Question

Prove that:
$\cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos \frac{3\pi }{15}\cos \frac{4\pi }{15}\cos \frac{5\pi }{15}\cos \frac{6\pi }{15}\cos \frac{7\pi }{15}=\frac{1}{128}$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\cos \frac{\mathrm{\pi }}{15}\cos \frac{2\mathrm{\pi }}{15}\cos \frac{4\mathrm{\pi }}{15}\cos \frac{3\mathrm{\pi }}{15}\cos \frac{5\mathrm{\pi }}{15}\cos \frac{6\mathrm{\pi }}{15}\cos \frac{7\mathrm{\pi }}{15} \\ =\cos \frac{\mathrm{\pi }}{15}\cos \frac{2\mathrm{\pi }}{15}\cos \frac{4\mathrm{\pi }}{15}\left(\cos \frac{3\mathrm{\pi }}{15}\cos \frac{6\mathrm{\pi }}{15}\right)\times \left(-\cos \frac{8\mathrm{\pi }}{15}\right) \\ =-\frac{1}{2}\left[\cos \frac{\mathrm{\pi }}{15}\cos \frac{2\mathrm{\pi }}{15}\cos \frac{4\mathrm{\pi }}{15}\cos \frac{8\mathrm{\pi }}{15}\right]\times \frac{1}{2}\times \left(\cos \frac{3\mathrm{\pi }}{15}\cos \frac{6\mathrm{\pi }}{15}\right) \\ =-\frac{1}{2}\times \frac{2^3}{2^4\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}\left[2\sin \frac{\mathrm{\pi }}{15}\cos \frac{\mathrm{\pi }}{15}\cos \frac{2\mathrm{\pi }}{15}\cos \frac{4\mathrm{\pi }}{15}\cos \frac{8\mathrm{\pi }}{15}\right]\times \frac{2}{2^2\times \sin {\displaystyle \frac{3\mathrm{\pi }}{15}}}\left(2\sin \frac{3\mathrm{\pi }}{15}\cos \frac{3\mathrm{\pi }}{15}\cos \frac{6\mathrm{\pi }}{15}\right) \\ =-\frac{2^3}{132\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}\left[\sin \frac{2\mathrm{\pi }}{15}\cos \frac{2\mathrm{\pi }}{15}\cos \frac{4\mathrm{\pi }}{15}\cos \frac{8\mathrm{\pi }}{15}\right]\times \frac{2}{4\sin {\displaystyle \frac{3\mathrm{\pi }}{15}}}\left(\sin \frac{6\mathrm{\pi }}{15}\cos \frac{6\mathrm{\pi }}{15}\right) \\ =-\frac{2^2}{32\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}\left[2\sin \frac{2\mathrm{\pi }}{15}\cos \frac{2\mathrm{\pi }}{15}\cos \frac{4\mathrm{\pi }}{15}\cos \frac{8\mathrm{\pi }}{15}\right]\times \frac{1}{4\sin {\displaystyle \frac{3\mathrm{\pi }}{15}}}\left(2\sin \frac{6\mathrm{\pi }}{15}\cos \frac{6\mathrm{\pi }}{15}\right) \end{gathered}$$
$$\begin{gathered} =-\frac{2}{32\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}\left[\sin \frac{8\mathrm{\pi }}{15}\cos \frac{8\mathrm{\pi }}{15}\right]\times \frac{\sin {\displaystyle \frac{12\mathrm{\pi }}{15}}}{4\sin \frac{3\mathrm{\pi }}{15}} \\ =-\frac{1}{32\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}\left[\sin \frac{16\mathrm{\pi }}{15}\right]\times \frac{\sin {\displaystyle \frac{12\mathrm{\pi }}{15}}}{4\sin {\displaystyle \frac{3\mathrm{\pi }}{15}}} \\ =-\frac{\sin \left(\pi +{\displaystyle \frac{\mathrm{\pi }}{15}}\right)}{128\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}\times \frac{\sin \left({\displaystyle \mathrm{\pi }-\frac{3\mathrm{\pi }}{15}}\right)}{\sin {\displaystyle \frac{3\mathrm{\pi }}{15}}} \\ =-\frac{-\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}{128\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}\times \frac{\sin {\displaystyle \frac{3\mathrm{\pi }}{15}}}{\sin {\displaystyle \frac{3\mathrm{\pi }}{15}}} \\ =\frac{1}{128} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$