hi folks welcome to the second part of this trigonometry recap for ib math aasl in the first part we looked at triangle trig we looked at bearings we looked at some trig identities in this part we'll be looking at trigonometric functions and trig equations this presentation is not produced or endorsed by ibo it's just one that i made for my students it's not an exhaustive list of what's in geometry and trigonometry but it's a pretty good overview there are time stamps and downloadable notes in the video description you can also find the link to the first part of the video there there are often many different ways to solve a particular problem so if you're doing something different than i am especially when it comes to trig equations keep doing what you're doing one of the best ways to get the most out of the presentation is to try the questions yourself with all that out of the way let's dive in the first thing we're going to look at is the anatomy of sinusoidal functions so that's either sine x or cosex those are both sinusoids and they both have a default period of two pi and when we say period we mean the width of one full cycle and really you can measure that cycle from anywhere as long as it's done its complete thing and then it starts a new cycle well then that's one period now these functions do continue on and on forever in both directions that's what makes them periodic they have a thing that they do and they repeat it same with the cosine keeps going forever and ever in both directions what i'm showing here is sort of the basic first positive cycle that you would see of each of these graphs without any transformations happening so a period is the width of one cycle the amplitude is this height right here from a max to what we call the midline because it's right in the middle we have great names in math sometimes the midline is called the principal axis but i like midline better and they'd go up to a maximum of one and a minimum of negative one that's true for sine and for cosine so on your basic sine or cosine graph the amplitude is one it's half the distance between a max and a min or a better way to think about it is it's that distance between a max and the midline the principal axis or midline is on the y equals 0 or x axis the default period is 2 pi and here's how i keep them straight a basic sine graph quote starts in the middle and moves up and a basic cosine graph starts quote high and moves downward that brings us to modeling with sine and cosine graphs or finding the equation for a situation so i've got some letters put into each of these functions a b c and d the choices of the letters are arbitrary you could use different letters and you probably will run into different letters for them but it's their position in the function that matters so whatever's in front of the sign controls the amplitude whatever's at the very end here the plus d or a number that's just on its own that's going to control the midline we already learned about general transformations of functions and all those rules that happen in general transformations of functions in the functions unit they all hold true for sinusoids as well it's just for some of these we have our own particular terminology when we're talking about trig functions so generally that a value would be a vertical stretch we call it amplitude when we're talking about trig functions or sinusoids in particular it can be found by taking the difference between the max and the min and dividing it by 2. i actually think a better way is to just take the max and subtract the midline but that means finding the midline first it's important to mention here that a negative a value will reflect the basic graph through the x-axis so whereas sine x looks like this negative sine x will look like this flip through the x-axis moving on to the quote b value so the b value controls the period but it does not tell you directly the period if b equals 1 then you would really wouldn't see a coefficient in front of x and the period would be 2 pi so this b is the coefficient of x generally the period of the graph is 2 pi divided by b so let's do a couple of quick examples here at 3 cosine 5x that would be a function with an amplitude of 3 and a period of 2 pi divided by this 5. if i had a function like this negative 5 sine of x over 4 its amplitude would be 5. the negative would mean that you'd have an x-axis reflection let's zoom in here a little bit sometimes it's confusing to see what's going on when we have an x over four so let's think of that as negative five sine of a quarter x that may help and we know that period is 2 pi divided by the coefficient of x so that would be 2 pi over a quarter ah that's the same as 2 pi times 4 over 1 or would have a period of 8 pi it's still ok for you to think of the b as a horizontal stretch by factor of 1 over b but it can be more helpful just to think of what it does to the period so period is 2 pi over b where b is the coefficient of x and also if you know the period you can find the b value just by rearranging this formula so that it says b equal so b is 2 pi over period the c value that looks like a shift when we think of our general transformation rules and we already know that an x minus something sends it to the right so cos of x minus pi over 4 would be shifted pi over 4 units to the right if you had the graph of y equals sine of x plus pi it would be shifted pi units to the left coming back to all these letters the