 In a previous video, we saw how you can simplify the composition of a trigonometric function with an inverse trigonometric function, but we saw things like sine of sine inverse of three-fourths. Because of the inverse function property, sine of sine inverse of three-fourths will just be three-fourths. They cancel each other out, the sine of the sine inverse. What happens if we have a mismatch? What if we have sine of tangent inverse? Now we're not going to throw up our hands and declare defeat. It turns out we actually are in a very appropriate place to approach this thing. And the reason we can is because all of the trig functions, all six of them are connected to each other by some type of right triangle diagram. So when you see an inverse trigonometric function, I want you to think of the following. Inverse trig functions are angles. Because with a trig function, you put it in angle, it spits out a ratio. Inverse trig functions then go the other way around. You're going to insert a ratio into the function and it'll output an angle theta. So tangent, so you see something like tangent inverse of three-fourths, I want you to think of that as an angle. And use the symbol theta to make us to queue in on that. This is an angle, right? That's the variable we usually use for angles. So if I were to then think of, well, what's the right triangle diagram associated to this thing right here? What's the angle here? So this is going to be a right triangle associated to the angle theta right here. What we know about theta here is that tangent inverse of three-fourths equals theta. But if tangent inverse of three-fourths equals theta, that actually means that that means that tangent of theta equals three-fourths. You can flip this thing around. And I'm like, okay, tangent theta equals three-fourths. Now that gives me something I could use like a Sokotoa argument. Tangent is opposite over adjacent, so we get three over four. Then if we use the Pythagorean equation, right? So let's label this side x, this side y, this side r, right? Every right triangle satisfies the relationship that x squared plus y squared equals r squared. So if we plug in three and four there, we're going to get four squared plus three squared equals r squared. We're going to get 16 plus nine. That's equal to 25 equal to r squared taking the square root. All right, we're going to get that the other side is five, which, you know, looking at this thing, this is one of our favorite Pythagorean triples three four five. Okay, what does this do for us? Well, we now have a right triangle diagram for the angle theta. With this triangle diagram in hand, I can now produce any of the trigonometric ratios, right? Because I'm trying to compute right now sine of theta, right? If tangent inverse of three fourths is theta, then I'm trying to compute sine of theta. And sine of theta with respect to this triangle will be the opposite side over the hypotenuse. And so this is going to equal three fifths. Wow, that doesn't seem so intimidating when you draw a triangle. That's going to be our strategy here. Let's look at another one. Let's do this time tangent of sine inverse of one third. So our strategy is the following. Think of the trig inverse function as an angle theta and draw for yourself a triangle associated, right triangle, associated to that angle theta, which what do we know about theta here? We know that since sine inverse of one third equals theta, sine inverse of one third equals theta. That means that one third is equal to sine theta using so Catoa, which is opposite over hypotenuse, we're going to get the opposite side is one, the hypotenuse is three. And then by the Pythagorean equation, we get that this unknown side squared plus this plus one squared is equal to three squared. So solving using the Pythagorean equation, we're going to get that the other side is the square root of three squared minus one squared. That is the square root. Why did I write three cubed there? My apologies. That should be a three squared. So we get the square root of nine minus one, that is the square root of eight, which we use that are you could write this as the square root of four times two, that is two root two, either one would be acceptable. It doesn't matter so much. But now that we know all three sides of the triangle, we can then calculate the thing we're trying to do. So we're trying to compute tangent of theta. Tangent, as we know, is opposite over adjacent. So we take the opposite side, which is one over the adjacent side, which can take the square root of eight, or you can take two root two, whichever you want. And we get that this would be the ratio right here. You can rationalize the denominator if you need to, which you don't need to, let's be honest. One over two root two is perfectly fine. One over the square root of eight would also be acceptable. Now, this doesn't, we don't get obstructed even when there are variables in play here. What if we want to take cosine of arc tangent of x? Well, our strategy is going to be to draw a right triangle diagram and make a so-catoa argument to resolve things here. So we have an angle theta and theta here is going to be just cosine of x. So what we need to find is we're looking for cosine of theta. That's the question we're trying to solve right here. So what do we know about theta here? Well, we know that arc tangent of x is equal to theta, then taking tangent of both sides, we're going to end up with x equals tangent of theta, which you might get a little bit stuck there. It's like tangent theta equals x. I need a ratio, right? Well, every number can be written as a ratio. You can write this as x over one, like, oh, okay, that's brilliant. That's bloody brilliant there. In which case, once you make that recognition, like, okay, the opposite side is x, the adjacent side is one. And then so by the Pythagorean equation, the hypotenuse would then be the square root of x squared plus one. For which case, then if we want to do cosine, cosine is adjacent, adjacent over hypotenuse. Well, the adjacent side is one. The hypotenuse is the square root of x squared plus one. And so we can then take these trigonometric statements and actually turn them into algebraic functions, one over the square root of x squared plus one. Let me give you one more example of something like this. Take the expression three absolute value of sine theta. We want to simplify this. What does that even mean? Simplify it if theta equals arc tangent of x over three for some real number x, okay? Well, the first thing I want to mention here is what we know about theta. So theta is actually, it's actually given to us this time, right? We don't have to insert it. Theta is equal to arc tangent of x over three. Now arc tangent as a function, it's going to be bounded above by pi halves. And it's going to be bounded below, likewise, by negative pi halves, which means theta has those same constraints. Theta is going to be between pi halves and negative pi halves, which if you think of the unit circle, right? Like so, you have your first quadrant, zero to pi halves, and you have your fourth quadrant, negative pi halves to zero. On this side of the y axis, cosine's always positive. Cosine's just the x coordinate, right? So one thing we know here is that cosine of theta for this, for this theta, it's going to have to be positive, which since secant is just one, is this the reciprocal of cosine? This is likewise going to be positive. So it turns out, the first thing we can do here is when you take three times the absolute value of secant of theta, it turns out this is just three secant of theta. Because of the, because of the restrictions placed on tangent and therefore arc tangent, we know that this is for this choice of theta, secant's going to have to be positive, okay? The other thing we know is that we could, we could run a little triangle diagram here, just like we did on the previous slide. And you know, once you get good at these things, you can probably start doing this without having to draw the triangle, but you know, feel free to draw the triangle to help you organize stuff. I highly encourage it here. So we have that tangent theta is going to equal x over three, so opposite is x, j's into three. By the Pythagorean equation, we're going to get the square root of x squared plus nine. So then what's the secant ratio? Secant is the reciprocal of cosine, so you can use that if you want to. So we're going to get three over cosine theta. Or how I like to think of it is you're going to get three times. Well, if, if cosine is adjacent over hypotenuse, secant is going to be the reciprocal, so you get hypotenuse over adjacent. For which case we're going to get three times the hypotenuse, which is the square root of x squared plus nine over three, for which you can see then the three on top cancels with the three on the bottom. And so therefore three times the absolute value of secant of theta is equal to the square root of x squared plus nine. We can take this trigonometric statement and turn it into something purely algebraic. And so truly we simplify the expression because we had this trigonometric substitution that theta equals tangent inverse of x plus three.