Find the value of sin left(frac{pi}{3}-sin ^{ | vedclass

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(C) Let $\sin ^{-1}\left(-\frac{1}{2}\right) = x$. Then,$\sin x = -\frac{1}{2}$. Since the range of the principal value branch of $\sin ^{-1}$ is $\le

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Let $\sin ^{-1}\left(-\frac{1}{2}\right) = x$. Then,$\sin x = -\frac{1}{2}$. Since the range of the principal value branch of $\sin ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,we have $\sin ^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$. Now,substitute this value into the expression: $\sin \left(\frac{\pi}{3} - \left(-\frac{\pi}{6}\right)\right) = \sin \left(\frac{\pi}{3} + \frac{\pi}{6}\right)$. $= \sin \left(\frac{2\pi + \pi}{6}\right) = \sin \left(\frac{3\pi}{6}\right) = \sin \left(\frac{\pi}{2}\right)$. Since $\sin \left(\frac{\pi}{2}\right) = 1$,the final value is $1$.

Find the value of $\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)$.

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