 In this video, I wanna show you how you can transform a trigonometric expression here. It's gonna be theta over two plus sine two theta over four. I wanna turn into an algebraic expression just using the algebraic variable X, not the trigonometric variable theta. And we're gonna do this using the trigonometric substitution X equals three tangent of theta. And believe it or not, in calculus this type of conversion from algebraic variables to trigonometric variables or our video here, the reverse, we switch from the trigonometric variable to the algebraic variable. It's very common practice. And so it's appropriate to practice this in a trigonometric setting. So how do you write this expression in terms of just the variable X right here? So let's write that out. Theta over two plus sine of two theta over four. So it's all based upon our identity right here. X equals three tangent of theta. And so this equation here is, you can manipulate, you can change things because it is an equation, right? If we divide both sides of the equation by three, you end up with tangent theta is equal to X over three. This is a very important observation because if tangent theta equals X over three, we can think of this, of course, as opposite over adjacent, just usual so Catoa type things. And then from here, we can create a right triangle diagram associated to, of course, the angle, whoops, the angle theta right there. In which case then we see that our angle in consideration here will be theta. It's a right triangle. And so the opposite side is X, the adjacent side is three. And so then by the Pythagorean equation, we're gonna take three squared plus X squared, take the square root, that gives us the square root of X squared plus nine. That's the hypotenuse of this. And so with this right triangle diagram, we could compute sine of theta, cosine of theta, tangent theta. We already know that one. Cotangent theta, secant theta, cosecant theta. But in all of these, it's with regard to the angle theta. The issue here is we have to compute sine of two theta. How do you deal with that? Well, we can actually use a double angle identity to help us out, makes life so much easier. So theta over two plus sine of two theta is equal to two sine theta, cosine theta over four. So we need to compute sine of theta and cosine of theta. So using our triangle right here, we're gonna get that cosine of theta is opposite over hypotenuse. So I can't fit that inside of the triangle there. So we're gonna say cosine of theta is equal to three over the square root of X squared plus nine. Sine is gonna be the opposite side over the hypotenuse. So we're gonna get X over the square root of X squared plus nine, like so. And so we can then substitute this in for cosine. We can substitute this in for sine and then try to simplify it from there. What do you do about the theta? We'll come back to the equation here, right? So we have X equals three tangent, which we found out that tangent theta equals X over three, for which if you take arc tangent on both sides, you're gonna get theta is equal to arc tangent of arc tangent of X over three. And so we make that substitution in for theta right here. And so then our expression is gonna look like one half arc tangent of X, arc tangent of X over three. Then we have this two force, which is the same thing as one half. So I'm just gonna write that as a one half there. We get sine, which sine is X over the square root of X squared plus nine. And then cosine was three over the square root of X squared plus nine. Notice that the square root of X squared plus nine shows up twice in the denominator. So we multiply that together, you're just gonna get X squared plus nine. So writing this one more time, we have one half arc tangent of X over three. Then we're gonna get a, excuse me, we're gonna get a plus three X over two times X squared plus nine. And so with that trigonometric substitution, X equals three tangent of theta, our trigonometric expression, theta over two plus sine of two theta over four is equivalent to the algebraic expression, one half arc tangent of X plus three plus three X over two X squared plus nine. And so this idea of trigonometric substitution, like I said, it's very, very useful in the calculus setting. And so it's good practice right now to be able to convert between trigonometric expressions and algebraic expressions using these trigonometric substitutions. And in this example, we do sometimes have to use a double angle identity or other type of identities to do these trigonometric substitutions. So it's a good idea to practice those identities now.