quote d value controls the vertical shift and the location of the midline it's halfway between the max and the min so you can find that d by going max plus min over two and honestly when i'm trying to find a model i often start with the midline i always start with the midline then i find the amplitude then i find b then i find c at the end you can do it in any order that you want that just has worked out well for me one word of caution about the midline sometimes the ib will present the function like this two plus three cos of i don't know five x we're used to seeing that as three cos five x plus two and we would recognize the midline as being at 2 in the second form but it's the same if you see it in the first form feel free to rearrange to something that feels comfortable for you though so now that we've defined all our parameters as a b c and d let's do a practice question where it uses none of those letters so we're given a sinusoid or a graph of one down below and essentially we're asked to find what its equation is now there are infinitely many ways to answer this question correctly but it is asking us for a specific form it's asking us for a cosine graph and it is telling us that p must be positive in this particular one there are still as it turns out infinitely many are values but there's one that's kind of low-lying fruit the one that we'll go for so when i'm doing these kind of questions here's the order that i look for things in i always look for the midline first so i've got a max here at two and a minimum at negative eight halfway between those is negative three and i can just draw that in and it looks like the middle to me and it is if i'm not sure what i can do is i could take this maximum value and this minimum value and add them together and divide by two essentially i'm finding the average of them so it'll be negative six over two the midline is at y equals negative three i'll worry about which letters those are at the end next thing i'm going to do is find the amplitude so the amplitude is this kind of distance right here and i can just count it up it's one two three four five blocks up so the amplitude is five the next thing i'm going to look at is the period so the period is how wide a full cycle is and it's kind of hard to tell here maybe i could measure it from here to here but i'd have to know that my midline is correct so what i'm actually going to do is i'm going to measure from here to here how wide is that and that is 4 units wide but it's only half a cycle or half a period so a full period is going to have to be eight units wide so just from the max of the min is only half of the cycle it would finish up its cycle later on up here and it's going to be eight units wide for a full cycle so the period is eight now what does that mean for the b value well the b value is two pi over period and i'm still using the letters from the last one we'll assign them their correct letters for this question at the end so that's going to be 2 pi over 8 or pi over 4. lastly i'm looking for the shift so a cosine graph starts high where is sort of the start of this cosine graph it's right here on a cosine graph the shift is going to be the x coordinate of a max so normally a cosine graph would start high right on the y axis but here it's been shifted over 3. so 3 to the right as an equation that's going to be y equals 5 as my amplitude cosine of b is pi over 4 x minus 3 from that red one right that is the shift and it's also moved down three so this would be the midline is negative three and then in terms of the letters that they're giving in this question well this is p p is five q is pi over four r is 3 right because it says x minus r so what are we minusing 3. and then plus s at the end that means that s is negative 3 okay just based on the form they're giving it to us in again you could have other r values but this is the simplest one now it says hence find all solutions to the equation f of x equals negative 6. so when we think about where f of x is negative six that's you know at these points over here f of x is negative six at three separate points right here right here and right here i want to get this to three sig figs so i'm going to head to my gdc i'm going to set one of the functions equal to 5 cos pi over 4 x minus 3 minus 3 and the other one equal to negative 6 and just see where they intersect i'm also going to make sure that i'm in radian mode and i'm going to choose good window settings i only want to see stuff from 0 to 10 and i want to see y values from negative 10 to 5. so this window setting is okay i hit graph one thing we can do at this point is just double check that our curve looks correct our model should be okay here because it looks like the curve that is given in the question and if it doesn't then maybe we need to do some tweaks all right i'm going to go second trace intersect and i have to choose some left bounds and right bounce so left bound for my first one right bound intersections at 0.181 we're going to have to find another intersection so 2nd calc intersect go to the left bound so just to the left of the next intersection and then just to the right it may be a little bit different on your calculator you may have to guess tells me 5.82 and there's one more intersection we can find it using intersect but i also know that this is a periodic function so if there's an answer at 0.18 then there's an answer at 8.18 okay but again we could just use the gdc lastly