Mark the tick against the correct answer in the following:
The value of $\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$ is
A. $\frac{2 \pi}{3}$
B. $\frac{5 \pi}{3}$
C. $\frac{\pi}{3}$
D. none of these
To Find: The value of $\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$
Now, let $x=\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$
$\Rightarrow \sin x=\sin \left(\frac{2 \pi}{3}\right)$
Here range of principle value of sine is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\Rightarrow x=\frac{2 \pi}{3} \notin\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Hence for all values of $x$ in range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the value of
$\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$ is
$\Rightarrow \sin x=\sin \left(\pi-\frac{\pi}{3}\right)\left(\because \sin \left(\frac{2 \pi}{3}\right)=\sin \left(\pi-\frac{\pi}{3}\right)\right)$
$\Rightarrow \sin x=\sin \left(\frac{\pi}{3}\right)(\because \sin (\pi-\theta)=\sin \theta$ as here $\theta$ lies in II quadrant and sine is positive)
$\Rightarrow x=\frac{\pi}{3}$
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