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The principal value of $\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right)$ is
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Verified Answer
The correct answer is:
$-\frac{\pi}{3}$
Let $\theta=\sin ^{-1}\left[\sin \frac{5 \pi}{3}\right]$
$$
\begin{aligned}
\Rightarrow & \sin \theta=\sin \frac{5 \pi}{3}=\sin \left[2 \pi-\frac{\pi}{3}\right] \\
\Rightarrow & \sin \theta=-\sin \frac{\pi}{3}=\sin \left(\frac{-\pi}{3}\right) \\
&(\because \sin (-\theta)=-\sin \theta)
\end{aligned}
$$
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 $\frac{\pi}{2}$.
$$
\begin{aligned}
\Rightarrow & \sin \theta=\sin \frac{5 \pi}{3}=\sin \left[2 \pi-\frac{\pi}{3}\right] \\
\Rightarrow & \sin \theta=-\sin \frac{\pi}{3}=\sin \left(\frac{-\pi}{3}\right) \\
&(\because \sin (-\theta)=-\sin \theta)
\end{aligned}
$$
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 $\frac{\pi}{2}$.
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