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Q)

The value of $\sin^{-1}\bigg(\sin\frac{3\pi}{5}\bigg)$ is ________________.

$\begin{array}{1 1} \frac{3 \pi}{5} \\ \frac{-3 \pi }{5} \\ \frac{2 \pi}{5} \\ \frac{-2\pi}{5} \end{array} $

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A)
Toolbox:
  • Principal interval of sin is $ [\large\frac{\pi}{2},-\large\frac{\pi}{2}]$
  • $\sin(\pi-\theta)=\sin\theta$
Ans : $\large \frac{2\pi}{5}$
 
$\large\frac{3\pi}{5}$ is not in the principal interval of sin.
Therefore, reduce it to its principal interval
 
$\large\frac{3\pi}{5}= \pi - \large\frac{2\pi}{5}$
$\sin\large\frac{3\pi}{5} =\sin(\pi- \large\frac{2\pi}{5})=\sin\large\frac{2\pi}{5} $
$\large\frac{2\pi}{5}$ is within the principal interval
 
$\Rightarrow\:\sin^{-1}\sin\large\frac{3\pi}{5}=\sin^{-1}\sin\large\frac{2\pi}{5}=\large\frac{2\pi}{5}$
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