Question

How do you find the exact value of `cos^-1(sin((7pi)/2))`?

Answer 1

`cos^-1(sin((7pi)/2)) = pi`

Explanation:

`cos^-1(sin((7pi)/2))` is

the `t` in `[0,pi]` with `cost = sin((7pi)/2)`

Since `sin((7pi)/2) = -1`, we need

the `t` in `[0,pi]` with `cost = -1`

So `t = pi`