what's up my pre-cal people I'm Michael PRACK and in this video we're going to do a quick recap of unit 3 for AP precalculus now unit 3 covers trigonomic in polar functions and it's a Well a really big topic so what we're going to do is not cover all the teeny teeny tiny details that's what you should be doing in class what we're going to do in this video is cover the Big Ideas the big skills the big Concepts that are going to enable you to prepare and pass the AP pre Calculus exam in May so let's dive into those big skills and Big Ideas right now so trick all starts out with a setup like this now all angles start in standard position where the initial side is the positive x-axis and the terminal side stops at a point p on a circle of radius R so we have an angle that gets created here and this angle right here has an initial side that's always the positive x-axis and it has a terminal side right here and that terminal side stops at a point technically that terminal side goes on forever but if we stop at a certain radius for a circle we get a point and on the rectangular coordinate plane that is a point x comma y now it's important to know that angles could be positive or negative positive is an angle that goes in the counterclockwise rotation and negative angle is any angle that goes in a negative or a clockwise rotation now keep in mind that angles can go around and around and around and around forever and ever and ever as many times as they possibly want so there's no limit to what an angle can be positive or negative so the domain for Theta or the the values that Theta is allowed to be are all values negative Infinity to Infinity now the radian measure of an angle in Center position is the ratio of the length of the Arc of a circle centered at the origin subtended by the angle to the radius of that same Circle okay I know that can be a little bit confusing so let me try to explain that so when we have this setup when we have a point that is at the terminal side of an angle starting in that Center position what we do is we create an arc and this length of this Arc is called s we'll just call it the length right imagine if we to lay a piece of string out on that Arc and then straighten it out and measure it that's the length of the arc well obviously a bigger angle is going to produce a bigger Arc a smaller angle is going to produce a smaller Arc but then we also incorporate the radius because how big the circle is depends on how big the arc is going to be as well so a radian measurement of that angle Theta is the ratio of the arc length s to the radius R so when the arc length and the radius are both the same whether it's 55 6 6 77 1/2 1/2 we get a radian of one we get one radian when the ark length equals the radius but otherwise if the Arc Length doesn't equal the radius whatever that ratio is between the arc length and radius is our radian measurement and hopefully you're fairly familiar with working with radians now let's talk about our trig function s cosine and tangent all based on this same idea so when we take an angle in standard position and we have that angle stop right here then what we do is we look at two things we look at the vertical displacement the vertical displacement is this value if I were to come straight down basically it's the y coordinate of our point and then we look at the radius of that AR or of that Circle as well and that's where we get sign sine of theta is the ratio of that vertical displacement the y coordinate divided by the radius that's it that's what sign is pretty simple now when we're on the unit circle our radius is automatically one so when we're on the unit circle we just get y because y / 1 is y so on the unit circle Theta is the angle it goes into sign and sign kicks back out the y coordinate from where that angle took us hopefully that makes some sense now let's talk about cosine here let's quickly review cosine cosine once again we have that angle Theta there Co sign is the horizontal displacement so how far over that point is from the Y AIS that horizontal displacement now that's the x value right the x coordinate of our point is that horizontal displacement and then of course we also look at the radius of that Circle so therefore cine of theta is that x coordinate from where our angle stopped divided by the radius and of course when we're on a unit circle with the radius of one then cosine is just X the x coordinate from that unit circle where that angle tolds to stop so again for S and cosign we have an angle angle goes in and that angle gets created and that angle leads us to a terminal side that intersects a circle of radius r at a point x comma Y and the cosine of that angle is the relationship between the x of that point and the radius of that Circle sign is relationship between the y-coordinate and the radius of that Circle but again we love being on a unit circle because our radius is one all right now here's the final trig function tangent tangent is the slope of that terminal Ray doesn't even matter what the radius is we can allow that that terminal Ray to go on forever the slope of that terminal Ray is tangent so you can also say well wait a minute wait a minute wait a minute how do we find slope well slope is rise over run so if we start at 0 0 which is of course where that Center Point always is the rise would be the y coordinate the Run would be the x coordinate therefore tangent of theta is y/x but again that is the slope of that terminal Ray so that's how we look at tangent it's the ratio between the y coordinate and the xord doesn't even matter what the radius is but that relationship is rise over run which is the slope of that value all right so all in all we end up with this basic scenario where we have an angle and all you got to do is think of that angle that's allowed to go around and around in positive or negative directions producing positive negative angles that angle's always going to stop at a point on a circle of radius R and that point in the rectangular plane is of course X comma y therefore s is the ratio of Y over R cosine is the ratio of X over R and tangent of theta is the ratio of Y overx if you could remember that you could remember a lot of the ideas that you need to know for trigonometry now obviously if we stop in a different quadrant like down here our x's and y's are both negative that's okay well we just got to incorporate the negative with the x's and the Y's and on tangent when your negatives you know Y and X are both negative then you're going to cause a positive now I'm not going to review the entire unit circle but I hope in class you've already understood the unit circle so here we have the very nice angles based on our 30690 triangle which in radians is a pi over 6 piun over 3 pi over 2 triangle and we should know these ratios very nicely when we're on the unit circle now I got lots of other videos to explain this in more detail but right now I'm just trying to review it with you so right here we see pi over 6 and at pi over 6 we see the point on the unit circle X is radical 3 over2 and Y is2 so if we were to say what is sine of pi/ 6 all we got to do is say well that's going to be the Y value2 divided by the radius which on the unit circle is 1 so 1/2 divided 1 is 1/2 so s of pi over 6 is 1/2 and you should just kind of remember these you just kind of have to know it's really important that you take the time to study and prepare you're not going to get a unit circle on the AP exam but you should know how to quickly create one on your own I like to count by six we got one six 2 six 3 six 4 six obviously it reduces 5 six 66 76 86 96 106 116 126 takes us right back to two Pi which is one full cycle one full rotation around the circle then we can keep going as many times as we want we could also count by fours we got 1/4th two fours three four four fors 54s 64s 74s and again 84s takes us all the way back to the full cycle or 2 pi but again once you know those basic ideas and you kind of have some of these values memorized you should be able to quickly find for example what is cine of 7 pi/ 6 all we got to do is find 7 pi/ 6 and cosine is the x coordinate which is going to be Ral 3/ 2 uh what