INSTRUCTOR: OK. In this video, we're going to talk about evaluating compositions of trig and inverse trig functions, specifically the kind where the normal trig function, not the inverse trig function, is the outer function, like the last one that you evaluate. So let's look at what we're talking about. First thing we got to understand is that you do trig functions-- so sine, cosine, tangent, et cetera-- of an angle, and that will give you a ratio. So you do the trig function of an angle, you get a ratio. 

On the other hand, you do the inverse trig function of a ratio, and it gives you an angle. They're inverses of each other, so that kind of makes sense. So we need to know that so we can do the problems. You should also know, because I'm probably going to say it at some point, I use the terms inverse sine and arcsine interchangeably, inverse cosine, arccosine? inverse tangent, arctangent. I'll try to say "inverse" throughout this, but sometimes I slip up. 

So let's look at a couple of these before we even do the compositions. We could have something like the sine of, let's say, a famous angle, so pi/3. That's from the unit circle. Sine of pi/3 is a famous ratio. The famous ratio is radical 3 over 2. We could also look at a famous inverse sine, so the sine inverse of a famous ratio, so, for example, radical 3 over 2. That's actually an angle that we know. That's the angle pi/3. So it's just a funny way of writing pi/3. 

We could also have something like the sine of an angle that's not quite famous, so 9 pi over 37. I don't know what the sine of 9 pi over 37 is, but I do know that that's a ratio. And the reason I don't know what it is because this is just not a famous angle. The famous angles are really the unit circle angles. So that's going to work out like that. We don't know what the ratio is, but we do know that it is a ratio. 

The same sort of thing can happen with inverse trig functions. We could try to do the sine inverse of 19/37. 19/37 is not a famous ratio. So I know that this is an angle, but I do not know what angle because it's just not a famous one. So it's not a unit circle angle. 

So we're going to deal with problems that really have to do with these, nonfamous ratios and nonfamous angles. Let's take a look at an example. So say we want to evaluate the tangent of the sine inverse of 7/25. So sine inverse of 7/25 is an angle. And we're looking for the tangent of an angle. So the answer to this is going to be a ratio. And I'm going to try to color code everything as we go through. 

So the first thing I do is I look at sine inverse of 7/25, and I say that could be theta. So theta, an angle, is the sine inverse of 7/25. If that's true, I can say that the sine of theta is 7/25. And now is the key to evaluating these sorts of things. If sine of theta equals 7/25, I can actually draw a right triangle with an angle theta for which that's true. So we're looking at-- theta is the inverse sine. And that means we're trying to evaluate tan of theta. 

Let's draw a triangle. So we draw our triangle. And I'm going to put theta in it. And then the sine of theta is 7/25. 7 goes opposite. 25 is hypotenuse. The missing side is 24. I know that kind of automatically because it's a 7-24-25 right triangle. I definitely recommend that you know the Pythagorean triples or at least a bunch of them. 

So there's 3-4-5, 5-12-13, 7-24-25, 8-15-17. Those are all good to know. If you don't remember those, you could have used the Pythagorean theorem. We knew a leg and hypotenuse. So we could have done the square root of the hypotenuse squared minus the leg squared gives us the other leg. 

So we get that. Back to the actual question, the question is, what's the tangent of theta, where theta is this angle in the triangle? And if you look at the triangle, tangent should be opposite over adjacent. So this all evaluates to 7/24. So what we just figured out was that the tangent of the inverse sine of 7/25 is just 7/24. We can do any number of problems like this. Sometimes we just combine them together so we'll get something that looks like this-- cotangent of the inverse cosine of the tangent of the inverse cosine of 7/9. 

So it looks to me like we're just going to do one problem and then another problem. First, start with this. So start with the innermost function. So that is an angle, right? Inverse cosine of a ratio gives me an angle. So that's an angle. So I know theta is the inverse cosine of 7/9, Which Means that cosine of theta is 7/9. And when you do a lot of these, you might jump right to writing cosine of theta is 7/9. That's totally fine. 

Now I'm going to draw a triangle based on this information. So cosine-- so I got to put a theta. Cosine is adjacent over hypotenuse. So the adjacent side is 7. Hypotenuse is 9. You can find the missing side using Pythagorean theorem. So it's going to be the square root of-- 9 squared is 81, minus 7 squared is 49. And that gives me square root of 32. It doesn't usually pay to simplify radicals along the way. So I'm just going to leave it because it's going to end up another side of a triangle, and then I'll probably have to square it anyway. So I like to just leave those. 

So let's look at the original problem. We're trying to find the tangent of this angle, so from this triangle. So let's rewrite cotan of the inverse cosine of-- so all of this together, we're looking for the tangent of the angle. Looking at the triangle, the tangent of the angle is opposite over adjacent, so radical 32/7. 

All right. And now we just do another problem. So we have a new angle. So this is our new angle, which I'm going to still call theta. You could call them theta sub 1 and theta sub 2 if you wanted to. You could call them alpha and beta, whatever you really want. You could actually call them x and y because those aren't in the problem. So I start with this. Then I go to cosine of theta is radical 32/7. 

At this point, we draw a triangle. And in this triangle, there's an angle theta. And we know that cosine of theta is radical 32. So adjacent is radical 32. Hypotenuse is 7. Use the Pythagorean theorem to find the missing leg. So it's going to be 49 minus 32. That's where you ended up squaring that radical. So if you had simplified, it wouldn't really have helped you. So we get this, and that's the square root of 17. 

So what we're trying to do is we're trying to find the cotangent of theta. So based on the triangle, tangent of theta is radical 17 over radical 32. So cotangent is the reciprocal of that. So we get radical 32 over radical 17. I'm perfectly happy with that answer, so I would box that and move on. But your teacher might want you to simplify radicals. So radical 32 is radical 16 times 2, which is 4 root 2. So we could do this over radical 17. 

But then you probably would also have to rationalize the denominator. So multiply by radical 17 over radical 17. And that'll give you 4. Radical 2 times radical 17 is radical 34-- and then over 17. So all three of those are equivalent. And that's how you do it. 

So the one really key thing when you're doing these-- draw a new triangle for every inverse trig function that you hit because every inverse trig function is a new angle, and the sides relative to that angle could change position, could change values. So every new inverse trig function you hit, draw a brandnew triangle. All right. I hope you found this helpful, and good luck with that.