Find the principal values of sin ^{-1}left(-f | vedclass

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(C) Let $y = \sin ^{-1}\left(-\frac{1}{2}\right)$.Then,$\sin y = -\frac{1}{2}$.We know that $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$,so $-\sin

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $y = \sin ^{-1}\left(-\frac{1}{2}\right)$.Then,$\sin y = -\frac{1}{2}$.We know that $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$,so $-\sin \left(\frac{\pi}{6}\right) = \sin \left(-\frac{\pi}{6}\right)$.Thus,$\sin y = \sin \left(-\frac{\pi}{6}\right)$.The range of the principal value branch of $\sin ^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.Since $-\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,the principal value of $\sin ^{-1}\left(-\frac{1}{2}\right)$ is $-\frac{\pi}{6}$.

Find the principal values of $\sin ^{-1}\left(-\frac{1}{2}\right)$.

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