about tangent of 7 piun 4 locate 7K 4 tangent is the Y ID the X that's going to be negative radical 2 over 2 ID radical 2 over 2 well that's going to be just 1 so again you should know those values for the most part I'm not going to walk you through I do all of them but you got us kind of have to memorize this unit circle that way you could quickly understand it when you're ask these questions on the exam all right let's start talking about the graphs of s and cosine here is the graph of s now if we think about it this graph should make complete sense so at zero radians we start out at a y-coordinate of zero go back and look at the unit circle to understand that then as our angle starts to get bigger the Y value starts to increase and increase until at pi over 2 which is 1/4 of a circle we reach the maximum y-coordinate of one then as we continue to rotate our YCO coordinate starts to drop and get closer to zero until at half of a cycle or half of a circle Pi radians we reach a s value or a y-coordinate of zero then our angle continues to get bigger but now our y-coordinate becomes negative and more and more and more negative until at 34s of a circle it reaches its absolute lowest point of negative one and then it continues to get bigger and the angle now is still negative the angle is not negative excuse me the y coordinate is negative but it's rising and getting closer and closer to back to one full complete rotation to Pi and that is of course going to take us back to a value of zero for our y-coordinate so this is what we call one full period of sign starts at 0 0 goes up to its Max down to its Min back to zero and that is one full cycle so one full cycle we call a period and the period length is 2 pi for our typical S curve now we also have something called the amplitude the amplitude is how much you go up and how much you go down now the amplitude is a distance so whether we're going up one or down one it's still a distance so the amplitude is one and then we also have something called the midline the midline is the halfway point you know it's it's right in the middle so the midline is going to be y equals 0 and then the amplitude is how much you go up and down from that midline so the midline is y equals 0 we go up one or we go down one but again the amplitude is a distance so it's positive one no matter what all right let's take a look at the graph for cosine now cosine is a little bit different it actually starts at zero radians it starts at its Max at zero radians the x coordinate is one then as our angle starts to get bigger that x coordinate immediately starts to drop and drop and drop and drop and drop until at Pi / 2 1/4 of the circle our x coordinate is zero and then as our angle continues to get bigger our x coordinates continue to get smaller but now they're actually getting more and more and more negative until reach the absolute lowest x coordinate which is -1 at 1 half circle or Pi radians then we start to increase our angle more but now our x coordinates start to rise they're still negative but they're getting less and less negative so they're getting closer to zero that is why at 34s of a circle 3 pi over 2 radians we are at zero we're back to zero and then we continue to get larger and larger and larger until we come full circle back to 2 pi back to that x coordinate of one so that's one full cycle of cosine starting at one going down to its Min and then going back up to its Max one full cycle once again is the period and the period length is still two pi to go all the way around the amplitude is how much we go up and down from that midline so the midline is once again y equals 0 and the amplitude is one it's that distance that we go up from the midline or down from the midline so not to too bad there when we're understanding the graphs of s and cosine now what we really want to focus on right now is understand the different transformations of those two graphs because this is where it could get a little bit tricky and there's bound to be some multiple choice questions and also a free response question dealing with transformations of s and cosine so we have a couple values here we have a that's out in front we have B that's inside we have C that's inside then we have D all four of these values play a different role in the different Transformations let's talk about each one of them right now so the a value out in front is going to affect your amplitude now the absolute value of a is your amplitude remember amplitude is a distance so it cannot be negative it's just the absolute value of a now this is a vertical dilation by a factor of a so if your a value is two you're going to stretch vertically by two or if your a value is a half you're going to shrink vertically by a half now if a is negative then that causes a reflection across the x-axis so if we think about our sign curve that normally looks like this if we were to reflect it across the x-axis it's going to go down first then up so it's just that reflection cosine here is our traditional cosine curve right there and then if it's a reflection it's going to start down at its Min go up to its Max back to its Min because if there's an a value that's negative so again the a value affects your amplitude it's going to stretch you up or Shrink you in by a factor of a pretty straightforward there all right let's jump to the D value out in the back this is going to be a vertical translation by D meaning that this is going to vertically slide you up or slide you down a positive D is going to go up a negative D is going to go down now this will affect your midline as well because remember your midline traditionally starts at y equals 0 but if you move the whole graph up three for example your midline is now y equals 3 if you move your whole graph down five then your new midline is going to be5 so yals D is your midline but transform you know think about Transformations this is a vertical translation by D all right next up we have the B value this is the one that causes the most confusion with kids so let's spend a little bit time on this one this is going to be a horizontal dilation by a factor of 1 over B okay so it's kind of weird when you have that a value in front that's a vertical dilation by a factor of a but on the inside we have a horizontal dilation it's not by B it's by a factor of 1 over B so now graphically speaking this is going to change our period length it's either going to stretch it out make it bigger or Shrink it and make it smaller now the most important thing you need to know here is the formula to find the period length you take 2 pi that's the original period length of a traditional s or cosine graph and you divide it by the absolute value of B this is going to tell you what your new period is now this is why when you have a b value of2 that's actually going to stretch you and make you have a bigger period because when you divide by a fraction you multiply by the reciprocal it's going to make you bigger and again that all goes back to this idea that it's actually 1 over B so if your B value is 1/2 you're actually multiplying by a factor of two one divided half is two hopefully you understand that all right so a lot going on there but just understand that b value directly affects your period length 2 pi ided by the B value is how you find your period length now additionally if that b value is negative again remember the period length is divided by the absolute value of B so it's why it's always going to be positive but if your B value has a negative attached to it that's going to cause a reflection across the y AIS keep that in mind as well and the final transformation here is C now C is going to cause a horizontal translation this is going to slide you left or slide you right now generically rewrite the formula with a plus c but it's opposite so this is going to be a horizontal translation by negative C so basically this means if you have X plus C that's going to actually shift you left C because your C value therefore is negative C so that means you're going left C if you have an x minus C inside the parentheses well that's minus a negative minus negative is a positive and that's going to shift