Question

Prove that:
$\cos \frac{\pi }{65}\cos \frac{2\pi }{65}\cos \frac{4\pi }{65}\cos \frac{8\pi }{65}\cos \frac{16\pi }{65}\cos \frac{32\pi }{65}=\frac{1}{64}$

Answer 1

$\mathrm{LHS}=\cos \frac{\mathrm{\pi }}{65}\cos \frac{2\mathrm{\pi }}{65}\cos \frac{4\mathrm{\pi }}{65}\cos \frac{8\mathrm{\pi }}{65}\cos \frac{16\mathrm{\pi }}{65}\cos \frac{32\mathrm{\pi }}{65}$

On dividing and multiplying by $2\sin \frac{\mathrm{\pi }}{65}$, we get

$$\begin{gathered} =\frac{1}{2\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\times 2\sin \frac{\mathrm{\pi }}{65}\times \cos \frac{\mathrm{\pi }}{65}\times \cos \frac{2\mathrm{\pi }}{65}\times \cos \frac{4\mathrm{\pi }}{65}\times \cos \frac{8\mathrm{\pi }}{65}\times \cos \frac{16\mathrm{\pi }}{65}\times \cos \frac{32\mathrm{\pi }}{65} \\ =\frac{2\times \sin 2{\displaystyle \frac{\mathrm{\pi }}{65}}}{2\times 2\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\times \cos \frac{2\mathrm{\pi }}{65}\times \cos \frac{4\mathrm{\pi }}{65}\times \cos \frac{8\mathrm{\pi }}{65}\times \cos \frac{16\mathrm{\pi }}{65}\times \cos \frac{32\mathrm{\pi }}{65} \\ =\frac{2\times \sin 4{\displaystyle \frac{\mathrm{\pi }}{65}}}{2\times 4\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\times \cos \frac{4\mathrm{\pi }}{65}\times \cos \frac{8\mathrm{\pi }}{65}\times \cos \frac{16\mathrm{\pi }}{65}\times \cos \frac{32\mathrm{\pi }}{65} \\ =\frac{2\times \sin 8{\displaystyle \frac{\mathrm{\pi }}{65}}}{2\times 8\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\times \cos \frac{8\mathrm{\pi }}{65}\times \cos \frac{16\mathrm{\pi }}{65}\times \cos \frac{32\mathrm{\pi }}{65} \end{gathered}$$

$$\begin{gathered} =\frac{2\times \sin 16{\displaystyle \frac{\mathrm{\pi }}{65}}}{2\times 16\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\times \cos \frac{16\mathrm{\pi }}{65}\times \cos \frac{32\mathrm{\pi }}{65} \\ =\frac{2\times \sin 32{\displaystyle \frac{\mathrm{\pi }}{65}}}{2\times 32\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\times \cos \frac{32\mathrm{\pi }}{65} \\ =\frac{\sin 64{\displaystyle \frac{\mathrm{\pi }}{65}}}{64\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}=\frac{\sin \left(\mathrm{\pi }-{\displaystyle \frac{\mathrm{\pi }}{65}}\right)}{64\sin {\displaystyle \frac{\mathrm{\pi }}{65}}} \\ =\frac{\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}{64\sin {\displaystyle \frac{\mathrm{\pi }}{65}}}\left[\because \sin \left(\mathrm{\pi }-\mathrm{\theta }\right)=\mathrm{sin\theta }\right] \\ =\frac{1}{64}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$