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If $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}$, then $\sin ^{-1}(\sin x)$ is equal to
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Verified Answer
The correct answer is:
$\pi-x$
We have $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}$
$\begin{aligned}
& \Rightarrow \frac{-\pi}{2} \leq x-\pi \leq \frac{\pi}{2} \Rightarrow \frac{-\pi}{2} \leq \pi-x \leq \frac{\pi}{2} \\
& \Rightarrow \sin ^{-1}\{\sin (\pi-x)\}=\pi-x
\end{aligned}$
$\begin{aligned}
& \Rightarrow \frac{-\pi}{2} \leq x-\pi \leq \frac{\pi}{2} \Rightarrow \frac{-\pi}{2} \leq \pi-x \leq \frac{\pi}{2} \\
& \Rightarrow \sin ^{-1}\{\sin (\pi-x)\}=\pi-x
\end{aligned}$
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