you to the right C so those are kind of Transformations that are typical and we talked about them in unit one with just regular functions so it kind of keeps in line with that same idea that when you have a plus c on the inside you're going left C if you have a minus C on the inside you're going right C so these are going to be translations that move you left or right right now let's actually look at how all of these different Transformations can actually play out in a function all right so what do we see here in this particular function well the first thing I see is I have an a value of two I have a d value of -2 and I have a b value of 2 pi over 3 it's whatever you're multiplying the X by in this case it's a lot 2 pi ID 3 so first my amplitude is going to be two that two out in front is going to affect my amplitude it's going to be a vertical dilation stretching me um by a factor of two now the D value in the back is going to affect my midline that's going to now move the entire graph down two so I have a vertical translation down two and my new midline is yal -2 now what about that period length well my new period is three because if I take my 2 pi that's the formula for finding the period Take 2 pi and divided by the B value which is 2 pi over 3 so instead of dividing by a fraction I multiply by the reciprocal 3 over 2 pi the 2 Pi's reduce and we just get a period there of three pretty simple so again the period is three so it stretches it out and the horizontal dilation is by a factor of 1 over B which is 3 over 2 pi so a lot going on there but hopefully we understand all of that nice and slowly and there's no horizontal translation there's no C value inside the parentheses that move me left or right all right let's take a look at another one here and one I actually do have an a value a b value a c value and a d value let's see how all these things play out so first we have an amplitude of two that's going to be the a value on front that's a vertical dilation by a factor of two then we have a midline of yal ne3 the minus 3 on the outside move the whole graph down three that's a vertical translation down three then we have a period of Pi my B value is two so if I take 2 pi divided by the B value of two the two is reduced and I just get Pi this is a horizontal dilation by a factor of a half now that actually should make a lot of sense remember the original period length of s is 2 pi and if I'm literally understanding that the horizontal dilation is by a factor of one over B which is2 that's going to cut into my period of a half and a half of 2 pi is one Pi so it should makes complete sense as to why my period is pi now we also have a horizontal translation here left Pi or two because I have a plus pi/ 2 for my C value and that's going to cause me to have that horizontal translation left pi/ 2 so a lot going on there but please make sure that you understand how to identify these characteristics based on looking at the function it's really going to come in handy on the multiple choice especially all right let's work backwards let's actually look at some graphs and try to identify the function that created these graphs all right so here is a function that's clearly a periodic fun function s or cosine now the cool thing is that every periodic function s or cosine could be written as s or cosine it all depends where you look so what I tell students to do is this find one full period whether it's s or cosine it doesn't even matter find one full period so the first thing that jumps out to me is right here I see one full period right here now this is going to be a sign curve right because it starts at the midline so wait a minute it starts at the midline yeah locate the midline the midline is going to be halfway between the Min and the max so the Min is down here at one the max is up here at 15 1 + 15 is 16 16 divid by 2 is 8 so we see we have a midline right there at 8 that's going to be my DV value so my D value is going to be eight because my midline is at y equal 8 now the amplitude is how far you go up and down from that midline so from eight I go down seven to get to one or I go up Seven to get to 15 so there's my amplitude or my a value okay now what else well the period length how long did that whole process take from beginning to end well it looks like it took eight units started at zero whole thing took eight units okay so this is where I'm going have to do a little bit of math because the formula to find my B value is 2 Pi / B but based on looking at the graph I know my period is eight so now I could solve for what the B value has to be to create that period so first thing I'm going do is multiply the B over so I get 2 pi = 8 B divide both sides by eight and I get a B value of pi over 4 is my B value that produces the eight now I call these three things the non-negotiables okay and what I mean by that is I cannot change the midline I cannot change the amplitude I cannot change my B value which affected my period now what happens here is I chose a S curve goes down to its Min up to its Max back to the midline that's a S curve but notice that this particular sign curve went down first that means my a value is going to need to have a negative on it because it got reflected across that midline so my final equation here is going to be this it's going to be7 S then I have my B value Pi over4 * Theta there's no C value because I started at the Y AIS I didn't move left or right at all I didn't shift it left or right so I don't need a c value and then I need Plus 8 for my midline so a little sloppy there I apologize but hopefully we understand how I built this but here's the kicker I could do a different function you say what what are you talking about you could do a different function uh yeah hold on a second watch this I can also say well wait a minute I want a cosine curve cosine curve starts at its Max goes all the way down to its Min and then all the way back to its Max so there's a cosine curve okay awesome let's talk about this cosine curve the midline guess what it's still eight I can't change the distance between the Min and the max so that's still going to be a midline of eight and I can't change the amplitude the amplitude is still up seven and down seven from that midline so that's still going to be seven and I can't change my period length because the period here started at -2 and it ended at six that's still an overall period of eight and then again going back to my work here solve for the B value that produces that eight I still get P four that's why I call those the non-negotiables because whether you look at this as a sign or cosine curve you cannot change the midline you cannot change the amplitude you cannot change your B value which stems from your period but now I actually want a cosine curve here because I started at my Max went all the way down to my Min back to my Max okay so now I need to create my cosine function here so let's go ahead and do that all right so I'm going to need cosine now it's not starting down it's starting up at it's Max that's what cosine does so I don't need a negative on my a value so I'm going to have seven for my a and then I have cosine and then I have my B value Pi over4 but wait a minute I have a c value traditionally cosine starts at its Max on the Y AIS but I started two units to the left so I need theta plus C but my C value is going to be two because I started it left two so I need a plus two which made me go left two and then on the outside of course I need that plus eight for my bid line so I know there's a lot going on there but hopefully you know you might have to slow down if you need to but hopefully you understand the process for how I came up with that function and again any of these functions can be written as s or cosine so if you're had a multiple choice be very careful all right let's try another one here so once again identify a s or cosine graph so maybe we say oh right here here's a a a sign graph starts at the midline goes up to its Max down to its Min back to the midline all right let's find the function that matches with that well first we have to identify what the midline is well we're going to look right halfway