 Welcome back to our lecture series, Math 1060, Trigonometry for Students at Southern Utah University. As usual, I will be your professor today, Dr. Andrew Misildine. In lecture 20, we're going to finish our discussion of trigonometric identities in chapter six in our lecture series. In doing so, we want to pick up a few things we haven't covered yet, some additional identities, so to speak. In this first video, what I want to do is actually show you how you can use trigonometric identities to help you calculate some more complicated trigonometric expressions, particularly trigonometric expressions that involve these inverse trigonometric functions. Now some of these things would be pretty easy. Like if you take sine of sine inverse of three-fifths, then sine inverse of three-fifths is the angle which, if evaluated on sine, would give you three-fifths. So as these are inverse functions, this is equal to three-fifths, no problem. And this is another one we've also done before. We did sine of tangent inverse of two. Like, so this one's not as obvious from the get-go, but the idea is if you draw a triangle, you can be like, oh, the tangent ratio is supposed to be two, which is two over one by the Pythagorean theorem, you can finish it. We'll do this all in just a second, so don't worry about that. You could get the evaluation from there, all right? But that's not the setting we have right now. We want to compute sine of sine inverse of three-fifths plus tangent inverse of two. So there's two separate inverse trigonometric functions inside of it, how do you handle that? Well, when it comes to inverse trigonometric functions, I want you to think of it as an angle, right? Because after all, sine inverse of three-fifths, this is the angle which, if sine were evaluated at that angle, you get three-fifths. And tangent inverse of two, this is the angle which, if evaluated at tangent, would give you two. And so whenever you see inverse trigonometric functions, think of them as angles. And we usually like to use Greek letters to help us with angles, right? So think of sine inverse of three-fifths, we're gonna call that one alpha. And when you see tangent inverse of two, I want you to think of tangent inverse of two as some other angle beta, or call it theta and phi, whatever you care about as your favorite variable, that's okay. And so then with sine inverse of three-fifths, we can, by the inverse function problem, we can move the sine inverse to the other side, it becomes a sine. And we get sine of alpha is equal to three-fifths. And with this statement in hand, we can then draw ourselves a right triangle, a diagram that we can complete to help us compute any of the trigonometric ratios. So this is the triangle associated to angle alpha, like so, for which sine of alpha is equal to three-fifths. So we see that opposite is three hypotenuses, five. And so by the Pythagorean relationship, the adjacent side would be four. So with that in mind, we can compute any angle with respect to alpha. We can do the same thing for beta, of course. Let's draw a right triangle, diagram, and that situation as well, like so. And so then let's label it. So this is the angle with respect to beta. What we've then discovered about beta so far, let me move this down a little bit. What we've learned about beta so far is that its tangent ratio, tangent of beta, is equal to two, or better yet, two over one. So we get opposite over adjacent is two and one. And so then by the Pythagorean relationship, one squared is one, two squared is four, so we get the other side is the squared of five. So with this right triangle diagram, we can then compute any trig ratio with respect to beta. So what does that have to do with all of this? Well, what I want you to think of is sine of sine inverse of three-fifths plus tangent inverse of two, this is just sine of alpha plus beta, right? We just have two angles. Angles, we don't know what they are necessarily. I mean, we can compute them with the calculator, but we're going to get some approximations here. We can do much better than that. Sine of alpha plus beta, if we were to apply the angle sum identities we've seen previously, remember that this is equal to sine of alpha, cosine of beta plus cosine of alpha times sine of beta. The angle sum identity for sine is the harmonious one. We get sine and sine, cosine and sine, like so. And then we can compute each and every one of these things, right? Sine of alpha, we already know that one, that's a three-fifths. We can compute that. Cosine of beta, we don't know what that is yet, but we can compute that. You get adjacent over hypotenuse. Cosine of beta is going to be one over the square root of five, like so. Now we need to do cosine of alpha. We don't know that, but we can compute it. Cosine of alpha, you're going to get adjacent over hypotenuse, so you're going to get a four-fifths, like so. And then sine of beta, again, we can compute it. We get two over the square root of five, like so. We get two over the square root of five. And so putting these things together, three times one is three, four times two is eight. This is above the common denominator of five times the square root of five, for which we then can add three and eight together. Of course, we get 11 over five times the square root of five. You can rationalize the denominator if you have that irrational compulsion to do so, we're just going to leave this as 11 over five times the square root of five. This is the exact value. So whenever you see these inverse trigonometric functions, again, think of these as angles. And so this became an angle sum problem. The angle sum identity helps us out so that we can compute this in terms of the angles alpha and beta directly. And so these right triangle diagrams also save the day when you have these inverse trigonometric functions. So that's what you want to think of. When you have inverse trigonometric functions, think of those as angles and use right triangles to help you compute the other trig ratios. Let's do another example of this before we end this video. Let's do sine of two times tangent inverse of X. Can we come up with an algebraic formula that doesn't involve trigonometry here whatsoever? So you'll notice here that if this was just sine of tangent inverse of X, this would be very similar to what we just did a moment ago without all of the extra identities though. But this is not sine of tangent of inverse of X. This is the sine of two tangent. And so the double angle identity has to come into play here. So let's do what we did before. Let's take theta to equal tangent inverse of X. Whenever you see a inverse trig function, think of it as an angle. So this tells us that tangent of theta is equal to X or better yet, X over one. Thinking of the associated right triangle, we draw that picture right here. This is our right triangle for theta. So mark it up, we get theta right angle right there. And so tangent will be X over one. The hypotenuse will be the square root of one plus X squared by the Pythagorean equation. And so now we're ready to go. So we have to compute sine of two tangent inverse of X, but tangent inverse X is just theta here, right? So you get sine of two theta. In which case this makes me think of the double angle identity for sine. So we're gonna get two sine of theta times cosine of theta, for which case sine of theta is gonna be opposite over hypotenuse. So we get X over the square root of one plus X squared. And then we need cosine, which cosine is adjacent over hypotenuse like so. So we get one over the square root of one plus X squared. So in the top, you get two times X times one, which is a two X. And then the denominator, you get the square root of one plus X squared times the square root of one plus X squared. So since you're squaring the square root, you just end up with a one plus X squared in the denominator. And so we've now simplified our fraction. I should say that we took the trigonometric expression sine of two tangent inverse of X, and it becomes the rational function two X over one plus X squared. And we can do these calculations involving inverse trigonometric functions, but when the inside of the trigonometric function, which is supposed to be an angle, when it's more than just an inverse trig function, you're still gonna use this right triangle technique, but you combine it with the appropriate identity, whether it's angle sums, angle differences, double angles, half angles, whatever you have to do, we can then use those identities we've already learned in chapter six to help us simplify these inverse trigonometric expressions.