Question:
What is the principal value of $\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right) ?$
Solution:
Let $y=\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
Then,
$\sin y=-\frac{\sqrt{3}}{2}=\sin \left(-\frac{\pi}{3}\right)$
$y=-\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Here, $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is the range of the principal value branch of inverse sine function.
$\therefore \sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}$
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