it says the equation f of x equals k has exactly one solution on the given domain write down the two possible values of k think about what we just did we have f of x equals negative six that's essentially where it intersects a horizontal line that says negative six if i had f of x equals negative nine there would be no solution if i had f of x equals negative one there'd be one intersection or one solution so one of the answers here is k equals negative eight if i move that horizontal line upwards i have two solutions two solutions oh three solutions three solutions three solutions now two solutions two solutions and finally just one solution okay what i have in green here is not part of the graph or it's outside the domain so we have one solution when k equals two so there's a bit of a problem-solving question that you could throw on top of this trig modeling comes up in all sorts of real-life examples involving periodic and circular motion so whether we're talking about ferrous wheels or tide heights or recurring temperatures all those things could have some periodic motion and therefore some sinusoidal stuff happening this brings us to our last topic in trigonometry which is trig equations now we actually just solve the trig equation using our gdc and if it's paper 2 i highly recommend that you use your gdc to deal with all trig equations if it's paper 1 then we need some analytic techniques the steps that i like are to isolate the trig ratio to find a reference angle determine quadrants and then write our solutions but in order to do that we need to have sort of a treasure trove of information to pick from so i'm going to write all the things that i sort of keep straight in my mind so this is how i organize my trig values and my conversions from degrees to radians i use the unit quarter circle which i can then sort of expand to the unit circle without having to write the whole thing out i remember that over here i'm showing that all of the trig ratios are positive in quadrant one only sine is positive in quadrant two only tangent is positive in quadrant 3 and only cosine is positive in quadrant 4. that's what the astc stands for i write down what the coordinates are on the axis and the big idea here is that x values on the unit quarter circle are cosine and y values are sine y over x is tan theta and i like to just write those on the outside so all those red numbers are my tangent values and that should be enough to get me through any of the trig equations because it deals with all the multiples of 30 and 45 degrees that i'm going to run into i also remember that pi radians is 180 degrees but i have a nice little way of converting right in the unit quarter circle you can organize this however you want though if you want to do a table great if you want to learn the hand trick great if you want to use special triangles equally great some people also just think of the graph i'm not trying to be prescriptive about how you need to keep track of this stuff but i am saying you need to know what say the sine of 30 degrees is and i go over making this table in the first trig video but there it is so that brings us to a trig equation two sine x equals root three so i'm going to isolate my trig ratio sine x is root three over two and once i know that i'm going to go ahead and i'm to think what would my reference angle be well i'm talking about a sine value of root 3 over 2. that means on my quarter circle that's a y value of root 3 over 2 and that means i've got a reference angle of 60 degrees or pi over three okay so it has something to do my answer has something to do with pi over three the next thing i like to do is determine what quadrants i might be in so sine is positive here i could be in this quadrant or this quadrant so it might look like pi over 3 in this quadrant or pi over 3 in this quadrant reference angle means with respect to the x-axis and that's basically a picture of my final answer one of those angles is pi over 3 so that's an answer here the other angle right here is a little less than pi so if i went all the way over here that would be 180 degrees or pi so this is pi take away pi over 3 or i could think of that as 3 pi over 3 minus pi over 3 okay that's 2 pi over 3. those are my two answers that's one way to work it out and it's the way that i often go to another way to think about it though would be to take this equation that's in purple and just graph sine x which would have a period of 2 pi this would be pi right here and say huh i wonder where it would be root 3 over two well root three over two is close to one so it would be right here and right here it would be a little bit on either side of pi over two and pi over three and two pi over three are a little bit on either side of pi over 2. so you could think it through graphically as well it's up to you now here comes another one we've got cos 2x is negative a half there are many ways to solve this you could if you really wanted to use one of those double angle formulas but i'm just going to think here i wonder what the reference angle is so we've got an cosine value of negative a half cosine let's go to that unit quarter circle cosine is x on the unit quarter circle so where do we have a half for x right here so i'm once again talking about a reference angle of 60 degrees or pi over three and i'm going to choose my quadrants so they're all positive in the first quadrant only signs positive in the second only tangent is positive