Question

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

Answer 1

Let $y={\sin }^{-1}\left(\frac{1}{2}\right)$
$\Rightarrow \sin y=\frac{1}{2}$
$\Rightarrow \sin y=\sin \left(\frac{\pi }{6}\right)$
Range of principal value of ${\sin }^{-1}$ is $[\frac{-\pi }{2},\frac{\pi }{2}]$
$\therefore y=\frac{\pi }{6}$
Now,
${\tan }^{-1}\left[2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right]={\tan }^{-1}\left[2\cos (2\times \frac{\pi }{6})\right]$
${\tan }^{-1}\left[2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right]={\tan }^{-1}\left[2\cos \frac{\pi }{3}\right]$
${\tan }^{-1}\left[2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right]={\tan }^{-1}[2\times \frac{1}{2}]$
${\tan }^{-1}\left[2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right]={\tan }^{-1}\left[1\right]$
${\tan }^{-1}\left[2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right]={\tan }^{-1}\left[\tan \frac{\pi }{4}\right]$
${\tan }^{-1}\left[2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right]=\frac{\pi }{4}$