Question

Find :$\cos \frac{\pi }{65}\cos \frac{2\pi }{65}\cos \frac{4\pi }{65}\cos \frac{8\pi }{65}\cos \frac{16\pi }{65}\cos \frac{32\pi }{65}$.

• A. $\frac{1}{64}$
• B. $\frac{1}{32}$
• C. $\frac{1}{16}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{64}$

Multiply by $\frac{1}{2\sin \left(\pi /65\right)}\times 2\sin \left(\frac{\pi }{65}\right)$

$\therefore \frac{1}{2\sin \left(\pi /65\right)}\times 2\sin \left(\frac{\pi }{65}\right)\left[\cos \frac{\pi }{65}\cos \frac{2\pi }{65}\cos \frac{4\pi }{65}\cos \frac{8\pi }{65}\cos \frac{16\pi }{65}\cos \frac{32\pi }{65}\right]$

$\sin 2\theta =2\sin \theta \cos \theta$

$\frac{1}{2^2\sin \left(\pi /65\right)}\times 2\sin \frac{2\pi }{65}\times \cos \frac{2\pi }{65}\times \cos \frac{4\pi }{65}\times \cos \frac{8\pi }{65}\times \cos \frac{16\pi }{65}\times \cos \frac{32\pi }{65}$

$\frac{1}{2\times 2^2\sin \left(2\pi /65\right)}\times 2\sin \frac{4\pi }{65}\times \cos \frac{4\pi }{65}\times \cos \frac{8\pi }{65}\times \cos \frac{16\pi }{65}\times \cos \frac{32\pi }{65}$

$\frac{1}{2\times 2^3\sin \left(2\pi /65\right)}\times 2\sin \frac{8\pi }{65}\cos \frac{8\pi }{65}\cos \frac{16\pi }{65}\cos \frac{32\pi }{65}$

$\frac{1}{2^6\times \sin \left(\pi /65\right)}\times sin\left(\frac{6\pi \pi }{65}\right)=\frac{1}{64\sin \left(\pi /65\right)}\times \sin (\pi -\pi /65)$

$\therefore \sin \theta =\sin (\pi -\theta )$

$\frac{1}{64\sin \left(\pi /65\right)\times \sin }\left(\pi /65\right)=\frac{1}{64}$

Hence proved