Prove that:
$\cos \frac{\pi}{65} \cos \frac{2 \pi}{65} \cos \frac{4 \pi}{65} \cos \frac{8 \pi}{65} \cos \frac{16 \pi}{65} \cos \frac{32 \pi}{65}=\frac{1}{64}$
$\mathrm{LHS}=\cos \frac{\pi}{65} \cos \frac{2 \pi}{65} \cos \frac{4 \pi}{65} \cos \frac{8 \pi}{65} \cos \frac{16 \pi}{65} \cos \frac{32 \pi}{65}$
On dividing and multiplying by $2 \sin \frac{\pi}{65}$, we get
$=\frac{1}{2 \sin \frac{\pi}{65}} \times 2 \sin \frac{\pi}{65} \times \cos \frac{\pi}{65} \times \cos \frac{2 \pi}{65} \times \cos \frac{4 \pi}{65} \times \cos \frac{8 \pi}{65} \times \cos \frac{16 \pi}{65} \times \cos \frac{32 \pi}{65}$
$=\frac{2 \times \sin 2 \frac{\pi}{65}}{2 \times 2 \sin \frac{\pi}{65}} \times \cos \frac{2 \pi}{65} \times \cos \frac{4 \pi}{65} \times \cos \frac{8 \pi}{65} \times \cos \frac{16 \pi}{65} \times \cos \frac{32 \pi}{65}$
$=\frac{2 \times \sin 4 \frac{\pi}{65}}{2 \times 4 \sin \frac{\pi}{65}} \times \cos \frac{4 \pi}{65} \times \cos \frac{8 \pi}{65} \times \cos \frac{16 \pi}{65} \times \cos \frac{32 \pi}{65}$
$=\frac{2 \times \sin 8 \frac{\pi}{65}}{2 \times 8 \sin \frac{\pi}{65}} \times \cos \frac{8 \pi}{65} \times \cos \frac{16 \pi}{65} \times \cos \frac{32 \pi}{65}$
$=\frac{2 \times \sin 16 \frac{\pi}{65}}{2 \times 16 \sin \frac{\pi}{65}} \times \cos \frac{16 \pi}{65} \times \cos \frac{32 \pi}{65}$
$=\frac{2 \times \sin 32 \frac{\pi}{65}}{2 \times 32 \sin \frac{\pi}{65}} \times \cos \frac{32 \pi}{65}$
$=\frac{\sin 64 \frac{\pi}{65}}{64 \sin \frac{\pi}{65}}=\frac{\sin \left(\pi-\frac{\pi}{65}\right)}{64 \sin \frac{\pi}{65}}$
$=\frac{\sin \frac{\pi}{65}}{64 \sin \frac{\pi}{65}} \quad[\because \sin (\pi-\theta)=\sin \theta]$
$=\frac{1}{64}=\mathrm{RHS}$
Hence proved.
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