Question

The value of $\cos \left(\frac{\pi }{7}\right)\cos \left(\frac{4\pi }{7}\right)\cos \left(\frac{5\pi }{7}\right)$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $-\frac{1}{8}$
• D. $\frac{1}{8}$

Answer 1

Answer: (D)

We have, $\cos \frac{\pi }{7}\cos \frac{4\pi }{7}\cos \frac{5\pi }{7}$
$=\cos \frac{\pi }{7}\cos \frac{4\pi }{7}\cos \left(\pi -\frac{2\pi }{7}\right)$
$=-\cos \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}$
$[\because \cos (\pi -\theta )=-\cos \theta ]$
$=\frac{1}{2\sin \frac{\pi }{7}}\left[2\sin \frac{\pi }{7}\cos \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\right]$
$=\frac{-1}{2\sin \frac{\pi }{7}}\left[\sin \frac{2\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\right]$
$\left(\because 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=\sin \theta \right)$
$=\frac{-1}{2^2\sin \frac{\pi }{7}}\left[2\sin \frac{2\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\right]$
$=\frac{-1}{2^2\sin \frac{\pi }{7}}\left[\sin \frac{4\pi }{7}\cos \frac{4\pi }{7}\right]$
$=\frac{-1}{2^3\sin \frac{\pi }{7}}\left[2\sin \frac{4\pi }{7}\cos \frac{4\pi }{7}\right]$
$=\frac{-1}{8\sin \frac{\pi }{7}}\left[\sin \frac{8\pi }{7}\right]$
$=\frac{\sin \left(\pi +\frac{\pi }{7}\right)}{8\sin \frac{\pi }{7}}$
$=\sin \frac{\frac{\pi }{7}}{8\sin \frac{\pi }{7}}=\frac{1}{8}$
$[\because \sin (\pi +\theta )=-\sin \theta ]$