that would have a 30° reference angle or you would also want to do 330 Dees that would have a 30 Dee reference angle as well so in each case s of those S of 210 degrees or S of 330° would be -24 * one2 would give you two in each case Okay so let's write these out we have 0 comma 0° we have 4 comma 270° we have 2 comma 210° and we have 2 comma 330° okay so let's grab these guys so 0 comma 0 degrees that's obviously at the pole so let's go ahead and highlight that then 4 comma 270° go out to a circle with a radius of four we're going to swing around okay to 270° so that's going to be right there we have two comma 210° so let's go out to a circle with a radius of two we're going to swing around to 210° this going to be right there and then 2 comma 330° go out to a circle with a radius of two and then we're going to swing around to 330° which is going to be right there okay so again if I have four points that's enough for me to come through here and sketch that guy as a circle okay let's look at another example so we have x^2 plus the quantity y - 4^ 2 = 16 again you're going to use this formula make sure it matches up so x^2 plus you have y - A being squared is equal to a 2 so you've got to look at this guy and this guy right so is this guy being squared giving you this guy in this case a is four right not negative4 you're taking this right here it's minus whatever this is so a is four if I squared four would I get 16 yes if this was 17 or 18 or 19 or 20 or something that was not 16 then you could not use this formula because it does not work okay so this guy is going to give you the circle that passes to the origin and it's centered on the y- axis again this converts over to R is Al to 2ti a * the S of okay so if I know a is 4 then R is equal to 2 * 4 which is 8 times the S of theta okay so if you know the formula and you get these a lot these are very common problems you can convert them almost right away okay so let me erase this and again I'll show you how to do this the long way so what I would do is expand this so I'd say x^2 + y^ 2 and then You' have minus 2 * this guy * this guy so 2 * 4 is 8 and then * y Y and then + 4 16 is equal to 16 you'll notice that again this is going to cancel so what you have here is X2 + Y2 which is R 2 and then I'm going to say is equal to I'm going to add 8 y to both sides and then y I can always replace with r okay I can always replace with r times the S of theta okay and again I can just divide both sides by R you get R is equal to 8 times the S of theta okay so pretty quick to do that way as well but the formula is actually faster okay so let's go ahead and get rid of this and let me write this a little bit more neatly so R equals 8 * the S of theta okay so let me go ahead make that Theta a little bit better so let me go ahead and get some points for us so R and then Theta okay so we know again if I choose a zero here so 0 degrees s of 0 degrees is 0o so this would be zero okay if I choose 90° well the sign of 90° is one so r would be eight okay and then I'm also going to deal with a half s of 30° is a half so if I put 30° here that's a half a half * 8 is 4 and then if I did 150° s of 150° is also a half 8 * a half is 4 okay so my points here would be 0 comma 0° and then 8 comma 90° and then 4 comma 30° and then 4 comma 150 degrees so let me grab this real quick and let's go ahead and paste this in so 0 comma 0° going to be right there if you want 8 comma 90° so we're going to go out to a circle with the radius we have 1 2 3 4 five six 7even and then eight and then I'm going to swing around to 90° so it's going to be right there and then I want 4 comma 30° so a circle with a radius of four and then I'm going to go to this angle here at 30° and then 4 comma 150° go out to a circle with the radius of four I'm going to swing around to 150 degrees so right there okay so again four points is usually enough if you need more you can approximate some of these other ones that are not going to be whole numbers in terms of the r values but you can get very close to where you can sketch an accurate graph all right so let's wrap up the lesson and look at some examples where we're not centered on the x-axis we're not centered on the Y AIS but we're still going to have our Circle pass through the origin so to set this up I'm going to be working with rectangular coordinates and so let me write this out we have x - h^ 2 plus we have y - k^ 2 = R 2 so this is back to basic algebra I'm going to be replacing things we can see that this passes through the origin okay which is the point in rectangular as 0 comma 0 we can see that the center okay right here is labeled with rectangular coordinates a comma B so the X location is going to be a so I can erase this and put an A and the Y location is going to be a b so I can erase this and put a b okay now again you've seen me throughout the Lon replace this typical r that we see in algebra for the radius with a because we don't want to get confused with the r that comes up in polar form so for right now I'm just going to erase this all together and I'm just going to put a blank here and squared okay so what would I need remember the radius is the distance from the center to any point on that Circle so the distance from right here to right here can be found with the distance formula if you want to use that so you could say the distance between these two points would be the square root of what we would subtract Aus 0 right you subtract the x coordinates so let's say a minus 0 would just be a and you're going to square that then plus you're going to subtract the y-coordinates so B minus 0 would just be B and that would be squared okay now if I put this in here and squared it I would just get a 2 + B2 so let me erase this and just put a 2 + B squar okay so that's really all there is to this formula when you see it in rectangular form when you convert it over into Polar form it actually works out quite nicely so let's grab this guy real quick let's copy it and let's paste it on this fresh sheet here let's see what we can do all right so from here what I'm going to do is put this into Polar form we're going to find this is really really useful when we're trying to convert back and forth between the two so I'm going to start by expanding this guy and this guy on the left hand side so we already know how to do this this is x^2 minus 2 * this guy time this guy so- 2 2 ax then plus this guy squar so a s then plus same thing over here so y^ 2 - 2 * this guy * this guy so - 2 * b y then plus this last guy b^ 2 and this equals a 2 + b^ 2 okay so we can see that this is on each side right so I can subtract it away from each side and it's going to go away from each side this is on each side so I can subtract away from each side it's going to go away from each side because everything is gone on the right hand side I'm going to put a zero there okay so over here what I can do now is rearrange things so I'm going to put the x^2 plus the y^2 next to each other again that's a common theme when you're working with these problems you're trying to get the polar so I'm going to say x^2 + y^2 again you can replace that with r 2 then what I'm going to do is I'm going to take this guy and move it over here and I'm going to move this guy over here as well so let me put my equals here I'm going to add 2 ax to both sides so of course it's going to cancel over here and then I'm going to add 2 by y to both sides so again of course it's going to cancel over here okay so this is my equation now and basically to convert it over to Polar X2 + y^2 is R 2 by definition and then everywhere I see an X I can always put R cosine Theta so I'll put 2 