Question

The value of $\cos \left[\frac{1}{2}{\cos }^{-1}\cos (-\frac{14\pi }{5})\right]$ is

• A. $\cos (-\frac{7\pi }{5})$
• B. $\sin \left(\frac{\pi }{10}\right)$
• C. $\cos \left(\frac{2\pi }{5}\right)$
• D. $-\cos \left(\frac{3\pi }{5}\right)$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\sin \left(\frac{\pi }{10}\right)$
C $\cos \left(\frac{2\pi }{5}\right)$
D $-\cos \left(\frac{3\pi }{5}\right)$

We have, $\frac{1}{2}co{s}^{-1}\left(cos\right(\frac{-14\pi }{5}\left)\right)$

$=\frac{1}{2}co{s}^{-1}\left(cos\right(\frac{14\pi }{5}\left)\right)$ ...(cos(-A)=cosA)

$=\frac{1}{2}co{s}^{-1}\left(cos\right(3\pi -\frac{\pi }{5}\left)\right)$

$=\frac{1}{2}(\pi -\frac{\pi }{5})$

$=\frac{2\pi }{5}$

Therefore

$cos\left(\frac{2\pi }{5}\right)$

$=cos(\pi -\frac{3\pi }{5})$

$=-cos\left(\frac{3\pi }{5}\right)$

Also $cos\left(\frac{2\pi }{5}\right)$

$=sin(\frac{\pi }{2}-\frac{2\pi }{5})$

$=sin\left(\frac{\pi }{10}\right)$