Prove that:

Question:

Prove that:

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

Solution:

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

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.