A Time R cosine of theta and then plus 2B everywhere I see a y I can put R and then sine of theta okay let me erase all this and essentially what you want to do here is solve this for R again the common mistake is to take take the square root of each side what you want to actually do is Factor this R out okay so Factor this R out so that you can divide both sides by R so I'll say this is 2 a * the cosine of theta and then plus 2 B * the S of theta okay and this equals the r squ okay so again the trick here is I'm going to divide both sides by R this is just multiplication here so this is perfectly legal cancel one of these with this cancel this with this okay and what I'm left with let me just write this above here okay because we're use this we'll say in polar form this is R is equal to you'll have your 2 a Time the cosine of theta okay and then plus your 2B times your s of theta so there will be a lot of problems that deal with this both in pre-calculus and in calculus again this is for a very specific case your circle is going through the origin okay and then your Center is not on the X axxis it's not on the Y AIS the center let me write this again so the center is is going to be at a comma B okay and then you want to check that your radius your radius again is the square root of a squ plus b^2 okay so if your circle fits this description okay in rectangular then you can quickly convert it over to Polar using this form here okay so you take your a from here and your B from here you can check that a squ plus b square is what you've got here is your number okay and then you can convert it over all right let's take a look at an example so we have R is equal to 6 * the cosine of of we have this theta plus pi 3 okay so the first thing I would do here is just simplify this using my cosine sum identity okay so I'm going to say R is equal to I'm going to put six you can use parentheses or brackets whatever you want but basically you want to expand this out so I'm going to say this is the cosine of theta and then times the cosine of pi over 3 and then remember when you use the cosine sum identity your signs are going to change this is plus so it's going to be minus so this is now going to be the S of this Theta and then times the S of pi over 3 okay close these brackets down we know pi over 3 is 60° and cosine of 60° is a half we know s of pi over 3 or S of 60° is square 3 over2 so let's just replace that real quick I'll say R is equal to 6 open up the brackets I'll say 12 * the cosine of th and then minus I'll say < TK 3 over2 times the S of theta okay now when you distribute the six into each you have a denominator of two 6 / 2 is 3 so you basically have 3 * cosine of theta and then You' have -3 * 3 * s of Thea okay so let's go ahead and say we have R is equal to again this would be 3 * cosine of theta and then minus 3 * < TK 3 * the S of theta okay so here's where we can stop I'm just going to erase all of this we don't need it anymore all right so to use the formula let's just write r equal our2 2 a * the cosine of theta then plus the 2 B * the S of thet this converts over to x - a^ 2 + y - b^ 2 is equal to a 2 + b^ 2 okay so if I'm trying to convert this over here I know that 2 a I just got to match these things up 2 a is equal to 3 and 2B okay is going to be equal to the negative of 3 * of 3 okay so be very very careful there because depending on your book you might get different formulas so I'm going to say that 2 a is equal to 3 divide both sides by two and a is equal to three Hales okay so that would get plugged in here and here okay so let me erase this and actually I'm just going to put this up here so a is equal to 3es and then similarly if I have 2B is equal to -3 * 3 we'll divide both sides by two and B would be equal to the negative of 3 * 3 over two okay make this three a little bit better okay and then again you're just plugging in so if I took that and I plugged in here so x minus 3es this quantity would be squared and make these parentheses a little bit larger then plus you would go y minus a negative is plus positive so y plus I'll go ahead and say 3 * 3 over 2 quantity squared is equal to so you've got a square a You've Got A square B and then you want to sum those results so a s 3 half squar is going to be 94s then plus if you square this guy the negative becomes positive 3 squ is 9 squar of 3 squ is 3 and 2 squar is 4 so basically this is 274 if you add these guys together because you do have a common denominator you're going to get 36 over 4 which is n okay so this right here would be how you can quickly convert and I say quickly but how you can more quickly convert this over to your rectangular form I realize this problem does take some time no matter how you do it okay so what I would do from here let me just erase this and I'm just going to do this the long way okay so when you see something like this and you're trying to convert it to rectangular remember R 2 is x^2 + y^2 and if you have R * cosine of theta it's X and R * s of theta it's y so the strategy here if you just have R is to multiply both sides by R this would give me R 2 is equal to 3 r cosine of thet - 3 * > 3 * R * the S of thet okay I just multiplied both sides by R now I can replace things so this is x^2 + y^2 okay and this equals this is going to be 3x and then minus this is going to be 3 * > 3 and then time y now if you feel more comfortable you can write the Y right here so you could also write this as -3 * y * 3 a lot of books will do this so that you don't get confused and think that the Y is underneath the radical there so we'll just kind of use this notation but the coefficient for y is -3 * 3 so we have to know that okay so what I want to do now is move things over so I want to move these guys over here okay and the reason I'm going to do that is so I can complete the square so if I subtract 3x away from both sides I'm going to have x^2 I'm going to have minus 3x then I'm going to put plus and I'm going to leave a space right because I want to complete the square then plus I'm going to have my y^2 I'm going to add this to both sides so Plus 3 y * < TK 3 okay then plus I'm going to leave a space and this equals I'm going to put zero so what I would do here is complete my Square again take the coefficient for your variable raised to the first power so in this case it's -3 cut in half so that would be3 Hales Square it you would get 94s okay so let me write 94s here again it's not legal to do this unless you add it to both sides so we got to put it over there okay then over here this one is a little bit more tricky but we just did it it's 3 * Square Ro of three cut it in half and I'm writing these threes terribly so let me put this in a little bit better form here cut it in half and then Square it we just saw this is going to be 274 so this is 27 over 4 okay so plus 274 we already know this becomes 364 and this is nine right because we basically just did this so this is nine okay and again if you go through and Factor these this is a perfect square trinomial and this is a perfect square trinomial so this factors into x minus your three Hales quty squar and then plus this is going to factor into y plus your 3 * your square of 3 over 2ty squar and this equals 9 so however you do this you're going to get the same answer I just want to give you multiple ways to attack the same problem okay let's erase all this in order to graph this on the Polar grid I'm going to use this form it's a little bit easier when you just have cosine involved versus having cosine and S because I can just plug stuff in here then add the 60° to it and then I can take the cosine of that you know so on and so