Question

How would one prove that:
arccos (-x) = π - arccos x for -1 ≤ x ≤ 1

Answer 1

Consider
`\arccos(-x)+\arccos(x)`, where
`x \in [-1,1]`.

This means there exists a
`\theta \in [0, \pi]` such that
`\cos \theta=x` and
`\cos (\pi -\theta)=-\cos \theta=-x`.

Thus, taking the inverse cosine of both of these equations, we get that:

`\arccos(x)+\arccos(-x)=\theta+(\pi-\theta)=\pi`.

Subtracting arccos(x) from both sides, we get the required result.