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Q.
The principal value of $\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right)$ is
Inverse Trigonometric Functions
Solution:
Let $\theta=\sin ^{-1}\left[\sin \frac{5 \pi}{3}\right]$
$\Rightarrow \sin \theta=\sin \frac{5 \pi}{3}=\sin \left[2 \pi-\frac{\pi}{3}\right]$
$\Rightarrow \sin \theta=-\sin \frac{\pi}{3}=\sin \left(\frac{-\pi}{3}\right)(\because \sin (-\theta)=-\sin \theta)$
Therefore, principal value of $\sin ^{-1}\left[\sin \frac{5 \pi}{3}\right]$ is $\frac{-\pi}{3}$, as
principal value of $\sin ^{-1} x$ lies between $\frac{-\pi}{2}$ and $\frac{\pi}{2}$.