 It's not actually very important to be able to find the inverse sign or cosine of a value since any time you really need the value you can use a calculator. It will be more important to be able to simplify an expression involving an inverse trigonometric function. To do that, remember, always draw the right triangle or the unit circle. For example, let's say we want to find the sign of the inverse tangent of three-fifths. So, remember, in a right triangle with acute angle theta, sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. And so this means inverse tangent of three-fifths is an angle whose tangent is three-fifths. So let's draw a right triangle with opposite side three and adjacent side five. This gives us a right triangle and we know two of the side lengths. We can find the hypotenuse. And again, in our right triangle with acute angle theta, sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. So, and since we want sine, that will be the opposite side three divided by the hypotenuse square root of 34. How about the tangent of the inverse cosine of one-third? So the inverse cosine of one-third is an acute angle whose cosine is one-third. Let's draw the unit circle this time. So, remember, if x, y is the intersection of the angle with measured theta and the unit circle, then sine of theta is the y value, cosine of theta is the x value, and tangent is y divided by x. And since cosine is one-third, that means on the unit circle, this corresponds to a point with x coordinate one-third, which indicates an angle in the first quadrant. And the x coordinate of this point is our cosine value one-third. Since this is the unit circle, we know that every point satisfies the equation x squared plus y squared equals one. Now, we need both x and y, so we have our x value, so we can solve for y, and we take the positive value since our point is in quadrant one. And so, remember, our tangent is going to be y divided by x. And we know both y and x, and so we see that the tangent of the inverse cosine will be, which we can simplify. How about the sine of the arc c into five? Well, let's draw a right triangle where c-cant of theta is equal to five. So, it's helpful to remember that c-cant is hypotenuse over adjacent, cosecant is hypotenuse over hypotenuse, and so on. And since five is five divided by one, we can make the hypotenuse five and the adjacent side one. So, drawing our triangle, we have two sides of a right triangle, so we can find the third side. Now, we want to find the sine, which is going to be the opposite, square root twenty-four, divided by the hypotenuse, five. What if we have a negative argument, so the tangent of the arc cosine of negative one-third? Drawing a right triangle where cosine of theta is negative one-third is impossible, but we can still represent this on the unit circle. So, remember that if x, y is the intersection of the angle with measure theta and the unit circle, then sine is our y-coordinate, cosine is our x-coordinate, and tangent is y divided by x. The other thing we have to remember is that if theta is the arc cosine, then theta has to be an angle between zero and pi, and so that puts us in quadrant one or two. So, our cosine negative one-third is an angle whose cosine is negative one-third. So, the cosine is the x-value, and since our angle has to be between zero and pi, this corresponds to an angle in quadrant two. So, let's draw our picture, and we know the cosine value, which means we know the x-coordinate of our point. Since this is the unit circle, we know that every point satisfies the equation x squared plus y squared equals one. So, our cosine value is negative one-third, that's our x-value, and so we can find our y-value, and we take the positive square root because our point is in quadrant two. And, again, since we want our tangent value, that's going to be y divided by x, and so we find the tangent of the arc cosine is, which we can simplify. What if you don't have any numbers at all? So, how about the sine of the arc tangent of x? Again, we can always draw our right triangle, so remember arc tan x is an angle whose tangent is x. So, remember in a right triangle with the acute angle theta, tangent is opposite over adjacent, and since x can be written as x divided by one, we'll draw a triangle with opposite side x and adjacent side one. Now, we have two sides of a right triangle, so we can find the third side, where we take the positive square root because this third side should be a length. And, again, remember that sine is opposite over hypotenuse, and so that means that the sine of the arc tangent will be...