Question

The value of $16\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\cos \frac{14\pi }{15}$ is

Answer 1

$16\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\cos \frac{14\pi }{15}$

Divide and multiply by $\sin \frac{2\pi }{15}$

$=\frac{8\left(2\sin \frac{2\pi }{15}\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\cos \frac{14\pi }{15}\right)}{\sin \frac{2\pi }{15}}$

=$\frac{8\left(\sin \frac{4\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\cos \frac{14\pi }{15}\right)}{\sin \frac{2\pi }{15}}$ [ Since,$\sin 2\theta =2\sin \theta \cos \theta$ ]

=$\frac{4\times 2\sin \frac{4\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\cos \frac{14\pi }{15}}{\sin \frac{2\pi }{15}}$

=$\frac{2\times 2\sin \frac{8\pi }{15}\cos \frac{8\pi }{15}\cos \frac{14\pi }{15}}{\sin \frac{2\pi }{15}}$

=$\frac{2\sin \frac{16\pi }{15}\cos \frac{14\pi }{15}}{\sin \frac{2\pi }{15}}$

=$\frac{2\sin (\pi +\frac{\pi }{15}).\cos (\pi -\frac{\pi }{15})}{\sin \frac{2\pi }{15}}$

=$\frac{-2\sin \frac{\pi }{15}.(-\cos \frac{\pi }{15})}{\sin \frac{2\pi }{15}}$

=$\frac{2\sin \frac{\pi }{15}.\cos \frac{\pi }{15}}{\sin \frac{2\pi }{15}}$

=$\frac{\sin \frac{2\pi }{15}}{\sin \frac{2\pi }{15}}$

=$1$