in between so here we have a netive one for the max A5 for the Min right in between between that is of course -3 add them together you get -6 / 2 you get -3 so my midline is at y = -3 that's going to make my D value -3 then I say okay what about the amplitude from that midline I go up to down two from neg3 I go up to the negative 1 down to the neg five so that means my a value is going to definitely be either positive or negative2 the negative just depends if there's a reflection which I went up so there's no reflection based on how I drew My Graph so that's going to be the amplitude 2 and then now I need my period length well I started at Pi / 2 and I ended at 3 pi/ 2 so how far is it from 1 12 pi to three halfes Pi well that's a total distance of Pi so my period equals Pi so now I have to solve for my B value 2 Pi / B equals Pi that's my period length now I'm to solve for the B value that creates it multiply the B value over and then I'm going to divide by pi the pi is reduced and I get a B value of two again I know it's a little bit sloppy I'm sorry I'm trying to rush to make this not too too too long of a video but that's what my B value is that produces a period of two hopefully that produces a period of Pi hopefully that makes sense all right so now the final thing I did is do I need a c value well in this situation yes because I'm not starting on the Y AIS I chose to start over Pi / 2 I sorted to the right Pi / 2 so when I go to write this sign function I need a two s and then I need my B value which is two and then in parentheses I need X+ pi/ 2 the plus pi over 2 is going to cause me oh actually I'm wrong on that I'm so sorry I need a minus pi/ 2 because I want to start right pi over 2 I went to the right Pi / 2 and on the outside I need my amplitude which is down three so minus three so again a lot going on there hopefully you learned all about this in class already I'm just doing a quick review of it so you're like oh yeah I already get all this and that's kind of the goals for me to just quickly go over it so you can understand but again don't hesitate I could have said oh I want to use this cosine graph starting at the max going to its Min back to its Max then you could rearrange everything but won't change is the midline the amplitude and your period length those are the non-negotiables they can't change but now you're going to use cosine and you're going to need it to go left pi over4 to cause that horizontal translation let's switch gears and talk about inverse trigonomic functions now so here's the idea we all know what s of X is Right X is an angle it's an angle there's infinite angles that could go into X and then the output for sign is the y coordinate from the unit circle okay we all know that we all love that so if we pull the old switcher which is going to cause an inverse switching our inputs and outputs we get the inverse sign which is sign with a little negative one that's a symbol for S inverse now the input is no longer an angle the input is a y coordinate from the unit circle the output is an angle okay makes sense so let's practice a little bit here so if we have regular sign and we say hey what is s of Pi / 6 so we locate Pi / 6 s is the y-coordinate 1/2 there is our one and only one answer because to be a function your input can only produce one output but if we use a s inverse we run into a problem if we now take that y coordinate of 1/2 and we plug it into sin inverse we say okay well gosh there's infinite angles that have a y-coordinate of 1/2 first off just looking at the picture of the unit circle from 0 to 2 pi we got pi over 6 we got 5 pi over 6 they both have Y coordinates of 1/2 and if we add 2 pi as many times as we want or subtract two pi as many times we want we get more and more solutions but there're it's no longer a function if one output leads to multiple outputs or excuse me one input leads to multiple outputs so to prevent this from not being a function and if we're going to call an inverse function it kind of has to be a function what we do is we limit the values that we could get out we limit our angles so we're only allowed to give angles going from 0o up to Pi / 2 or 0 down to negative pi/ 2 so we cause a limit or not a limit excuse me we limit our outputs from negative Pi / 2 to positive piun / 2 therefore we're only going to get one answer the only answer we're going to get here is pi over 6 because in that interval from pi over 2 to negative pi over 2 the only time we see a y coordinate of 1/2 is at pi over six so that's what we have to do for S inverse to make sure that we are creating a function let's look at two more examples here so the inverse sign of radical 3 over2 so again we're looking only from negative p/ 2 to pi/ 2 and we say where do we see a y-coordinate of radical 3 over2 right there at pi over 3 that's the output input was the y coordinate now the output's the angle simple what about Sin inverse of - one2 well right here we see a y-coordinate of - one2 and some people would say oh the answer is 11 pi over 6 and the answer is not 11 pi over 6 don't even think about putting that because 11 pi over 6 says let's go all the way past Pi / 2 past pi to 11 6 but we're not allowed to do that remember the restriction on inverse sign in order to allow us to be a function is from negative pi/ 2 to positive pi/ 2 so the answer we have to give is the answer that gets to this spot but gets there by going down not around so the answer is NE pi/ 6 it's co-terminal to 11 pi over 6 but negative pi over 6 is in that restricted area of negative power 2 to positive power two a lot of kids get that wrong so be very careful with that that all right let's now switch gears and talk about our buddy cosine here so traditionally the input for cosine is an angle infinite inputs could go in there and the output is that x coordinate from the unit circle and if we plug in an angle for example 2 pi over 3 the only x coordinate that we could possibly get is negative - one2 nice and simple but if we switch those inputs and outputs and now the input is the x coordinate and the output is the angle well once again we're not allowed to have multiple outputs so what we do is we restrict it but this time for cosine we restricted from zero to PI from 0er to Pi because from zero to Pi all of our X values are not repeated so we have positive X values and we have Negative X values nothing's repeated if we go down in quadrant three and four we repeat some of those X values and that's again we're not allowed to have those multiple outputs so let's look at two examples real quick here so we can see how this works inverse cosine of radical 3/2 so we're seeing where is an x coordinate R 32 we see an X radical 3 over2 right there at Pi / 6 and you say well wait a minute I see it right here at 11 Pi 6 up but wait a minute I'm only allowed to look from zero to Pi when we're working with inverse cosine what about cosine inverse of negative one2 where do I see an x coordinate of netive one2 right here at 2 pi over 3 nice and simple not overly complicated all right and let's finally do tangent here so again remember tangent the input is an angle and the output is the r Ratio or the slope of that terminal side the ratio of the Y to the X all right so now we're just going to reverse that now the input is that ratio Y overx and the output is an angle but once again we don't want any multiple outputs for every input we're only allowed to have one output so tangent repeats itself diagonal so any tangent value in quadrant one is the same in quadrant 3 any tangent value in quadrant 2 is the same in quadrant four so what we do is restricted very similar to S we restricted from piun / 2 to positive Pi / 2 now the only difference is we don't include p/ 2 because at p/ 2 and at negative p/ 2 tangent is actually undefined so we wouldn't want to include it anyway so it kind of works a lot like inverse sign so here's a quick example where do we see a tangent value of radical 3 over2 well that's going to happen right here at pi over 3 if we take the Y and divide by the X we get pi over or we get radical 3 which happens at pi over 3 and then where do we see a tangent value of - one2 