 We're now going to talk more about inverse trig functions. So first, let's find the value of the following. Previously, whenever we had two trig expressions within one, we were able to cancel them out usually. However, cosine and inverse sine do not cancel out. So the moral of the story is anytime the inner trig function is an inverse trig function, we're going to make a triangle. So make a triangle with an angle theta, and then we'll apply cosine to that angle theta. So the goal is let's focus on the inside. Let's make a triangle. So when I make my triangle, I have an angle theta, and it's important to note that sine of theta is equal to negative 1 third. So 1 third, 1 over 3 opposite over hypotenuse, 1, 3. Now I need this missing side, which I can find by using Pythagorean Theorem. So we'll just call this side, call it x. I know that x squared plus 1 squared equals 3 squared. I can use Pythagorean Theorem because remember I am dealing with right triangles. x squared plus 1 equals 9. x squared equals 8. So x equals, remember 8 breaks up in the square root of 4 times square root of 2. That would be 2 square root of 2. So this missing side measure would be 2 square root of 2. Now the question is what quadrant does this angle theta lie in? Well, first off, because of the way inverse sine is designed, theta is in, has to be in quadrant 1 or quadrant 4. Because inverse sine is only defined for quadrant 1 and quadrant 4. Now, since sine is negative, since sine is negative, our angle theta is in quadrant 4. Our angle theta is in quadrant 4. So keep that in mind as we evaluate cosine. So now it's time. I used my inner portion to create a triangle. Now let's supply cosine to that angle theta. Cosine of theta, adjacent over hypotenuse. Json is 2 square root of 2, hypotenuse is 3. Now this is positive or negative. Remember theta is in quadrant 4. Cosine is positive in quadrant 4. So answer is positive since cosine is positive in quadrant 4. So 2 square root of 2 over 3 is the answer. Let's do another one. Inverse trig function on the inside. We will make a triangle. We will make a triangle from this inside portion here. So I hear if here cosine of theta is equal to negative 1 half. Now inverse cosine. Inverse cosine is only defined in quadrant 1 or 2. So theta is in quadrant 1 or quadrant 2. Which is it? You could guess or you could recall that, hey, the answer to cosine is negative. So since cosine is negative, theta is in quadrant 2. So we are in quadrant 2. And I'll go ahead and write out since we are in quadrant 2 and our final answer will be taking the tangent of an angle. Tangent is negative in quadrant 2. So we will slap a negative sign on our final answer. Let's draw our triangle. Use that inner inverse trig function to draw us a triangle with an angle theta. Cosine 1 over 2 adjacent over hypotenuse. So I have 1, 2. This is actually a 30, 60, 90 triangle. So without even using Pythagorean theorem, I know the other side is square root of 3. So I want to find tan of theta. Tan of the angle theta opposite over adjacent square root of 3 over 1. But remember tangent will be negative. So throw a negative sign out front. The answer is negative square root of 3. Negative square root of 3. How about another one? Except this time the inverse trig functions on the outside and the regular functions on the inside and no, we can't cancel them out. There are two different types of trig functions here. So the best thing to do here is let's focus on the inside here. Let's evaluate. We know from previous sections how to evaluate cosine of 7pi over 6. So keep in mind that 7pi over 6 is in quadrant 3. 7pi over 6 is in quadrant 3. So I mean since we're in quadrant 3, I'm dealing with cosine. Cosine of 7pi over 6 will be a negative answer. Well, I need to find a reference angle. Reference angle is if an angle is in quadrant 3, how do I move it to quadrant 1? So take away 180 degrees or take away pi radians. So my reference angle will be 7pi over 6 minus pi. That would be 7pi over 6 minus 6pi over 6. Which is pi over 6. So it turns out my expression is actually sine inverse of cosine of pi over 6. We'll be throwing a negative sign there shortly. What is cosine of pi over 6? Cosine of pi over 6. We have this 30 degree angle here. Across from it is 1. Bottom side is square root of 3. Hypotenuse would have to be 2. Cosine of pi over 6 adjacent over hypotenuse. Square root of 3 over 2. And it's got to be negative. So cosine of pi over 6, the original angle was in quadrant 3. So we do need a negative answer here. Now this looks just like something we saw in a previous section. My goal is now to figure out sine of what angle. Sine of what angle is negative square root of 3 over 2? Sine of what angle is negative square root of 3 over 2? Now