Hey, everyone. We've been evaluating composite trig functions using the unit circle, but what if you were given a composite trig function like this one: the sine of the inverse tangent of three-fourths? Three-fourths is not a value that is directly on our unit circle. So, how can we go about evaluating this? Well, you may be worried that this sort of problem is going to be really tricky, but we can actually just solve this by drawing a right triangle, and then we can use the absolute basics of right triangles that we already know in order to come to a solution. So here, I'm going to walk you through exactly how to solve these problems step by step, and soon, you'll be able to evaluate any composite trig function, whether you can use the unit circle or not. So let's go ahead and get started here.
Now, here in this example, we're tasked with evaluating the composite trig function, the sine of the inverse tangent of three-fourths. Now remember, for any composite trig function, we're always going to start with that inside function, which here is the inverse tangent of three-fourths. So we can also think of this by remembering how we think about inverse trig functions, as the tangent of what angle theta gives me a value of three-fourths. So, for what angle is this true? Well, we can actually just make this angle by drawing a right triangle. Now, here, my right triangle is in quadrant 1, which makes sense because we're taking the inverse tangent of a positive value. So if I draw my angle theta right here and I want the tangent of this angle theta to be equal to three-fourths, I can just label my opposite side as 3, my adjacent side as 4, and the tangent of it is three-fourths. Now you'll notice here that I don't actually care about the value of the angle, but just that its tangent is equal to three-fourths.
Now from here, I'm just left to find the sine of this angle and evaluate that outside function. Now, in order to get the sine based on SOHCAHTOA, I know that I need my opposite side and my hypotenuse. So, I first need to find my hypotenuse, which I can do using the Pythagorean theorem. Or here, you might also notice that since my leg lengths are 3 and 4, that means my hypotenuse has to be 5 because this is a 3, 4, 5 triangle. Now, from here, I can take that opposite side 3 divided by my hypotenuse 5 in order to get my final solution. The sine of the inverse tangent of three-fourths is equal to 3/5.
Now let's recap how we got there. Here, we took our inside function, used it to draw a right triangle, and then found our missing side length, allowing us to evaluate that outside function and come to a final solution. Now, not all of these problems will be quite as straightforward. So here, I'm going to walk you through some steps that will work for any composite trig function for which you can't use the unit circle.
So let's come down to this example here. Here, we're asked to evaluate the expression the sine of the inverse cosine of negative 5 over 13. Now, you'll notice here that our argument is negative. And that's kind of what complicates this a little bit. But no worries. We're going to walk through some steps here. Now, if you're ever unsure whether or not you should use the unit circle or draw a right triangle, if you look at your composite trig function and you see that your inside function is an inverse function and your argument is kind of a strange number that you don't recognize as being on the unit circle, that's usually a good hint that you should draw a right triangle, because the unit circle is not going to help you here.
So here, seeing both of those things are true, let's go ahead and get started with our steps and get to drawing this right triangle. Now, of course, we're going to start by using our inside function. This is true of any composite trig function. So, in step 1, we're going to take that inside trig function, and we are going to use our interval in order to determine what quadrant we will eventually draw our right triangle in. So here, my inside function is the inverse cosine. And for the inverse cosine, I know that my values have to be between negative one and one, and my angles between 0 and pi.
Now looking at my coordinate system based on that angle interval, I know that my angle has to be in quadrant 1 or quadrant 2 because this is my interval from 0 to pi. Now here's where we want to consider the sign of our argument. And here we're taking the inverse cosine of a negative value. Our argument is negative here. Now based on our intervals, we're either in quadrant 1 or quadrant 2, but cosine values can only be negative in quadrant 2. So that's where my triangle is going to go, and we've completed step number 1. Quadrant 2 is where we'll draw our triangle, moving on to step 2.
So we're actually going to draw this triangle, and then we're going to use our argument in order to label our angle and our two sides. So here in quadrant 2, I'm going to go ahead and draw my right triangle. Now, the orientation of your right triangle might change just depending on what quadrant you're in. But you're always going to draw it with one side length against the x-axis so that your angle can be measured from that x-axis. So here, this is where my angle theta is. And looking at my argument negative 5 over 13, this tells me that the cosine of my angle that I labeled has to be equal to negative 5 over 13 in order to satisfy this inside function. So using SOHCAHTOA, I know that that means that my adjacent side has to be negative 5 and my hypotenuse has to be 13.
Now, it might be kind of weird here that I labeled this side length as being a negative value. But if we just think about where we're located on our coordinate system, my x values are negative here in quadrant 2, so it makes sense that this side length is negative. So always consider your location on the graph when you're looking at the signs of your values. So now we've completed step number 2. We can move on to step number 3 and use the Pythagorean theorem in order to find that missing third side. Now, here, our missing side is a leg length, so we're either solving for a or b. It doesn't matter which one. So I'm going to go ahead and set up my Pythagorean theorem, a squared plus b squared equals c squared. And I'm going to be solving for a here. So I have a squared. I'm going to plug in that other side length of negative 5, square that, and that is equal to my hypotenuse, which is 13 squared. Now from here, we can just do some algebra. A squared plus negative five squared gives me 25, and that's equal to 169. Now subtracting 25 from both sides cancels on that left side, leaves me with a squared equals 144, having subtracted that 25 from 169. Now, from here, my last step is to take the square root, giving me my side length of 12, which I can go ahead and label up here on my triangle that missing side length of 12.
Now, we've completed step number 3, and we can move on to our final step where we're going to use our triangle in order to evaluate that outside function. So moving on to that outside function, which here is the sine. So here we want to find the sine of our angle theta that we have labeled on our triangle here. And I know that using SOHCAHTOA, that means I want to take my opposite side, divide it by my hypotenuse, and I have all the information I need. So my opposite side here is 12, and then my hypotenuse is 13. So that gives me my final answer of 12 over 13. So that tells me the sine of the inverse cosine of negative 5 over 13 is 12 over 13.
Now, you may notice here that our argument started out as negative. We were taking the sine of the inverse cosine of a negative value, but then we ended up with a final answer that's positive. But again, think about where we're located on the graph. In quadrant 2, my sine values are going to be positive, so this makes perfect sense. So now you should be able to evaluate any composite trig function, whether you can use the unit circle or not. Let's continue practicing together. Thanks for watching, and I'll see you in the next one.