forth so I'll say R and then Theta here I've got to think about the fact that whatever I plug in there I'm adding 60 degrees to it so if I put in 0 degrees I add 60 degre to that I get 60° cosine of 60° is a half a half time 6 is 3 then if I did something like let's say 30° well if I add that to 60° I have 90° cosine of 90° is 0 0 * 6 is 0 let's keep going with two more points again I'm just trying to get some whole numbers here so let me also do something like 240° 240° plus 60° would be 300° that has a 60° reference angle right and it's also in Quadrant 4 so it's going to be positive so the cosine of 300° is a half 6 * a half we already know is going to be three right so if this was 240° okay again adding 60° gives me 300° cosine 300° is a half 6 * half is three all right the last thing I want to consider we know that the cosine of 0° is 1 okay but again I'm adding 60° to this so let me go ahead and put 300° in here okay that's another one we can try so 300° plus 60° would be 360° cosine of 360 degrees is the same as cosine of 0 degrees it's going to be one so 6 * 1 would just be six okay so this is four points again if you need more you can always use some approximations I'm going to go ahead and say 3 comma 0° I'm going to say 0a 30° I'm going to say 3 comma 240° and then I'm going to say 6 comma 300° like this I'm going to grab these guys copy them let's paste this in here like that so we have it so again 3 comma 0 de well basically from the poll I'm just going to step out to a circle with a radius of three so I'm just going straight to the right like that and then 0 comma 30° again if this is zero here for the radius not going anywhere right so I'm going to be right there on the pole 3 comma 240° I'm going to step out to a circle with a radius of three swing around here to 240° so it's going to be right there and then you have 6 comma 300° so I'm going to go out to a circle with a radius of six I'm going to swing around to an angle of 300° let's go right there so again if you need some points in between these to make a more accurate sketch you can use some approximations okay that's up to you generally speaking if you do this for a homework assignment you show four points and you draw your circle through those points you're pretty much good to go all right so the last example here that we're going to do we have x + 3^ 2 plus we have y + 1^ 2 = 10 so for this again I'm going to write this in the form of x - A qu^ 2 plus we have y - b qu^ 2 is equal to a^ 2 plus b^2 so you first want to check to see if this matches up so if you have plus some number you have it here and then here right as minus a Nega so I'll say x - A -3 and thenan 2 plus you have y - -1 quantity squar is equal to in this case you'd have -3 being squared so -3 being squared plus you'd have -1 being squared so first you want to check to make sure that this works out if this was let's say 11 you could not use this formula because it doesn't work out -3 s is 9 - 1 SAR is 1 9 + 1 is 10 not 11 right or if it was 12 or 14 or anything other than 10 this is not going to work out but since it does follow this format what I can do is just grab these values so I can say this is a and I can say this is B and then you can just plug in right so you can say R is equal to well I have 2 * a so 2 * -3 time the cosine of theta then plus I have my 2 * b b is 1 * the S of theta okay and that's all you need to do so let's come down a little bit so this would be R is equal to we'll have 2 * -3 which is6 * the cosine of theta then we have 2 * 1 which is -2 so I'll just put min-2 and and then times the S of theta okay so this is converted into Polar form again just for the sake of completeness I'm going to do this one more time the long way just so you get some practice if I come back up here and I expand this this is x^2 plus 2 * this guy * this guy so 6X plus this guy squared so 9 plus this guy is y^ 2 plus 2 * this guy * this guy so 2 y plus this guy squar which is 1 and this equals 10 okay so if you sum nine and one you get 10 and again you can subtract that away from each side so I'm going to go ahead and line these out and line this out and put a zero right just like I subtracted 10 away from each side so then from here what I'm going to do is do my x^2 + y^2 I'll just go ahead and say that's R 2 and I'm going to say this equals I'm going to move these guys over to the right hand side so I'm going to subtract 6X away from each side and I'm going to subtract 2 y away from each side so then I'll have R 2 is equal to -6 * this is R cosine of theta then minus 2 * this is r s of theta okay and then from there again you can always Factor the r out so you have it here and you have it here so let me write r^ 2 is equal to R * the quantity -6 * the cosine of theta then minus 2 * the S of theta okay and then let me divide both sides by R so divide both sides by R and one of these will cancel with this this will cancel with this and you're back to that form that we found so R is equal to -6 * the cosine of theta then- 2 * the S of theta all right so graphing this guy is a little bit more challenging even if you were on the rectangular coordinate plane you would basically graph the center which would be at -3 comma 1 and then you would move up by sare < TK of 10 units you would move to the right by square of 10 units you'd move to the left by square of 10 units and you move down by the square Ro of 10 units that would give you your four points on the circle and then you could sketch your circle but here because 10 is not a perfect square we're going to have to do a lot of estimation let me go ahead and put my R and my Theta and I'm going to first start with 180° and the reason for that is I want something that's going to cancel this out completely and I want something that's going to multiply by this -6 and be Nega 1 so the cosine of 0 degrees is going to be one but the cosine of 180° because we're flipped over now we're going to be at negative 1 okay so if I plugged in 180° there for Theta this is gone that's zero right and then this right here would be -6 * 1 which is 6 so that's one whole number value that we can get so let me put 6 comma 180° and then another one that we can get if we do 270° the cosine of 270° that's going to be zero so this will be gone and the sign of 270 degrees is 1 so -2 * 1 is 2 so if I do 270° then basically I would have two now for the other points let me go ahead and put this down here as 2 comma 270° I'm going to go ahead and approximate some stuff so again you just got to pull your calculator out I'm going to go with 135° I think this one's going to be a little bit easy for us to work with so if you plug this in here and here and you punch that into your calculator you're going to get 2 * of two so 2 * squ of two okay so 2 * < TK of Two And this is 135° if you want an approximation for that you could say 2 * > two is approximately 2 let's just go ahead and say 8 three okay so that's close enough for me maybe you want to get something that's a little bit closer so you can just include more decimal places now let me get one more guy going here again you take as many as you need I'm going to go ahead and go 210 degrees here that works out kind of nicely so if I punch this in my calculator I'm going to end up with 3 * < 3 then plus one okay so 3 * 3 + 1 and then 210° okay so this one let me go ahead and write this over here I think I'm going to run out of rum but I'll just put it right here so 3 * 3 + 1 is approximately let's go ahead and say 6.2 okay so I'll say 6.2 so let me grab these guys real quick I'm going to copy this and let's come over here I'll paste this in again I've already drawn this