where do we see that ratio between the Y and the X be negative 1 and that's going to happen right here that's where we see when we take the Y divided by the X we get a tangent value of negative 1 but don't say 7 pi over 4 because you will be marked wrong because 7 pi over 4 goes all the way around to 7 Pi 4 we can't go over there remember we are restricted to only looking at negative p over 2 to positive pi/ 2 therefore we have to go down to get there so our answer is going to be the co-terminal value of negative pi over 4 but that is in our restricted value or in our restriction area that we're allowed to go to so hopefully that's not too difficult but just be careful with inverses as well they do come up on the AP exam from time to time we want to make sure we know how to handle them all right now the next thing we want to look at and we're just kind of kind to briefly go over this in this video again I can't dive too much into every little detail here but it's solving tricky equations now when we're solving a trig equation the most important thing we understand is we want to isolate the trig function once the trig function has been isolated we consider how to solve it so let's talk about this in two ways though real quickly so on the left hand side here we see a trig function it's actually already isolated for us s of xals one2 now there are infinite answers for this x value there are infinite answers that are going to produce a sine value or a y-coordinate from the unit circle of negative one2 so where do we see this happen this is going to happen at 7 piun over 6 at 7 piun over 6 we have a y coordinate of2 but it's also going to happen at 11 piun over 6 but there are millions and millions of more answers because anytime we add or subtract a full circle to these values we're going to get another iteration of that same co-terminal angle that produces that same y-coordinate of NE one2 but if we were to pull the old switch R using an inverse now the moment you are having an inverse as we just got done talking about then you only got one one answer sin inverse of2 the only one answer that could ever be to this problem as we just talked about is going to be neg piun over 6 at negative pi over 6 we have our y coordinate netive one2 and because we're using inverse sign we can only have that one answer so when you have that inverse you see it you can only have one answer when you don't you could have multiple multiple answers infinite answers in fact when you have to think about that so let's dive into that just a little bit more deeper here so this is where you really want to read the direction if the directions say to only give solutions from 0 to 2 pi well then you just got to consider one full circle 0 to 2 pi and just as we just mentioned on the unit circle this is going to happen at 7 pi over 6 that's where the Y Co will be negative one2 and it's also going to happen at 11 pi over 6 those are the two answers from 0 to 2 pi that produce that y-coordinate but if we want all solutions not just a couple like all of them we have to make a slight adjustment here now we know that anytime we add 2 pi we're going to get another answer and we're going to add it K times and that K just represents an integer I can add 2 pi one time two times three times four times five times six times as many times as I want I'm going to get more and more answers but since K is an integer could also be negative which means I can subtract 2 pi subtract 3 Pi or excuse me subtract four Pi subtract six pi and get more and more answers so to represent all answers there's infinite of them I can't possibly write them all we write 7 pi over 6 that's the first answer we see and then we add 2 pi K where that K is an inter value that allows us to add 2 pi or subtract 2 pi as many times as we want and then I have to do the same thing with the 11 pi/ 6 I have to add 2 pi K and again that K is a positive or negative integer that allows me to go around the unit circle as many times I want producing more and more answers so really read the directions do we want all solutions or we just want solutions from 0 to 2 pi all right now the other thing we got to be careful of when we're solving is what happens when there's a value inside the parentheses like this well okay we just kind of think of that as our angle I know right now it's 3x but just think of it as our angle so that angle which right now is called 3x needs to equal 7 pi over 6 once again if that angle 7 pi over 6 s of 7 pi over 6 produces a y coordinate of negative one2 or that angle could also be 11 pi over 6 as we just got done discussing but now because the x value has a three next to it now we just have to do a little bit more solving but we can't do anything with that three until after after we figure out what the angle is now I could go ahead and multiply both sides by 1/3 and I get a final answer for X as 7 pi over 18 so if I plug in 7 pi over 18 it times it by three changes to 7 pi over 6 then s of 7 Pi 6 is2 same thing over here I'm going to multiply by 1/3 multiply by 1/3 and we get x = 11 pi over 18 for our second answer so please be very very careful with that hopefully that's not too too too difficult and here's another one involving square roots a often times a lot of kids say I don't know what to do with the square there remember the goal is to isolate the trig function always isolate that trig equation so I need to get sign I don't want side squar so how do you get rid of a square take a square root all right so when I do this the two important things I have to keep in mind first when you create a square root through the solving process you got to put a plus or minus now I got to deal with the square roots well the square root of one on top is one the square on the bottom is is two now you might not think that looks familiar but if we rationalize the denominator multiplying top and bottom by radical 2 we get plus or minus radical 2 over two which every single one of you should easily recognize from a very easy value on that unit circle so if I make a quick Glimpse that unit circle that's going to happen at pi over 4 3 pi over 4 5 Pi 4 or 7 Pi over4 because I'm allowed to be positive or negative that means I'm going to have four answers so the answers are pi over 4 3 pi over 4 at those two places is its positive radical 2 over two then I have 5 pi over4 and then I also have 7 pi over 4 and then if it asks for all solutions I would need to take those values and add 2 pi K to every one of them to represent the fact that there could be multiple multiple Solutions all right now let's dive into the trigonomic identities trigonomic identities you know is a big topic but there's a lot in it so I'm going to try to go over as much as I can here as briefly as I can but why do we need the trigonomic identities first we use them to help simplify or rewrite trigonomic functions in fact on the fourth frq you're going to be asked to use the tri Iden identities to take a trigonomic function and rewrite it in simpler forms that's why we need these identities and they're also used to help us solve Advanced trig equations because sometimes we're trying to isolate that trig function as I just got done talking about we might need the identities to make that a little bit easier so let's briefly go over the identities the first we have the reciprocal identities these are the pretty easiest ones to memorize so we have sin = 1 over cosecant coseal 1/ s cosine is 1 over secant secant is 1 over cosine tangent is 1 over cotangent and cang one over tangent now these are pretty easy to memorize as long as you remember s and cosecant go together cosine and secant go together and tangent and cotangent go together then we have the quotient identities tangent is s over cosine remember tangent is y overx but s is y and cosine is X there for tangent s over cosine coent is cosine over s and then we have the most famous One a Pythagorean identity now the main Pythagorean identity the one you need to know is cosine square plus sin sare equal 1 but note that this could be manipulated in many ways you know for example I could subtract s^ squ over and I get cine s of theta = 1us sin squ of theta