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प्रश्न
Prove that: \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64}\]
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उत्तर
On dividing and multiplying by \[2\sin\frac{\pi}{65}\] , we get
\[= \frac{1}{2\sin\frac{\pi}{65}} \times 2\sin\frac{\pi}{65} \times \cos\frac{\pi}{65} \times \cos\frac{2\pi}{65} \times \cos\frac{4\pi}{65} \times \cos\frac{8\pi}{65} \times \cos\frac{16\pi}{65} \times \cos\frac{32\pi}{65}\]
\[ = \frac{2 \times \sin2\frac{\pi}{65}}{2 \times 2\sin\frac{\pi}{65}} \times \cos\frac{2\pi}{65} \times \cos\frac{4\pi}{65} \times \cos\frac{8\pi}{65} \times \cos\frac{16\pi}{65} \times \cos\frac{32\pi}{65} \]
\[ = \frac{2 \times \sin4\frac{\pi}{65}}{2 \times 4\sin\frac{\pi}{65}} \times \cos\frac{4\pi}{65} \times \cos\frac{8\pi}{65} \times \cos\frac{16\pi}{65} \times \cos\frac{32\pi}{65}\]
\[ = \frac{2 \times \sin8\frac{\pi}{65}}{2 \times 8\sin\frac{\pi}{65}} \times \cos\frac{8\pi}{65} \times \cos\frac{16\pi}{65} \times \cos\frac{32\pi}{65}\]
\[= \frac{2 \times \sin16\frac{\pi}{65}}{2 \times 16\sin\frac{\pi}{65}} \times \cos\frac{16\pi}{65} \times \cos\frac{32\pi}{65}\]
\[ = \frac{2 \times \sin32\frac{\pi}{65}}{2 \times 32\sin\frac{\pi}{65}} \times \cos\frac{32\pi}{65}\]
\[ = \frac{\sin64\frac{\pi}{65}}{64\sin\frac{\pi}{65}} = \frac{\sin\left( \pi - \frac{\pi}{65} \right)}{64\sin\frac{\pi}{65}}\]
\[ = \frac{\sin\frac{\pi}{65}}{64\sin\frac{\pi}{65}} \left[ \because \sin\left( \pi - \theta \right) = sin\theta \right]\]
\[ = \frac{1}{64} = RHS\]
\[\text{ Hence proved } .\]
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