sin ^{-1}left(sin frac{7 pi}{6}right) = | vedclass

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Question

(C) We know that the principal value branch of $\sin ^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Given expression is $\sin ^{-1}(\sin \frac{7 \pi}{6

Vedclass · Accepted Answer

$-\frac{\pi}{6}$

Vedclass · Answer

(C) We know that the principal value branch of $\sin ^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Given expression is $\sin ^{-1}(\sin \frac{7 \pi}{6})$.We can write $\frac{7 \pi}{6}$ as $\pi + \frac{\pi}{6}$.So,$\sin(\frac{7 \pi}{6}) = \sin(\pi + \frac{\pi}{6}) = -\sin(\frac{\pi}{6})$.Now,$\sin ^{-1}(\sin \frac{7 \pi}{6}) = \sin ^{-1}(-\sin \frac{\pi}{6}) = \sin ^{-1}(\sin(-\frac{\pi}{6}))$.Since $-\frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the value is $-\frac{\pi}{6}$.

$\sin ^{-1}\left(\sin \frac{7 \pi}{6}\right) = $

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