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The value of $\sin^{-1}\bigg(\cos\frac{33\pi}{5}\bigg)$ is

$\begin{array}{1 1} \frac{3 \pi}{5} \\ \frac{-7 \pi }{5} \\ \frac{\pi}{10} \\ \frac{-\pi}{10} \end{array} $

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  • \(cos(2n\pi+\theta)=cos\theta\)
  • Principal interval of cos is [0,\(\pi\)]
  • \(cosx=sin(\frac{\pi}{2}-x)\)
  • Principal interval of sin is \([-\frac{\pi}{2},\frac{\pi}{2}]\)
Ans (D) \( \frac{-\pi}{10}\)
Reduce the angle \(\frac{33\pi}{5}\)into the principal interval of cos
\(\frac{33\pi}{5}=6\pi + \frac{3\pi}{5}\)
\( sin^{-1} \bigg( cos \bigg( \frac{33\pi}{5} \bigg) \bigg)\)
\( = sin^{-1} \bigg[ cos \bigg[ 6\pi + \frac{3\pi}{5} \bigg] \bigg] \)
By taking \(\theta=\frac{3\pi}{5}\), and n=3 in the above formula, we get
\(cos \big( 6\pi + \frac{3\pi}{5}\big)=cos\frac{3\pi}{5}\)
\( \Rightarrow\: sin^{-1}cos \frac{3\pi}{5} \)
\(= sin^{-1}sin \big[ \frac{\pi}{2}-\frac{3\pi}{5} \big] =\frac{\pi}{2}-\frac{3\pi}{5}\)
\( = -\frac{\pi}{10} \)
answered Feb 18, 2013 by thanvigandhi_1
edited Mar 8, 2013 by rvidyagovindarajan_1
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