sin ^{-1}left(sin frac{23 pi}{6}right) = . | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We know that the range of the principal value branch of $\sin ^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Given expression is $\sin ^{-1}(\sin \frac{23 \pi}{6})$.First,simplify the angle $\frac{23 \pi}{6}$:$\frac{23 \pi}{6} = \frac{24 \pi - \pi}{6} = 4 \pi - \frac{\pi}{6}$.Since $\sin(4 \pi - 	heta) = -\sin 	heta$,we have:$\sin(\frac{23 \pi}{6}) = \sin(4 \pi - \frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = \sin(-\frac{\pi}{6})$.Now,$\sin ^{-1}(\sin(-\frac{\pi}{6})) = -\frac{\pi}{6}$,which lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.Thus,the correct option is $B$.

$\sin ^{-1}\left(\sin \frac{23 \pi}{6}\right) = $ . . . . . . .

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