with a computer which nowadays pretty much everyone does that so I'm just going to go ahead and plot these the best that I can so 6 comma 180° so I'm going to go out to a circle with a radius of six and then swing around to 180° it's going to be right there and then 2 comma 270° go out to a circle with a radius of two I'm going to swing around to 270° this going to be right there and then 2 * < TK two we said was about 2.83 okay so 2.83 so let me come out to about right there I'd say and then swing around to 135° that's going to put us about let's just say right there again you can't get this perfectly we're just trying to sketch a circle get an idea of what this looks like if you need something perfect you can use your ti 83 or 84 or you can use a graphing calculator online then we have 3 * 3 + 1 we said that was about 6.2 so let's go out to this about 6.2 right here and then we said that it was 210 degre so let me swing this around to 210° so that's going to be about right there okay so from those points again you could sketch your circle it won't be perfect but it'll be close enough to get your assignment completed in this lesson we want to talk about the inclination of a line the angle between two lines and also the distance between a point and a line all right so what I want to do is start out by talking about the angle of inclination so I'm going to go through a few scenarios with you generically and then we'll look at some examples so this is what you'll probably see in your textbook and there's just a few things to go through here as essentially every nonhorizontal line is going to impact the x-axis somewhere so in this case we're looking at a positively sloped line okay and it's going to impact our x axis right there okay so we're saying the point is x sub one that's your x value and then zero for the yvalue because again it's on the x- axxis you can label this as y sub one we're going to need that in a moment now when we talk about this angle of inclination we're considering the angle formed from the xaxis here okay measured counterclockwise to this line so we're going to name this with Theta okay so that's going to be our angle so with this guy right here the first thing is I would consider the slope of this line so remember if you have two points in a line so we have this point and then we have this point we can use our slope formula so M which is classically used to denote slope is equal to remember it's rise over run so typically we see y sub 2 - y sub 1 over your x sub 2 - your x sub 1 okay now in this case we know that y sub one has a value of zero so you can just plug that in okay and then essentially you can simplify this in this particular case let me make this a little bit neater I'm going to say that M our slope is going to be y sub 2 over x sub 2 minus X sub1 okay so that's pretty easy to understand now the other thing we can consider here is that we could form a right triangle okay so what I'm going to do is have this side right here and the length of this side which is adjacent to to our angle Theta is going to be x sub 2 - x sub1 so this x coordinate here whatever it is minus this x coordinate here whatever it is would be the length of this side then let's do the opposite side as well so this is opposite of our angle Theta and this guy has a length of Y sub2 whatever that is because you have y sub 2 minus your zero there because this point right here again is on the x- axis so I want you to notice that the tangent of th the tangent OFA is equal to what well it's opposite over a adjacent so this is y sub 2 over x sub 2 - x sub 1 okay so do you see the relationship between the slope of the line and the tangent of theta well they're going to be equal right M the slope is equal to the tangent of theta because they both have the same definition here so I can say that M our slope is equal to the tangent of theta okay but like all things in trigonometry it's not always so straightforward to get your answer all right let's start out with the easiest scenario this is where you have a positively sloped line like we just looked at here you're going to have an acute angle for Theta okay so your angle of inclination will be greater than 0 degrees and less than 90° so in this case you can just plug straight into to your inverse tangent function okay so if you have a positively sloped line that's all you need to do you'll have the correct answer okay now the harder scenario occurs when basically you have a negatively sloped line like we see here so now you're going to have an obtuse angle so so Theta is going to be greater than 90° and less than 180° and for this guy you have to do more work because when you use your inverse tangent function basically you're going to get something in Quadrant 4 it's going to rotate clockwise like this and you're going to get this angle measured clockwise okay so what you're going to have to do is you're going to have to take that angle you get and add it to 180° so that you can get this obtuse angle here that's going to be in quadrant 2 some other special case scenarios obviously you'll have a horizontal line and a vertical line you should know that basically if you have a slope of zero right something like Y is equal to K that's a horizontal line the slope is zero so Theta there is just going to be 0° now lastly you could also get a vertical line we know Theta here you're basically going to have a right angle so Theta there is going to be 90° and we know that tangent of 90° is not defined right we know that with a vertical line something like X is equal to K you have an undefined slope with all of that being said let's go ahead and look at an example so what I'm going to do is take this 4x + 5 y = 20 and put it in the slope intercept form so y = m the slope * X plus b the Y intercept and so I'm going to subtract 4X away from each side of the equation that would cancel I'll have my 5 Y is equal to -4x + 20 divide everything here by 5 and this is going to give me that Y is equal to we'll have -4 fths okay that's going to be my slope time x and then + 4 okay now a lot of times when you first start doing these you might want to graph it just to understand what's going on but essentially you know that if you have a negative slope which we have here M is-4 FS that essentially you have an obtuse angle right so you have to do additional work with this guy and to verify that let's just look at the graph real quick so the slope is4 fs and the Y intercept is going to be at 0 comma 4 so let's look at the graph so my Y intercept again is at 0a 4 so that's there and my slope is 45ths so drop 1 2 3 4 to the right 1 2 3 4 5 they'll give you that point right there okay so we're looking at this angle right here Theta we're trying to figure out what that guy is you can see just by inspection that this guy is going to be an obtuse angle okay so we know we're going to be in quadrant two so again what you want to do is you want to say the tangent inverse of this guy so -4 fths just punch that into your calculator you're going to get approximately I'd say - 38. 