Question

How do you find the exact value of `arccos(cos(pi/3))`?

Answer 1

`0 <= pi/3 <= pi`

So `pi/3` is in the range of `arccos` and `arccos(cos(pi/3)) = pi/3`

Explanation:

If we restrict the domain of `cos` to `[0, pi]` then it is a one-one function onto its range `[-1, 1]` with inverse `arccos`.

So `arccos` is defined to have range `[0, pi]`,

If `theta in [0, pi]` then `arccos(cos(theta)) = theta`

If `theta in [-pi, 0]` then `arccos(cos(theta)) = -theta`

If `theta = varphi + 2n pi` for some `n in ZZ` then `arccos(cos(theta)) = arccos(cos(varphi))`