Question

Find the value of : $\cos [\frac{\pi }{6}+2{\tan }^{-1}\left(1\right)]+\sin [3{\sin }^{-1}\left(\frac{1}{2}\right)+2{\cos }^{-1}\left(\frac{1}{2}\right)]$

Answer 1

We have;

$\cos [\frac{\pi }{6}+2{tan}^{-1}\left(1\right)]+\sin [3{\sin }^{-1}\left(\frac{1}{2}\right)+2{\cos }^{-1}\left(\frac{1}{2}\right)]$

$=\cos [\frac{\pi }{6}+\frac{2\pi }{4}]+\sin [3\times \frac{\pi }{6}+2\times \frac{\pi }{3}]$

$=\cos [\frac{\pi }{2}+\frac{\pi }{6}]+\sin [\frac{\pi }{2}+\frac{2\pi }{3}]$

$=-\sin \left(\frac{\pi }{6}\right)+\sin \left(\frac{7\pi }{6}\right)$

$=-\frac{1}{2}+\sin (\pi +\frac{\pi }{6})$

$=-\frac{1}{2}-\sin \left(\frac{\pi }{6}\right)$

$=-\frac{1}{2}-\frac{1}{2}=-1$

Hence;

$\cos [\frac{\pi }{6}+2{tan}^{-1}\left(1\right)]+\sin [3{\sin }^{-1}\left(\frac{1}{2}\right)+2{\cos }^{-1}\left(\frac{1}{2}\right)]=-1$