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$\sin^{-1}(\sin x) = x$ holds for $x$ in which interval?
A$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
BAll real $x$
C$[-\pi, \pi]$
D$[0, \pi]$
Answer & Solution
Correct answer: A. $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
$\sin^{-1}$ is the inverse of $\sin$ restricted to its principal-value range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. Inside that interval, $\sin^{-1}(\sin x)$ returns $x$ as expected.
Outside that interval, $\sin^{-1}(\sin x)$ folds back to whatever principal value matches: e.g. $\sin^{-1}(\sin \pi) = \sin^{-1}(0) = 0$, not $\pi$.
Identity $\sin(\sin^{-1} x) = x$ holds for all $x \in [-1, 1]$ (the domain side), no such restriction.
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