in the third and only cosine in the fourth so where is cosine going to be negative cosine will be negative in these two quadrants because it's positive in the a and c ones that means i've got some values that look like this i can figure out what those angles are i mean this one right here would be pi minus pi over 3 because it's a little less than 180 degrees so that's 2 pi over 3. and this one here would be pi plus pi over 3 because it's a little more than 180 degrees or 4 pi over 3. here's the rub though if you have a situation where it's not just x here the angle here is two x then it's not x equals two pi over three it's two x is two pi over three and two x is four pi over three in this technique this diagram in standard position it's not a picture of the answer but it helps you get there if you'd add cos x then then yeah it would be a picture of the answer but we've got one more step to do here if i want to get x on its own then i'll have to divide both sides by two so divide this side by two and divide this by side by two that will be x is pi over three and this one over here if i divide by two that will give me x is 2 pi over 3 and those are my answers so i do have to deal with any operations on the angle after the fact now let's think about this one graphically cos of 2x is a cosine graph with a period of 2 pi over 2 or just pi so if i were to graph it would look something like this there's pi there's pi over 2. where is it going to have a value of negative a half well a little bit on either side of pi over 2. and again pi over 3 and 2 pi over 3 are a little bit on either side of pi over 2. it's a coincidence that it's the same answers in both of these and here the domain means that we've found all the solutions all the solutions in the first cycle so recapping the technique isolate the trig ratio find your reference angle determine the quadrants and write the solutions deal with any operations on the angle like the 2x over here let's take a minute here just to think a bit about the periodic nature of these solutions so investigating graphical solutions on the last two equations reveals that on an unrestricted domain solutions recur according to the period of the function in the equation so that's pretty dense let's check it out on the first one this would just keep going over and over again and we'd end up with more solutions but they'd just be jumped over two pi or one period from where we were the first time around on this function we'd have the same sort of thing happening but since the period on this one is tighter the solutions would recur every one pi units so it says in the first equation they'd repeat every two pi units because that was the period of the function and the second they would repeat every pi units because that's the period of the function that was involved in the second equation a little typo here general solutions to trig equations are currently not included in the aasl curriculum but an informal understanding of the periodic nature of trig solutions may prove useful that whatever solutions you find are going to recur every period whatever that period is now that we've dealt with some common trig equations let's deal with a few special cases so in this one we've got a quadratic trig equation when i look at that whole thing it makes me think of something like this 2 n squared minus n equals 1. and i know if i wanted to solve that i would make one side zero and i would factor it so same sort of thing with a quadratic equation that has trig in it will make one side zero i'll notice here that mercifully there's nothing weird happening to the angle and the domain is just the first regular revolution so i'm not going to have to do anything fancy with dealing with periodicity of solutions and now i'm going to factor this just like i would if it were a regular quadratic so 2n squared minus n minus 1 would factor like this 2n and n a minus 1 here a plus 1 here if we have sine x involved well it's going to be 2 sine x plus 1 and then sine x minus 1. and i could distribute that or foil it all out and make sure that it does indeed get us back to our first step and now each of those get their own little elementary equation so 2 sine x plus 1 equals 0 is going to yield some solutions so is sine x minus 1 equals 0. and it becomes twice as much work we've got two little elementary trig equations to solve but not twice as difficult so two sine x equals negative one where sine x is negative a half and we'll go through this whole process again we have to find a reference angle so sine x is a half right here at 30 degrees or pi over 6. reference angle is pi over 6 i have to choose some quadrants sine is negative in this quadrant and in this quadrant the reference angles are pi over six and i'll have two different solutions from that i'll have pi plus pi over six which would be seven pi over six and i'll have this one over here which would be two pi minus pi over six because it's a little less than 360 degrees which would be eleven pi over six there are some solutions that come from sort of that elementary equation the other elementary equation is this sine x equals 1. so again you could think of this graphically but whenever i have sine or cosine of 0 or 1 or negative 1 i'm just going to think of the full unit circle and say hey i wonder where the y value is 1 and that would be right here at pi over 2 as an angle or 90 degrees and that gives us our three solutions in that first revolution