Question

The value of $\cos \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \frac{8\pi }{5}$ is equal to

• A. $\frac{1}{16}$
• B. 0
• C. $\frac{-1}{8}$
• D. $\frac{-1}{16}$

Answer 1

Answer: (D)

We have,$\cos \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \frac{8\pi }{5}$
$=\frac{1}{2\sin \frac{\pi }{5}}2\sin \frac{\pi }{5}\cos \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \frac{8\pi }{5}$
$=\frac{1}{2\sin \frac{\pi }{5}}\sin \frac{2\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \frac{8\pi }{5}$
$=\frac{1}{4\sin \frac{\pi }{5}}\sin \frac{4\pi }{5}\cos \frac{4\pi }{5}\cos \frac{8\pi }{5}$
$=\frac{1}{8\sin \frac{\pi }{5}}\sin \frac{8\pi }{5}\cos \frac{8\pi }{5}$
$=\frac{\sin \frac{16\pi }{5}}{16\sin \frac{\pi }{5}}$
$=\frac{\sin \left(3\pi +\frac{\pi }{5}\right)}{16\sin \frac{\pi }{5}}$
$=\frac{-\sin \frac{\pi }{5}}{16\sin \frac{\pi }{5}}$
$=-\frac{1}{16}$