 In this video, we present the solution to question number 11 for practice exam 2 for math 1060, in which case we're asked to evaluate the trigonometric expression sign of sine inverse of 3 fifths plus tangent inverse of 2. And we need to write the evaluation in exact form. So just throwing this in your calculator and getting some number doesn't do you a bit of good. So don't do that. So what we're going to do is we're going to first treat each of these trigonometric functions as angles. I'm going to call this first one alpha and the second one beta. You can call it A and B if you prefer. It doesn't really matter. So we're trying to compute sine of alpha plus sine of beta. Using the angle sum identity for sine, we say this is equivalent to sine of alpha cosine of beta plus cosine of alpha times sine of beta. In which case, then we have to do sine of alpha, cosine of alpha, cosine of beta, sine of beta. So we need to think of these angles. How do we, how do we compute these other ratios, right? So starting with sine inverse of three fifths here, start with alpha. Notice we've said that alpha equals sine inverse of three fifths. That equates to being, to mean to us that sine of alpha is equal to three fifths. So we're going to construct a triangle associated to exactly that value right there. So we have a right triangle for which then we identify that alpha is the triangle we're considering here. Sine is of course the opposite over hypotenuse by the Pythagorean equation. The other side is going to equal four. So we know that sine of alpha is three fifths. Then we also know that cosine of alpha, which we need to know is going to be four fifths. Great. Let's move on to tangent inverse of two, which is beta. So we have that beta is equal to tangent inverse of two. This tells us that tangent of beta is equal to two. And I'm going to write that as a fraction two over one. And so if I think of my triangle here, you're going to want to draw two triangles to help you out. Like so. This will be the triangle associated to angle beta. So tangent is opposite over hypotenuse. And then by the Pythagorean equation, the other side will be the square root of five. So we know tangent of beta. We need to know sine and cosine of beta. So sine of beta is going to equal two over the square root of five. You don't have to bother rationalizing the denominator or anything. Cosine of beta is going to equal then one over the square root of five. And so using these values, we can plug it into the formula above. So sine of alpha, remember was three fifths. Cosine of beta was one over the square root of five. Cosine of alpha was four fifths. And then cosine, excuse me, sine of beta was two over the square root of five. So multiplying these things together, you're going to get three over five root five. And then you're also going to get eight over five root five. For which you don't have to worry about rationalizing the denominator, but they do have a common denominator, so we can add those together. So the final answer is going to be 11 over five times the square root of five. Now you are allowed to calculate in this problem. What you can do is you can check with your calculator that 11 divided by five times square root of five is equivalent to this number if you had plugged it in into your calculator earlier, of course. And it doesn't actually matter if it's in degrees or radians in that regard because it'll just get a number in the end. But for full credit, you need to have this as the exact form 11 over five root five or something equivalent to that. And you also need to show these type of right triangle diagrams and how you unravel this thing to get 11 over five root five.