Question

Show that:

$16\cos \frac{2\pi }{15}.\cos \frac{4\pi }{15}.\cos \frac{8\pi }{15}.\cos \frac{14\pi }{15}=1$

Answer 1

$16\cos \frac{2\pi }{15}.\cos \frac{4\pi }{15}.\cos \frac{8\pi }{15}.\cos \frac{14\pi }{15}=1$
$\frac{2\sin \frac{2\pi }{15}}{2\sin \frac{2\pi }{15}}\left(16\cos \frac{2\pi }{15}.\cos \frac{4\pi }{15}.\cos \frac{8\pi }{15}.\cos \frac{14\pi }{15}\right)$
$\frac{16\times \sin \frac{4\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\cos \frac{14\pi }{15}}{2\sin \frac{2\pi }{15}}$
$\frac{16\times 2\sin \frac{8\pi }{15}\cos \frac{8\pi }{15}\cos \frac{14\pi }{15}}{8\sin \frac{2\pi }{15}}$
$\frac{16\times 2\sin \frac{16\pi }{15}\cos \frac{14\pi }{15}}{16\sin \frac{2\pi }{15}}$
$\frac{\sin 2\pi +\sin \frac{2\pi }{15}}{\sin \frac{2\pi }{15}}=\frac{0+\sin \frac{2\pi }{15}}{\sin \frac{2\pi }{15}}=1$