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

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

Can you answer this question?

Toolbox:
• $sin\large\frac{\pi}{6}=\frac{1}{2}$
• $cos\large\frac{\pi}{3}=\frac{1}{2}$
• $tan^{-1}1=\large\frac{\pi}{4}$
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
edited Mar 15, 2013