that's just another version of the Pythagorean identity I can also subtract one over to the left side subtract a cosine over to the right side and I get sin s thetus 1 = negative cosine s of theta again the point is there's multiple different ways that I could morph that around and have a different Pythagorean identity all stemming from the same Pythagorean identity but the Pythagorean identity also leads to two more identities 1+ tangent squar equals secant squar and coent squ + 1 equal cosecant squar now there's not going to be a formula sheet on the AP exam that's going to have these on it you are going to be expected to remember them but they're not too too difficult just spend some time making not cards of them going over them and kind of you know late in bed at night before the exam studying them and I think you'll be okay but these identities are definitely going to come up a lot and they're definitely going to be used all right next up we have the double angle identities this is when we have a double angle inside of our sign or inside of our cosine now for S of a double angle we have one pretty easy Formula 2 s of theta cosine Theta not overly difficult to remember but for coign of a double angle we actually have three options not one of them is more correct than the other we just have three different identities that all are the same 2 cosine squ of theta minus 1 1 - 2 sin s of theta or cosine squar minus sin squar so make sure we utilize all of these identities again make no cards do something that you could kind of study with these to remember those different identities and then finally probably the hardest ones to remember but honestly from what I've seen these aren't going to come up a ton I mean I'm not going to say they're not going to be on the AP exam at all but um you never know but these are the harder ones to memorize these are the sum and difference identities so this is when you're adding two angles inside of s or cosine so here are those forms I'm not going to read them to you you could see them but you just got to memorize them and then for a difference it's just subtracting two angles inside of s or cosine so just a couple things that I kind of use to help remember these with cosine it's always cosine cosine s s cosine cosine s sign that's kind of easy to remember but the tricky part is if you're adding you subtract or if you're subtracting you add so don't you know going to got to mess that up a little bit that's easy to get confused on and then when working with sign it's goes s cosine then it alternates cosine s switching the angles but the good news is if it's plus it's plus if it's minus it's minus it kind of holds true it's supposed to so these formulas again honestly I haven't seen a ton of them in use there might be like one multiple choice question that could possibly come out of this and if you don't have it memorized maybe you get it wrong but you know put them on not card spend some time trying to memorize them all right that's a lot of information on trigonomic identities I know that I mean there's just countless examples I'm not going to do a ton of examples in this video I have other videos that you can watch that go over the use of trigonomic identities but again we use them to simplify trig functions or to help us solve some more difficult trigonomic equations now one last thing I want to leave you with before we move away from trigonomic identities if you are trying to reduce simplify manipulate a trigonomic function by using the identities I have two kind of tips first think to use a pythagorean identities when you see trig function squared so when you see a cosine squar a s squar a tangent squar secant squar Cent squar or cosecant squared look to use these identities because every one of these identities uses those values being squared now especially if there's a number one nearby like maybe you see a one minus sin squar you could easily use Pythagorean theorem to convert that to cosine squar or maybe you see a see squ minus one that can instantly be changed to a tangent squar so it stinks that you kind of have to memorize these formulas I mean some teachers will say that you should have to memorize them you should be able to deduce them or figure them out based on the original one whatever you need to know them but if you see something squared and a one nearby look to use these trigonomic identities that's probably going to be where you want to go now my second big tip is this if all else fails look to convert all of your trig functions to either s of X or cosine of x for example a tangent can be turned into s of X over cosine X or a secant can be turned into one / cosine and once everything is in terms of s or cosine then you may see some things reduce or simplify and then things start to work out really really well so those are my two big tips whether you're working with trigonomic identities and trying to manipulate them and simplify them all right let's move on to the final Topic in this unit polar coordinates now first let's quickly remind you what a polar coordinate is so a polar coordinate grid is a completely new system it's not this standard XY X is left or right Y is up or down it's a whole new grid centered at 0 and and then we have radiuses so we have circles built for the radius and then the angle tells you where that radius goes so polar coordinates look like this um we have a radius in that first spot and a Theta that's going to be an angle in radians in that second spot now it's important to understand that it's kind of backwards really it's the angle that we do first that's kind of the independent variable the angle and then the radius goes based on where that angle is so lot of kids kind of get this confused especially because a lot of kids are learning this for the first time but let's take our time with this so again the r is the radius the absolute value of our is the radius and then Theta tells you the angle in standard position positive you're going to go counterclockwise negative you're going to go clockwise so let's look at a quick example here so here is the polar coordinate for 4 comma pi over 3 so you see that I went to Pi over3 here is pi over 3 if we're thinking about starting in that initial position and rotating to pi over 3 and then the radius of four tells you what Circle right what Circle that that point needs to go on so here we have one here we have two here we have three and right there is that radius of four so four is the distance from that Center 0 0 and pi over 3 is the angle where that goes now let's look at another one here this time we have a negative angle so again we're still going to start in that uh standard position we're going to go all the way to - 5 piun over 6 there that angle is and then our radius is three which means that at that angle we want to be three units away from that Center which is going to put us right there now the only thing that gets a little bit weird is when you have a negative radius so first we always start with the angle so the angle is pi over 6 pi over 6 is right here now normally you think oh but wait a minute why is our Point not on pi over six well because it's negative so the negative says here's where that point was supposed to be with a positive two but because it's negative that's going to cause us to reflect diagonally across to right here so here is that point -2 comma piun / 6 now it's important to note that every polar coordinate could be written in infinite different ways to be to be honest for example this polar coordinate right here that we're working with -2 comma Pi 6 it could be written a different way we could say all right well let's do -5 pi/ 6 that would be my angle and then if I'm using 5i 6 and I want this to be my point then I need a positive two for the Rus there would be no reflection there so those two points are technically the exact same location on that polar um coordinate grid but again there's infinite many more points I could even do 7 pi over 6 I can go all the way around this way to that angle of 7 pi over 6 and then once again I I want to stay right where I'm at in that quadrant so I'm going to use a positive two so again lots of