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

1 Answer

5 votes

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.

User Kohei TAMURA
by
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