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.