different options for these angles and these radiuses just as long as you know how to graph them now a big part of working with polar coordinates Now is working with polar functions so we got to understand how we could take a set of these polar coordinates and make a graph so what we're working here is we're still working with a function that we've actually seen before for example 2 cosine of theta but this time we're not letting the output represent an x coordinate from the unit circle which is typically what cosine is we're letting the output F of theta represent a radius so for example if I'm going to start at zero zero right here on that um this value right here the starting point right if I'm going to start at an angle of zero and I say okay what is 2 cosine of 0 well cosine of 0 is 1 1 * 2 is 2 so I plugged in zero and I got an output of two but that two now represents a radius in polar coordinates so that would be right here there's that point right there hopefully that makes sense now then we can say well what about Pi / 2 okay so we have 2 cine of > / 2 cosine of piun / 2 is 0 0 * 2 is zero but that zero represents a radius so at pi/ two we're looking right up here everybody knows where pi over two is the radius is zero that's going to put us right here and then let's just do one more what about um Pi over3 so we have 2 cine of Pi over3 and cine of pi over 3 is 12 so 2 * 1/2 is 1 okay so at Pi over3 which is right here our radius is one so it creates this Arc that's going to kind of go like this and then if I would keep going it' come over here so hopefully that makes sense in terms of following those values um but again that's what it does it creates radiuses when you're looking at these trig functions and polar coordinates the output is a radius the input's an angle that tells you where that radius needs to go so again I know it's a really quick overview of polar coordinates but hopefully that's um a pretty good job of getting you to understand exactly what polar coordinates are now what do we like to do with polar coordinates you know what could come up on the AP exam well we could convert rectangular coordinates which is our standard XY to Polar coordinates and this conversion is actually really easy all we need to know are these two simple formulas that you should already be familiar with x^2 + y^2 = r^2 so you could use the X use the Y and then you could solve for R then tangent of theta equals y /x so what you have to do is use your y use your X and then you're going to analyze okay what angle would produce you know tangent of what angle excuse me would produce that ratio and that's going to tell you what your Theta is so you really got to think about how to utilize these two formulas to determine a rectangular coordinate to a polar coordinate so for example if we have 5 comma five in rectangular coordinates we want to converted the first thing we're going to do is we're going to find R doing X2 + y^2 = r^2 so x^2 is 25 y^2 is also 25 = r^ 2 that's going to be 50 = R 2 square < TK square root and we get um the sare < TK of 50 is 25 * 2 that's going to be 5 radical 2al R now R could be positive or negative all depending where you want to be so we'll think about that in a second but the next thing we got to do is figure out what Theta is so we have tangent of theta equal y5 overx POS 5 that's ne1 so now I'm thinking okay where is tangent 1 oh that's should be one you guys are pretty familiar with Tangent is -1 at 3 pi over 4 or at 7 pi over 4 okay so then what I got to do is I got to think about where is the point 5 comma ne5 so I think about rectangular coordinate if we go right five and then down five it's going to be somewhere down here now that would be a perfect match with 7 pi over 4 so if I have my radius be 5 radical 2 and then my angle be 7 pi over 4 in polar coordinates that would land me right there again think of that angle 7 piun over 4 and then radius of 5 radical 2 but I could also use the 3 Pi over4 I could use the 3 pi over 4 if I wanted to but that's in quadrant two so I would therefore need to make my radius negative to reflect me back into Quadrant 4 where clearly my original point was so those would be two different answers that both would represent that rectangular coordinate all right next up is converting rectangular or excuse me convering polar coordinates are comment Theta two rectangular coordinates this is also fairly simple all we have to do is use these two equations we're going to know R we're going to know Theta from the point and all we got to do is do R * cosine of theta and that's going to produce our x r * s of theta and that's going to produce our y I mean it's really really that simple so for example if we have a polar coordinate of 3 comma pi over 6 and I say okay what's the rectangular coordinate well to find my x value I'm going to take the radius three multiply by cine of PK / 6 and cosine of piun 6 is radical 3 over 2 so we get 3 radical 3 / 2 for my X Y is going to be the R value 3 * sine of pi/ 6 sine of pi/ 6 is 12 so I have 3 * 1/2 that means my y value is three halves so my rectangular coordinate would be 3 radal 3 over 2 comma 3es I mean it's really that simple so easy to convert them all right the last thing we want to talk about is polar graphs now here's what I know so far and I'm not 100% so don't quote me on all this but there's not going to be a ton of polar graph questions on the exam definitely not any of the frqs but could come up on a multiple choice so I just want to quickly talk about different polar graphs so you could kind of recognize them now there are multiple different families of polar graphs but the first ones we have are circles now circles are going to be produced when we have a * cosine of theta or a * s of theta now a is going to determine the the radius of your circle and then Theta well if let me say this cosine is going to be a circle surrounding the x-axis if a is positive it's going to be surrounding the positive x-axis if a is negative it's going to be surrounding the negative x-axis if you have a sine Theta and if a is positive you're going to be surrounding the positive y AIS if a is negative you're going to be surrounding the or excuse me the negative Y axis that's kind of easy way to determine those two values there not too bad then we have a family called lemons these are a whole family all different types of really cool polar graphs and they are all going to be based on a plus b cosine of theta or a plus b s of theta now to determine the different type of lisan you are all you need to do is consider the ratio of A over B that ratio is going to determine what type of graph you are now I'm going to give you some more details when we look at these graphs so for example if a over B is less than one if the ratio from A to B is less than one you're going to have an inner loop we call that a lamison with an inner loop now what that looks like is exactly what these pictures look like we see like a circle but that it kind of gets looped in now a couple comments about this first if you take a plus b right you add your A and B together that's going to be the maximum value all the way to here all the way to here all the way up or all the way down now now how do I know if I'm going left right up or down well cosine goes left or right if B is positive it's going to go to the right if B is negative it's going to go to the left sign goes up or down if B is positive it's going to go up if B is negative it's going to go down now we also have the in Loop the length of that inner loop is going to be the difference between a and b so this value here this this length of that inner loop is going to be the difference of your a and your B value so hopefully that makes a little bit of sense there as well again I'm just trying to quickly go over these hopefully you learn a lot more detail in class now if the ratio between your A and B is equal to one like 1 and one 5 and 5 4 and four 9 and 9 you're going to get what we call a cardioid this is going to kind of look like a heart that why gets the name cardioid and that's where you don't have an inter Loop it actually goes into z0 that's kind of creates this like little