Find tan⁻¹{2cos(2sin⁻¹(1/2))}

Evaluate \( \tan^{-1}\left\{2\cos\left(2\sin^{-1}\left(\frac{1}{2}\right)\right)\right\} \)

Solution:

Let

\[ \theta = \sin^{-1}\left(\frac{1}{2}\right) \Rightarrow \theta = \frac{\pi}{6} \]

Now,

\[ 2\theta = \frac{\pi}{3} \]

\[ \cos(2\theta) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

Thus,

\[ 2\cos(2\theta) = 2 \cdot \frac{1}{2} = 1 \]

Hence,

\[ \tan^{-1}(1) = \frac{\pi}{4} \]

\[ \tan^{-1}\left\{2\cos\left(2\sin^{-1}\left(\frac{1}{2}\right)\right)\right\} = \frac{\pi}{4} \]

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