if this were paper two we would absolutely solve this using the gdc and graphical methods although tangent functions recur more frequently than sine and cosine functions when we solve the equations we can treat them pretty much the same in this question we'll deal with what i'd call a sneaky tangent equation so it's one that doesn't look like a tangent equation to begin with but sometimes an equation will contain a sine x and a cosex and no constant term and it looks like there's no way to solve it but if you can build this identity sine x over cos x you can turn it into a tangent equation so i'll show you what i mean i could move that root 3 cos x to the other side and now what i'm going to do is divide both sides by cos x so it'll be gone over here and on the left hand side what i'll end up with is three sine over cosine oh that's tan x and now i could divide both sides here by three and end up with this tan x is negative root three over three and we made the note in the first video that root three over three and one over root 3 are the same thing also root 2 over 2 and 1 over root 2 are the same thing so we're going to go ahead and just look for a reference angle where is the tangent root 3 over 3 and the answer to that is right here because root 3 over 3 is the same as 1 over root 3. so we're talking about a reference angle of pi over 6 or 30 degrees reference angle of pi over 6 all silly turtles crawl and so we have tangent is negative in this quadrant and it's negative in this quadrant and it'll work out with a basic tangent equation that they're always 180 degrees apart from each other but we don't necessarily need to memorize that we can just draw the answer and then figure out what these angles are this one would be pi minus pi over six and this one would be two pi minus pi over 6. so we have solutions of 5 pi over 6 and also of 11 pi over 6. if i were to think about the graph of tan x and wonder where it's negative root 3 over 3 i mean it's going to be at a couple of points that are a little before pi and a little before 2 pi so that's consistent with what we saw in the diagram in standard position i will concede that the titles to these slides are giving us more and more despair the last one was sneaky equations this one is messy equations the only thing worse than having two different trig functions is having two different angles in the equation so in this equation we've got cos 2x and then we've got some cosec so we've got some cos squared x and sine squared x it's a mess ideally we want to have a single angle and a single type of trig function so what are we stuck with here i think we're really stuck with the negative 3 cos x the cos2x it has a bunch of options in the formula booklet and so i have to make a choice here which of those three cos2x identities i'm going to use there's probably more than one correct way through this whole thing but if it were me i'd be choosing this one right here to get myself down to cos of x yeah there might be a quadratic but we can probably deal with that in fact we have dealt with a quadratic trig equation already the other problem here is that we have sine squared x and that makes me think of this identity sine squared plus cos squared equals one that can be manipulated so sine squared equals 1 minus cos squared of x and this gives us a couple of substitutions we can make so i'm going to replace sine squared with 1 minus cos squared x and i'm going to replace cos2x with 2 cos squared x minus 1. let's do that now that is still admittedly a mess but maybe it's one that we can clean up a little bit so i'm just going to collect like terms on the left hand side i've got a 2 co-squared x and a minus cos squared x so those go together i have a minus 3 cos x i have minus 1 minus 3 so that's negative 4. and then i have this it's a quadratic so we're going to want to make one side zero so i'm going to add co-squared x to each side that'll give me 2 co-squared x over here minus 3 cos x and i'm going to subtract 1 from each side so i get minus five and finally zero here i'll factor and if i'm confused by all the trig stuff here i'll think of it as two n squared minus three n minus five and hopefully i can factor that one and i'll get two n minus five by n plus one similarly this is going to factor as two cos x minus five and cos x plus one and then we get elementary equations from each of them so two cos x minus five equals zero where cos x equals five over two and this one over here says cos x plus 1 equals 0 or cos x equals negative 1. and i've got to solve both of these now here's the good news cos x equals 5 over 2 is not going to have any solutions the maximum value that cosine can go up to is one and we know that from the graph of cosex we also know it from the unit circle there's nowhere on the unit circle where you're going to get a x value of 5 over 2 or 2.5 the biggest you're going to get is 1. now where is cosine negative 1 well let's think about that unit circle again where is the x value negative 1 right back here at pi so x equals pi is the only solution on this domain on a question like this if you were lucky you might have a show that part up front that would ask you to do some of the manipulation but if not you just play around with your identities until you can get something that is workable that brings us to the end of the trigger geometry material for math aasl i hope this has been helpful good luck with the material and take care folks