66 okay let me put degrees there to be specific now once you have this again what you want to do because you're looking for your angle in quadrant 2 this is giving you something in Quadrant 4 rotating clockwise okay so what I want to do is I want to say okay 180° plus this 38.6 6° is going to give me I'll say about 14134 4° okay so this would be my angle of inclination obviously this is an approximation and I'll say Theta is approximately 141.5 4° now what I would do when I first start this is go back to your picture if you made a graph and just say okay is that about 141.5 4 degrees yeah that looks about right so I'm going to put approximately 14134 de all right let's try another one so we have -2x + 3 y = 6 again I'm going to put this in SL intercept form so y = mx + b so basically I would have if I add 2x to both sides then what I would have is this canceling over here I'll have 3 Y is equal to 2x + 6 then I'm going to divide both sides by three okay and I'll have y is equal to 2/3 okay that's going to be my slope * x + 6 ID 3 is two okay so your slope let me get rid of this your slope m is going to be 2/3 and your Y intercept your Y intercept is going to occur at 0 comma 2 okay so if we look at the graph of this guy you would say okay well your Y intercept is at 0 comma 2 your slope is 2/3 so up one two to the right one 2 three okay so right there up one two to the right one two three so right there okay so you have some points you can draw that line again we're looking for this angle Theta you can see this is an acute angle right so basically I would want to say the tangent inverse of let's go ahead and say 2/3 this is approximately I'll say 3369 de and we don't need to do anything else here because this guy is in quadrant one right so basically I can say this is approximately positively sloped line like this it's really really easy just plug this into your inverse tangent function and you're done you don't have to do any extra work okay let's look at one that is in Reverse so now we're given our angle of inclination our Theta is 135° and we're given a point on the line which is7 comma 2 and essentially we want to write the equation of the line and so if we wanted to do this the first thing is we need to figure out the slope right so the slope is going to be what remember m is equal to the Tang OFA well I havea so if I take the tangent of 135° what would that give me so my tangent my tangent of 135° is equal to I'll go ahead and say m which is equal to1 okay so I know the slope is negative 1 so let me erase this and put negative one here you can get rid of this now if I have my slope which in this case is Nega 1 and I have a point of the line let's call this X sub1 and then y sub one well I can plug into something called point slope right so y - y sub 1 is equal to M * the quantity x - x sub 1 so here Yus my y sub 1 is 2 is equal to M which is -1 * the quantity x - x sub 1 is -7 minus 7 is+ 7 now you can solve this for y and put it in slope intercept form you might also be asked to put it in standard form so just get with your instructions so y - 2 is equal to this would Bex and then - 7 let me add two to both sides of the equation let me erase this and I'll go ahead and say that our equation is y is equal to we'll say that it's X and then we'll go ahead and say minus 5 okay so that's your slope intercept form of the line if you want to look at this graphically your Y intercept would occur at 0 comma 5 and your slope would be Nega 1 so your Y intercept would be at 0 comma 5 so that's right there and your slope is negative 1 so if you wanted to get points going this way you would go up one and to the left one up one to the left one up up one to the left one up one to the left one up one to the left one so you get to this point here and you can see we have this angle Theta here which is going to be equal to 135° all right so what we want to do now is talk about the angle between two lines so if you have two lines in a plane they're either going to be parallel lines in which case they'll never intersect or there'll be non-parallel lines and in that case they will intersect at some point okay so the case we would talk about here today will be non-parallel lines obvious viously because we need some point of intersection and also they're not going to be perpendicular lines because we don't want them to meet at 90° angles okay so we're talking about the other situation so I have a little image here which I want to focus on so you'll see that this point of intersection here forms two pairs of opposite angles so you have these acute angles so you have this one and you have this one right so those would have the same measure and then you have these obtuse angles so you have this one and then you have this one right right so those would be of the same measure as well what we're going to focus on when we say the angle between two lines we're focused on this acute angle right here which I'm calling Theta okay so we're going to be focused on how to find Theta so Theta is equal to what well to come up with a little formula we're going to start off with this scenario and this won't work in every scenario so we're going to have to adjust our formula just a little bit okay so we're going to say that this blue line here is a negatively sloped line okay and so the angle of inclination here Theta sub 2 we know that that would be an obtuse angle right so we can say that Theta sub 2 is greater than 90° and less than 180° so we know this then this orange line is positively sloped so we know Theta sub one the angle of inclination there is an acute angle so Theta sub 1 is greater than 0 de and less than 90° So based on this we know that Theta sub one in this case is going to be less than Theta sub 2 okay so we can say say let me erase this I'll just say that Theta sub 1 is less than Theta sub 2 okay so that's for this case now another thing we can say because these lines are not going to be perpendicular I can say that Theta sub 2 which is the bigger angle here minus Theta sub one which is the smaller angle is going to be strictly less than 90 de because if this was equal to 90° these guys would be perpendicular okay so we're talking about these two assumptions to make what I'm about to say work if these two assumptions don't hold we're just going to change our formula up just a little bit we're going to do this with using the absolute value operation but we'll get to that later on okay so what I'm going to say here is that if I want this angle here okay I can find it by first getting this larger angle okay think about that and then just subtracting away the smaller angle so in other words I'm just finding this piece right here so it's going to be the Theta sub 2 okay minus the Theta sub 1 so this is equal to we'll say Theta sub 2 minus Theta sub one okay so that's for this case right here we're going to use this to derive a little simple formula and then I'll show you how to correct it for other scenarios okay so let's go down and say that we have Theta is equal to Theta sub 2 remember this was our bigger angle minus our Theta sub one which was the smaller angle okay so what we're going to do here is we're going to take the tangent of each side so I'm going to say the tangent of theta is equal to we're going to say the tangent of this Theta sub 2 minus this Theta sub 1 remember you cannot distribute this in you have to use an identity so if we go back to our sum and difference identities we see the tangent of a minus B is the tangent of a minus the tangent of B over 1 plus the tangent of a * the tangent of B okay so we're just going to follow that and we're going to say that this is basically equal to we'll do the tangent of this first guy Theta sub 2 minus the tangent of this second guy Theta sub 1 this is over we're going to do 1 plus plus we're going to multiply the two together now your book is going to usually change the order of this this is up to you so I'm going to go tangent of theta sub 1 times the tangent of theta sub 2 okay with your identity handout it's tangent of theta sub 2 * tangent of theta sub one but you can multiply in any order you'll get the same answer so it doesn't matter okay so let's go down a little bit we're going to keep this going from here I'm