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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} $

Can you answer this question?
 
 

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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
 
\(\Rightarrow\:sin^{-1}sin\large\frac{3\pi}{5}=sin^{-1}sin\large\frac{2\pi}{5}=\large\frac{2\pi}{5}\)

 

answered Feb 18, 2013 by thanvigandhi_1
edited Mar 16, 2013 by thanvigandhi_1
 
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