Question

Evaluate: $\cos \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \frac{8\pi }{5}=$

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

Answer 1

The correct option is D $-\frac{1}{16}$

$\cos \frac{\pi }{5},\cos \frac{2\pi }{5},\cos \frac{4\pi }{5},\cos \frac{8\pi }{5}$

Let, $\frac{\pi }{5}=\theta$ $\because \pi =5\theta$

$⟹\cos \theta .\cos 2\theta .\cos 4\theta .\cos 8\theta =\frac{(2\sin \theta .\cos \theta .\cos 2\theta .\cos 4\theta .\cos 8\theta )}{2\sin \theta }$

$⟹\frac{\sin 2\theta .\cos 2\theta .\cos 4\theta .\cos 8\theta }{2\sin \theta }=\frac{2\sin 2\theta .\cos 2\theta .\cos 4\theta .\cos 8\theta }{4\sin \theta }$

$=\frac{\sin 4\theta .\cos 4\theta .\cos 8\theta }{4\sin \theta }$

$⟹\frac{2\sin 4\theta .\cos 4\theta .\cos 8\theta }{2\times 4\sin \theta }=\frac{2\sin 8\theta .\cos 8\theta }{16\sin \theta }$

$⟹\frac{\sin 16\theta }{16\sin \theta }=\frac{\sin (3\times 5\theta +\theta )}{16\sin \theta }$

$⟹\frac{-\sin \theta }{16\sin \theta }=\frac{-1}{16}$