going to make some other substitutions I want you to recall that we say that our slope of a line M is equal to the tangent of our angle of inclination which we call Theta okay so now let's say I have two different lines so I have two different slopes and two different angles of inclination okay so I'm going to say that this is M sub one and this is going to be Theta sub one and this is M sub 2 and this is going to be Theta sub 2 so what I'm going to do now is just plug some things in since m sub1 is equal to tangent of theta sub one well where I see a tangent of theta sub one I can just plug in an m sub1 and then similarly where ever I see a tangent of theta sub 2 so here and here I can just plug in my M sub 2 okay so that's all I need to do so let me slide this over a little bit so I can fit everything on the screen I'm going to say that this is my M sub 2 and then minus my M sub1 okay and then over 1 plus I'm going to multiply them together so M sub 1 time M sub 2 Okay so let's talk about correcting this essentially because I want to end up with an acute angle okay and I don't want to have to worry about which slope is going to get put as M sub one or which slope is going to get put as M sub 2 I can just use absolute value bars here okay the reason this works is that basically if you do this in the wrong order you're going to end up with a negative inside of here okay and so you don't want that right remember if you do the inverse tangent with a negative you're going to get something in Quadrant 4 okay if you add 180 de to that you're going to get the obtuse angle that's between these guys and you want the acute one okay so you just want to make sure you use the absolute value bars so that whatever answer you get just make it positive okay and when you make it positive when you do your inverse tangent function you're going to end up with the answer in quadrant one which is your acute angle between the lines all right so we have 3x - y = 1 and x - 3 y = -6 so I like to put these in slope intercept form I also like to graph them so let's go ahead and say this is going to be what this is negative y and then I'll say equal -3x + 1 I just subtracted 3x away from both sides and then basically I could multiply everything by NE 1 so that would change this sign this sign and this sign okay so in slope intercept form this is just going to be Y is equal to 3x - one then for this one you would have -3 Y is equal to I'm going to subtract X away from each side so I have Min - x over here and then- 6 I'm going to multiply both sides by - 1/3 so this would cancel If I multiply this by- 1/3 this would be positive 1/3 and then - 1/3 multiplied by -6 would be positive 6 / 3 which is pos2 okay so this is my slope intercept form there so if you wanted to graph these two this one has a y intercept of 0 comma 1 and a slope of three and then this one has a y intercept of 0 comma 2 and a slope of 1/3 let me grab this real quick let's look at a graph real quick let me paste these in so we have some reference for where this stuff came from so we have 0 comma 1 so basically right here and the slope for this one is three so up one two three to the right one okay so there's a point there then up one two three to the right one so there's another point there okay so that's this line right here with a slope of three okay then the other guy is going to be 0 comma 2 so we have this guy right here okay the slope is 1/3 so up one to the right one 2 three so there's a point up one to the right one two three so there's a point right there okay so you see how we get the graphs of these guys and what's important is to understand what you're looking for you're looking for this angle right here this acute angle Theta okay now in some cases you're going to want to find this angle right here and basically once you found this angle right here you can subtract 180° minus that angle it'll give you this angle right here so let's go back and let's think about how we can find I'm going to say that M sub 1 is equal to 3 I'm going to say m sub 2 is = to 13 and then I'm just going to plug into the formula that I had so I'm going to say that tangent of theta is equal to remember you're going to do M sub 2 actually I should probably write this out so the absolute value of M sub 2us M sub1 so that goes up here and then down here you're going to have 1 plus M sub 1 * m sub 2 okay so my tangent of theta my tangent of theta is going to be equal to we're going to do the absolute value of let's go ahead and do 1/3 then minus 3 okay and then down here I'll do 1 plus you'll have 3 * 1/3 okay can't really fit this all on the screen so let me cut this away let's go to this blank sheet here and see if we can just crank this out so this equals what what is 1/3 minus 3 well you get a common denominator there three would be 9/3 basically 1 - 9 is8 so this would be 8/3 let me put my absolute value bars in there so don't forget them and then basically down here 3 * a 3 is 1 1+ 1 is 2 so this is divided by two and then what I could do here is say okay this is the absolute value of 8/3 since I'm dividing by two you can think about this as 2 over one just flip that guy multiply so times a half okay like this so from here I can cancel this eight with this two and get a four here okay so this would be 4/3 but we're taking the absolute value so we end up with 4/3 okay that's important because again we want to get an acute angle so when we do our inverse tangent we want this to be positive so we get something in quadrant one okay and let's bring this up here so tangent of theta equals 4/3 what I'd want to do is say the inverse tangent so the inverse tangent of 4/3 is going to be equal to what and I'll do an approximation here I'll say this is about 53.13% 3° again because this right here okay this angle right here would be a straight angle all you'd have to do is take 180° minus this angle here if you wanted this angle right here okay so if you wanted this guy right here you would say 180° minus 5313 de that would give you 12687 approximately degrees okay so sometimes they'll ask you to get this one usually you're asked to find this one but that's how you find both okay pretty easy overall let's go ahead and take a look at another example so for this one again I'm just going to put these in slope intercept form so I would say this is 4 Y is equal to I'm going to subtract 3x away from each side and I'm going to divide everything by 4 so that would give me Y is equal to I'll have -34 * x + 3 so my slope here my M let's call it m sub1 is equal to - 34s and I'm going to write the Y intercept just so we graph it and see what it looks like again not necessary so this is 0 comma 3 just doing it to get a full view of what's going on so let's erase this and let's drag this up here okay let me change my ink color for this one I would do what -4 Y is equal to subtract X away from each side Sox + 8 divide everything by -4 okay and that's going to give me that Y is equal to 1/4x so let's put this as 1/4x and this would be min-2 right so min-2 so M sub 2 let's say this is 1/4 here and my Y intercept is going to occur at 0 comma -2 again let's copy this real quick let's look at the graph so the Y intercept is at 0 comma 3 so that's going to be right there and then the slope is34 so down 1 2 3 to the right 1 2 3 four so that's that point right there then the other guy is 0 comma -2 so just drop by two so that's right there and then basically the slope is 1/4 so I can go up one to the right 1 2 3 4 okay there's a point then up one to the right 1 two 3 4 there's a point okay so we see where these graphs came from and again what are we trying to figure out we're