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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Using principal value, evaluate the following : \[ sin^{-1} \bigg( sin \frac{3\pi}{5} \bigg) \]

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  • Principal interval of sin is \([-\frac{\pi}{2},\frac{\pi}{2}]\)
  • \(sin(\pi-\theta)=sin\theta\)
\(\frac{3\pi}{5}\notin\:[-\frac{\pi}{2},\frac{\pi}{2}]\) which is principal interval of sin
Therefore, reduce it within the principal interval.
We know that \(sin(\pi-\theta)=sin\theta\) By taking \(\theta=\frac{3\pi}{5}\) we get
\(sin\frac{3\pi}{5}=sin(\pi-\frac{3\pi}{5})\)
\(=sin\frac{2\pi}{5}\)
\(\frac{2\pi}{5}\in[-\frac{\pi}{2},\frac{\pi}{2}]\)
\(\Rightarrow\:sin^{-1}sin\frac{3\pi}{5}=sin^{-1}sin\frac{2\pi}{5}\)
=\(\frac{2\pi}{5}\)
answered Mar 19, 2013 by rvidyagovindarajan_1
 

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