Question

If $\sin x\ne 0$, prove that $\cos x\cos 2x\cos 4x\cos 8x=\frac{\sin \left(2^{4}x\right)}{2^4\sin x}$ and hence prove that $\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\cos \frac{16\pi }{15}=\frac{1}{16}$.

Answer 1

$\begin{array}{c}\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\cos \left(\pi +\frac{\pi }{15}\right)\\ \cos \frac{2\pi }{15}\cos \frac{\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\\ 2\sin \frac{\pi }{15}\cdot \cos \frac{\pi }{15}\cdot \cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\\ 2\sin \frac{\pi }{15}\\ =\frac{-\sin \frac{2\pi }{15}\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}}{2\sin \frac{\pi }{15}}\left[divide\&nby2\right]\\ \frac{-2\sin \frac{2\pi }{15}\cos \frac{\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}}{2\times 2\sin \frac{\pi }{15}}\\ \frac{-\sin \frac{4\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}}{4\sin \frac{\pi }{15}}\times \frac{2}{2}\end{array}$

similarly, process is continuous, then we get

$\begin{array}{c}=\frac{-\sin \frac{16\pi }{15}}{16\sin \frac{\pi }{15}}=-\frac{\sin \left(\pi +\frac{\pi }{15}\right)}{16\sin \frac{\pi }{15}}\\ \frac{\sin \frac{\pi }{15}}{16\sin \frac{\pi }{15}}=\frac{1}{16}\end{array}$