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Find the value of \( \tan^{-1} \bigg[ 2\cos \bigg(2\sin^{-1}\frac{1}{2}\bigg) \bigg] \)

$\begin{array}{1 1} \frac{\pi}{4} \\ \frac{\pi}{2} \\ \frac{\pi}{8} \\ \pi \end{array} $

1 Answer

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  • \( sin\large\frac{\pi}{6}=\frac{1}{2}\)
  • \( cos\large\frac{\pi}{3}=\frac{1}{2}\)
  • \(tan^{-1}1=\large\frac{\pi}{4}\)
Given $tan^{-1} \bigg[ 2cos \bigg(2sin^{-1}\large\frac{1}{2}\bigg) \bigg]$:
Let \(x=sin^{-1}\large\frac{1}{2}\). Then \(sinx= \large\frac{1}{2}\Rightarrow\:x=\frac{\pi}{6}\)
The given expression becomes \( tan^{-1} \bigg[ 2cos \bigg( 2.\large\frac{\pi}{6} \bigg) \bigg]\) \(= tan^{-1} \bigg[ 2cos \bigg(\large\frac{\pi}{3} \bigg) \bigg]\).
Given that \( cos\large\frac{\pi}{3}=\large\frac{1}{2}\), the expression reduces to $tan^{-1} \bigg[ 2.\large\frac{1}{2} \bigg] = tan^{-1}1$
We know that \(tan^{-1}1 = \large\frac{\pi}{4}\)


answered Feb 22, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1

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