heart once again left right that's going to be your cosine positive B value to the right negative B value to the left and then for sign we're going to have a positive B value is going to go up like we see here negative value is going to go down like we see here but the cool thing is that we see that it goes all the way into 0 0 and that's because of the ratio from A to B being equal to one now how do you know the maximum distance of that heart whether it's left right up or down that maximum distance is going to be again adding your A and B values together now next up we have what's called a a dimple with no inner loop so imagine like a balloon and you just kind of push your finger in on one side the balloon and kind of create a dimple so this is going to happen if your ratio between A and B is between one and two all right so again this is going to be left or right is going to be a cosine up and down is going to be a sign positive so this graph right here was a positive cosine that means my B value is positive this graph down here is going to be a b value negative with a sign now the maximum distance to the right to the left to the up to the down is going to be your a plus b the minimum distance the the shorter side to the left the right up or down the shorter side that's going to be again the difference between your A and B value so just kind of noting noting those kind of little details is going to really help you understand when you're trying to choose a graph for a particular function or vice versa so again you just kind of have to memorize these different outcomes for the ratio and what the graph could look like whether it's a cardioid in Loop or in this one now is a dimple but just keeping in mind that adding the A and B Valu is going to be that maximum distance whether it be up down left or right so Trum is going to be that minimum distance and there's one more type here this is called No dimple no inter Loop so this is if I have a balloon and I I don't push in creating a dimple I just kind of flatten one side out I know it's kind of weird but notice these graphs here one side's just kind of flattened out so this is If the ratio between your a and your B is greater than or equal to two so once again does it go left does it go right does it hug the x axis does it hug the y- axis that's going to be cosine or does it hug the y- AIS up or down that's going to be sign now once again that maximum distance of your value the the bigger side is going to be adding your A and B whether that bigger side goes up down left or right and then the smaller distance between that flat side in the center is going to be the difference between your a and your B so I know like it's a little bit confusing and again I hope that your teacher went over this way more detailed in class I'm just trying to give a quick overview here if you remember these different types of lamons pretty cool graphs don't expect you know 18 questions about these on the AP exam but there might come up with one dealing with the graph so hopefully you could use this Quick Knowledge to recognize what type of graph it is and you can oh I got got to add my A and B to determine how you know big or small those values are all right and the final type of graph we have for Polar coordinates are called Rose curves these are actually really cool graphs as well now a rose curve is when you have a * cosine of n * Theta or a * s of n * data now we saw circles circles were when we had an N value of one that's when we had those circles the first type of graphs so if this n value is anything other than one it's going to produce these really cool Rose curves now let's talk about a couple of the details here first the number of Petals on the rose curve depends on N if n is odd there's going to be n pedals so three you're going to have three pedals five you're going to have five petals if n is even there are double that many petals so if you have an N value of two there's going to be four petals n value of eight is going to produce 16 petals then the length of each petal is your a value that's out in front now a sine Rose Curve will never have a petal on the xaxis so if you're looking at the graphs that's how you can know if it's a sign or a cosine if there is a pedal on the xaxis surrounding the x-axis it's going to be a cosine row curve if there is no pedal on the x-axis you're going to be a S curve so here's a couple graphs we're going to look at here to understand this so this first one here notice we have an odd value for our end so odd is how many petals there are so we see three petals the a values the length of those petals so each petal goes out to a radius of three and notice that one of the petals right here is on the x-axis that's why it's a cosine all right for this next one here once again oh we see a petal on the x- axis that's going to be a cosine we have an odd value for n so that's going to be five pedals and we see five petals and the length of each one of those petals is going to be four now for the sign ones over here notice we don't see any pedals on an X axis that's where or going straight horizontal because I guess you should argue that you know in a polar coordinate there is no x axis so you know looking at the horizontal line left or right maybe that'd be a better way of saying it somebody yells at me here but here we see um three so three pedals seven so seven pedals five and three are the lengths of each of those pedals but again those are signed because there's no pedal going straight horizontal left or right all right now here here are some with even uh values inside so two is going to produce four petals four is going to produce eight petals once again notice that we do see pedals on that horizontal plane those are definitely coign and then over here on the right hand side we see that there are no pedals on that horizontal plane so those are going to be signs and then again two is going to produce four pedals six th is going to produce 12 petals and once again the length of those petals is that a value out in front so not too bad hopefully that makes sense but those was just a quick overview of the most common different type of polar graphs now I know that those lisons can be a little bit tricky so just want to go over this kind of one more time with you so if you have a polar function like this 5 + 5 cosine of theeta you say okay what what's this graph look like well first the ratio of your A to B is one that's going to produce that cardioid and it's cosine and with a positive B value so that that it's a positive cosine so that means it's going to go to the right now 5 + 5 is 10 so that means that that Max distance all the way over here is 10 and 5 - 5 is zero which is why it's going to curve in all the way to zero and then all the way back out there and I know that that is a terribly ugly graph but that's going to at least allow you to choose the graph that's correct based on what you see there for that lamison there so kind of thinking about those values so little bit weird not too bad let's actually do one more let's do an in Loop one so if we have F of theta equals 2 minus s let's do 2 - 3 sin of theta so once again the ratio 2/3 is less than one so that's going to produce an inner loop okay so here's what I'm thinking it's a negative sign so means it's going to go down 2 + 3 is five so the maximum distance it's going to go down is five 2 minus 3 is one so that inner loop only goes to 1 and then the outer loop goes to five so again absolutely terrible drawing I you know I failed R class when I was in high school but let me try to make it a little bit nicer so this inner loop is going to go to one that would be a value of one the bigger Loop surrounding it is going to go all the way to five and I don't even know if that's a better drawing at all but again I'm just trying to get you guys to see the basic qualities that we notice here with these um polar coordinates all right so that's it for unit 3 lot of information unit 3 so hopefully this was kind of a nice quick video to just kind of Hit the big points that way when you're studying you can prepare for what you might see out of unit three on the AP pre-cal exam