$\cos^{-1}(-x) = $
A$-\cos^{-1}(x)$
B$2\pi - \cos^{-1}(x)$
C$\pi - \cos^{-1}(x)$
D$\cos^{-1}(x)$
Answer & Solution
Correct answer: C. $\pi - \cos^{-1}(x)$
For $x \in [-1, 1]$: $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$.
Reason: $\cos(\pi - y) = -\cos y$. Setting $y = \cos^{-1}(x)$ gives $\cos(\pi - \cos^{-1} x) = -x$, so applying $\cos^{-1}$ to both sides (both LHS and target lie in $[0, \pi]$) gives the result.
Compare $\sin^{-1}(-x) = -\sin^{-1}(x)$ (option A would apply here). The odd-function form works for $\sin^{-1}$ but not $\cos^{-1}$ because their principal ranges differ.
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