trying to figure out what is this angle right here and I'm just going to plug into my formula so let me box this off so the tangent of theta again hopefully you have this written down it's the absolute value of you're going to do M sub 2 which is a 4th minus M sub 1 which is -34 and I'll just leave that like that for now I'll just simplify as I go then it's going to be 1 plus the product of the two so you'll have your - 34s and then times your 1/4 okay so 1 4 minus a 34s is like 1/4 plus 34s there's already a common denominator so it's just 1 plus 3 which is four over four which is basically just going to give us one okay so that part is very simple down here if you multiply these together you would end up with - 316 so this down here is min - 316 the one there can be written as 16 over 16 and of course if you subtract you would get 13 over 16 so 13 over 16 now this is already positive so you can go ahead and drop the absolute value bars you don't need them and then basically if I take one and divide by this fraction all I need to do is flip the fraction right so essentially this ends up being 16 over 13 so 16 over 13 so let's punch up in our calculator tangent inverse of 16 over 13 and see what we get so we get about 50 point I'll go ahead and say 91° so if we go back so this angle Theta is about 50 and I'll say 91° again if you're looking for this angle here okay remember this guy right here if you consider this is a straight line right so this whole thing is 18 ° I know we're flipped upside down so it's a little bit hard to see but this angle right here if you think about it would just be 180° minus 5091 de so it would be about 129.0 n° if you need that angle all right so the last concept we're going to talk about is how to find the shortest distance from some line that you're given to a point that is not on the line okay so what I'm going to do here is give you a simple formula that you can plug into instead of going through and deriving this specific formula I'll show you how you can go through a similar thought process with our actual example okay I think it's a little bit easier to comprehend so let's say that we start with some point x sub1 comma y sub one okay and it's not on this line that we're given which has the format of ax plus b y + C equals z okay if you want the shortest distance from this point to this line then you can plug in to our little formula here where we say d the distance is equal to you have absolute value bars on the top you're going to need them to make sure that you get a result that's going to be positive you can notice that your a your B and your C here are coming from your equation of the line right here's your a it's the coefficient for X here's your B it's the coefficient for y and here's your C it's your constant okay then when you see that down here it's the same thing okay now when we talk about the x sub one and the Y sub one this is coming from your point that you're given okay so let's just try this out really quickly on an example and then I'll show you where this comes from so let's say we're given the point 5 comma 3 and we're given the line y = 3x + 4 okay the first thing is you want to rewrite this you have to have this in the form of ax plus b y and then plus C is equal to Z okay people often make the mistake of writing this as ax + b yal c okay that's what you're used to for standard form but we need to put it into this form in order for that formula to work okay so what I'm going to do is I'm going to just subtract y away from each side of the equation and so that would give me let me line this up 3x and I would have minus y and then I would have plus 4 and this equals zero okay so if you want to line things up here you would say a is three so a is equal to 3 then B is equal to what you have a negative y I could put a negative one there to make that clear here that the coefficient is -1 and then C your constant term is going to be four okay then when we talk about the X sub1 it's the x coordinate from the point so that's five and then your y sub one is the y-coordinate from the point so that's three okay so this is all you really need to do just figure out how you can put this into this format and just grab the values that you need so now I'm just going to plug into the formula so my distance is equal to remember you want the absolute value bars and then let me put this over the bottom part's easy so square root of a 2 so a 2 + b^ 2 up here remember you would take your a times your x sub 1 plus your B times your y sub one and then plus your C okay so that's a pretty easy thing to remember after you do it a few times let me go ahead and see if I can move this up and let's go ahead and paste that right there let's erase that make this border a little bit better okay so if we plug in here we'll have D is equal to the absolute value of we have a which is three let me write this out so 3 * your x sub 1 which is 5 and then plus your B which is -1 times you're going to have your y sub 1 which is three and then plus your C which is four okay close that down and this is over the square root of You' have a 2qu so 3^ 2 plus b squared so -1 wrap that in parenthesis squared okay scroll down and get a little bit of room going we can work on this so D is equal to 3 * 5 is 15 so let me write this 15 -1 * 3 is -3 so minus 3 and then we'd have plus4 okay so if we do 15 minus 3 that's 12 and then 12 + 4 is going to give me 16 okay so you have the absolute value bars you can wrap it in that but this is going to be positive so you really don't need them in this case this is basically just going to give you positive 16 up here okay then in the denominator you have 3 squ which is 9 and you have - 1^2 which is 1 so 9 + 1 is 10 so this is the square root of 10 okay now we know that we don't leave radicals in the denominator so let me fix this by multiplying the numerator and the denominator by the square root of 10 right so what we'd have here is that You' have 16 * the < TK of 10 over 10 * of 10 is 10 now you can simplify this further because between the numerator and denominator you have a common factor of two so this divided by two is 8 this divided by two is 5 so the final answer is going to be 8 * the < TK 10 over 5 okay so now let me give you a little bit of insight into where this formula comes from I'm not going to use the generic notation because every time I teach that I find that students get lost so using an actual example is a little bit easier to comprehend so the first thing is I want to think about if I want the shortest distance between this line and this point which is not on the line what's going to happen is I'm going to need to find Let Me Go to a little image here I'm going to need to find the perpendicular line segment that's going to join this point here and this line here okay so what you could draw is a little line like this okay and I'll show you how to get this line in a moment but basically you're looking for the distance from this point to this point right here okay if you just took this part of the line okay that's the line segment that connects this point right here which is the point of intersection okay and this point right here which is 5 comma 3 that is the shortest distance from this point to this line okay you can see we have a little right angle that's formed because these guys are going to be perpendicular so let's go back having viewed this let's think about a few things first and foremost perpendicular lines have slopes whose product is Nega 1 so if the slope here is three so if my first slope m is three let's call this m sub1 then M sub 2 would have to be what we can go through and set up an equation