Question
Evaluate the following:
$\sin^{-1}\Big(\sin\frac{5\pi}{6}\Big)$

Answer

We know
$\sin\big(\sin^{-1}\theta\big)=\theta$ if $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$
We have
$\sin^{-1}\Big(\sin\frac{5\pi}{6}\Big)=\sin^{-1}\Big\{\sin\Big(\pi-\frac{\pi}{6}\Big)\Big\}$
$=\sin^{-1}\Big(\sin\frac{\pi}{6}\Big)$
$=\frac{\pi}{6}$

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