with regard to this, inverse sine is only defined for negative quadrant 4 and quadrant 2. So theta is in the interval negative pi over 2. Since sine of theta is negative, this means that theta would have to be in quadrant 4. We want negative quadrant 4. Because that's where inverse sine is defined. So what are we going to do about this? Well, we're going to draw a triangle. That's what we're going to do with a mystery angle. Sine of the mystery angle is square root of 3 over 2. That would be opposite over hypotenuse. Square root of 3 over 2. So what is this question mark angle? What is this reference angle who is always across from square root of 3? That would be 60 degrees. That would be pi over 3. But we need to move it to quadrant 4. We need to move it to negative quadrant 4. Move to negative quadrant 4. Therefore, theta would be negative pi over 3. To move an angle from quadrant 1 to negative quadrant 4, you just make it the opposite sine. Make it negative. So theta is negative pi over 3. That is the value of this exciting inverse trig expression. So negative pi over 3 is the answer there. Well, now we're going to focus on inverse cosecant. Inverse cosecant and inverse cotangent. Inverse cosecant can only give you angles that are in negative pi over 2 and pi over 2 in that interval. Inverse cosecant can give you angles in quadrant 1 or 2. That's 0 to pi. And inverse cotangent can only give you angles in quadrant 1 or 2. 0 to pi. That being said, let's find cotangent inverse of square root of 3. What that means is cotangent of what angle is square root of 3 over 1. Alright, inverse cotangent is defined in quadrant 1 or 2. Since cotangent is positive, we are in quadrant 1. Our final answer, theta would be in quadrant 1. Let's draw a picture. We have our mystery angle or reference angle here that we're going to find, which that is our actual angle theta because this will be a quadrant 1 angle here. This question mark will be, and we keep the angle in quadrant 1. So cotangent is adjacent over opposite. So adjacent, opposite. Who is always across from the side with measure 1? The angle is 30 degrees pi over 6. And we want to keep that angle in 1. That's where we want our answer to be. So theta is pi over 6. The answer to my inverse trig expression would be pi over 6. Next, inverse secant. Secant of what angle is negative 2 over 1? Alright, so secant inverse is defined where? What angle measures? It's defined in quadrant 1 or 2. But you'll notice since secant is negative, our theta, our angle, our final answer is in 2. We'll be in quadrant 2. So we draw our picture with our mystery angle, which once we find our mystery angle, we will move it to quadrant 2, and that'll be the answer. Secant is the reciprocal of cosine. Cosine is adjacent over hypotenuse. Secant is hypotenuse over adjacent. So 2, 1, which means our other side would have to be square root of 3. Who is always across from the square root of 3? Who is always across from the square root of 3? What is my reference angle? What is my mystery angle? Well, that would be 60 degrees. That would be pi over 3. But we have to move this angle to quadrant 2. Move to quadrant 2. How do we do that? Well, our final answer, our quadrant 2 angle, will be pi minus the reference angle, minus the mystery angle that we found. This would be 3 pi over 3 minus pi over 3. Let's give you an answer of 2 pi over 3. Theta is 2 pi over 3. The answer to our inverse trig expression is 2 pi over 3. In this example, I'm doing cosecant of negative 1, inverse cosecant of negative 1. The issue is cosecant has the hypotenuse in it. And if I was to write 1 over 1, you can't have a hypotenuse the same length as a leg. So a recommendation here. Cosecant of theta equals negative 1. I'm going to change this into sine. 1 over sine equals negative 1. This would mean that after rearranging, 1 equals negative sine theta, which means sine theta equals negative 1. So where does sine equal negative 1? Well, first and foremost, we need to lay some ground rules here. Inverse cosecant is defined for or defined in quadrant 1 or quadrant 4. That would be negative quadrant 4. Since cosecant is negative, our theta, our answer, is in quadrant 4. Negative quadrant 4. All right, so when I draw my unit circle, I have 1, 0. I have 0, 1. I have negative 1, 0. I have 0, negative 1. Where is sine negative 1? Well, where is the second coordinate negative 1? That'd be down here at the bottom. This would be not 3 pi over 2, because I want negative quadrant 4. It would be negative pi over 2. In this example, theta is equal to negative pi over 2. So the answer to my inverse trig expression is negative pi over 2. So you got some more examples of how to evaluate inverse trig functions. I hope you enjoyed. Thank you for watching.