for that but you really don't need it three time what would give me -1 3 * 1/3 right it's always the negative reciprocal of the number right so flip the number you get 1/3 take the negative of it you get Negative 1/3 okay so now what I can do is I can say okay well I know my slope let me just get rid of this I'll just say I know my slope over here my m is - 13 but how do I get the equation of the line remember I know this point right here is on the line so if I have a point and I have the slope then I can use point slope formula okay so I can say y - y sub 1 is equal to M * the quantity x - x sub one I know it's been a while since we used that but basically I would plug in for x sub one and Y sub one so my x sub one is five let's plug that in my y sub one is going to be three plug that in and my M my slope is 1/3 so negative 1/3 okay so we can get rid of this and basically from here I can distribute this so let's say y - 3 is equal to- 13x and then plus 5/3 you could solve for y by adding 3 to both sides of the equation let's call this 9/3 so we can have a common denominator let's get rid of this from over here so I'll put Y is equal to 1/3x plus this will be 14/3 okay so this will be 14/3 so let's come over here and say Y is equal to- 1/3x and then + 14/3 okay so now that we have that done if we go back to the two graphs you can see that the slope of this line is negative 1/3 right if I started let me erase all this I've kind of marked it up if I started let's say right here okay which is a point on the line if I drop one and I go to the right one two 3 I'm back on the line okay so you can see that's where the slope is the Y intercept is unfortunately not in an integer value but you could check other points on the line to make sure you got it correct now how can we find this point right here the point of intersection remember this is nothing more than just solving a system of linear equations right we're trying to find the point that basically satisfies both so if we go back and we think about this we have each one solved for y so what I could do is I could actually set this guy right here equal to this guy right here why I'm doing nothing more than just the substitution method if Y is equal to this well then I can just plug this in for y right there right that's all I'm really doing so I could say 3x + 4 is equal to - 13 * x + 14/3 okay and then I just want to solve for x figure out what that's going to be so let's multiply everything by three get rid of the denominators this is 9x + 12 is equal to we'll put X and then + 14 okay so now this is going to make a little bit easier let's go ahe and add X to both sides of the equation let's subtract 12 from both sides of the equation and so this is going to cancel and this is going to cancel so you have 10x is equal to we'll say two divide both sides by 10 you'll get X is equal to 1 15 okay so the x value that we're looking for is a fifth so let me put that in here so the x value X is equal to5 so the Y value is equal to what remember if the x value is 1 it's 1/ here and it's also 1/5 here so I can plug it into either original equation and solve for the corresponding y value again this is going to be the point that solves both okay so we'll go ahead and say Y is equal to I'll say 3 * 1/ and then plus 4 okay so 3 * a 5th is going to be 3 fths so let's say this is 3 fths I need a common denominator so let's write this as 205s right 20 divid 5 is 4 so basically this is going to be 235s so Y is 23 fths okay so not the nicest point that you could come up with but still you can work with it so the point of intersection for these two lines is going to be 1/5 comma 235s and if you go back that looks about right so we can say this guy right here would be 1/5 okay that's the x coordinate comma 235s that is your y coordinate okay so now I just need to find the distance between this point and this point here okay that's all I want to do and that's just the regular distance formula so the distance is equal to remember the square root of you would take the difference in X values so I can go ahead and say this is let's say my second point right here so I'll say this is my x sub 2 comma y sub 2 and remember we're working with this point right here because this is the point we're trying to find the distance to so we'll say this is x sub one we'll say this is y sub one okay so you want your x sub 2 minus your x sub 1 so 1/5 minus your 5 this is going to be squared plus you're going to take your 23 fths so you're 23 fths and this is going to be minus 3 and then this is going to be squared okay so I know a lot of you don't like working with fractions but it's really not that bad so the distance is equal to the square root of so I need a common denominator here let's write this as 25 fs and let's go ahead and write this as 15 15 over 5 okay so 1 - 25 is going to be -4 so this is -4 fths this is going to be squared then plus 23 - 15 is going to give me 8 so this would be 8 fths and this is going to be squared okay so the distance is equal to the square root of if you square five obviously you get 25 if you square ne4 you're going to get positive 576 so then plus this would obviously be 64 over 25 if you want to add these two together because you do have a common denominator you would have 576 + 64 which is 640 so the distance is equal to the < TK of 640 over 25 so just simplifying this the distance is equal to the square root of 640 divid 5 is 128 and then 25 divid 5 is 5 okay so this is a little bit more manageable something you could probably do without your calculator so you know 128 is 64 * 2 and 64 is a perfect square right it's 8 * 8 so I could say that the distance is equal to I'll go ahead and say first let me break this up and say this is the square < TK of 128 over theare < TK of 5 okay so we have this established then let me write up here let me make a little border I'll actually do this in a different color I'll say that the distance is equal to this is going to be 8 * aunk of two and then over you have the square root of five and then I'm going to rationalize the denominator okay just like we did before and basically I'm going to end up with a distance that is 8 * the < TK of 10 over 5 okay so the same answer we got before obviously it took way longer but hopefully this gives you a little bit of insight into where that formula came from all right let's take a look at one more example we'll do this the short way using the formula obviously you can pause the video and try this the long way that's up to you so we have 2x - y = 1 so I want to write this as 2x - y + 1 = 0 so very important to do this this is your a this is your B it's going to be negative 1 again and then this is your C okay then your point here 4 comma 1 this is your x sub one this is your y sub one okay so the distance is equal to go ahead and plug in so you have the absolute value of remember it's a which is 2 * x sub 1 which is 4 then plus you have B which is Nega - 1 then times your y sub one which is one then plus you're going to have your C which is one okay that's your numerator then over the denominator which is the square root of you have your a which is two being squared so let's just go ahead and write that as four then plus you have your B which is NE - 1 being squared so write that as one okay so let's come down here and I'll go ahead and say the distance is equal to 2 * 4 is 8 then you would basically have minus1 + one so let's just go ahead and say this is the absolute value of 8 which is 8 and then over the square OT of five you can rationalize the denominator there and we'll go ahead and say this is going to be 8 * theare of 5 over five as our answer