Asap now will mark brainiest lots of 22. explain why an sin^-1(sin(3pi/4))not equal 3pi/4 when y=sin x and y=sin ^-1 are inverses. 25. given f(x)=sin x and g(x)= cos x, find f(g(pi/2))=0. show all your steps.

22. On its own, is not invertible because it's not one-to-one because it's periodic. For instance, we can always find more than one value of for which ; this happens when .

But we can restrict the domain so that it can become invertible. If we only allow values of within , for example, then each will only be associated with a single value of . This is how the standard inverse sine is defined. With (restricted domain of sine), we guarantee that (range). So the domain of the inverse is , and the range of the inverse is .

is not in this restricted domain. But still exists as long as we take the standard domain (the entire real line), and . But then because this is the only value of for which .

In short: and are NOT inverses of one another, but rather one is an imperfect inverse of the other.

25. Not much to say about this:

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3