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प्रश्न
Prove that: \[\cos\frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos\frac{6\pi}{15} \cos \frac{7\pi}{15} = \frac{1}{128}\]
उत्तर
\[LHS = \cos\frac{\pi}{15} \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{3\pi}{15} \cos\frac{5\pi}{15} \cos\frac{6\pi}{15} \cos\frac{7\pi}{15}\]
\[ = \cos\frac{\pi}{15} \cos\frac{2\pi}{15} \cos\frac{4\pi}{15}\left( \cos\frac{3\pi}{15} \cos\frac{6\pi}{15} \right) \times \left( - \cos\frac{8\pi}{15} \right)\]
\[ = - \frac{1}{2}\left[ \cos\frac{\pi}{15} \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \right] \times \frac{1}{2} \times \left( \cos\frac{3\pi}{15} \cos\frac{6\pi}{15} \right)\]
\[ = - \frac{1}{2} \times \frac{2^3}{2^4 \sin\frac{\pi}{15}}\left[ 2\sin\frac{\pi}{15}\cos\frac{\pi}{15} \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \right] \times \frac{2}{2^2 \times \sin\frac{3\pi}{15}} \left( 2\sin\frac{3\pi}{15}\cos\frac{3\pi}{15} \cos\frac{6\pi}{15} \right)\]
\[ = - \frac{2^3}{132\sin\frac{\pi}{15}}\left[ \sin\frac{2\pi}{15} \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \right] \times \frac{2}{4\sin\frac{3\pi}{15}} \left( \sin\frac{6\pi}{15} \cos\frac{6\pi}{15} \right)\]
\[ = - \frac{2^2}{32\sin\frac{\pi}{15}}\left[ 2\sin\frac{2\pi}{15} \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{8\pi}{15} \right] \times \frac{1}{4\sin\frac{3\pi}{15}} \left( 2\sin\frac{6\pi}{15} \cos\frac{6\pi}{15} \right)\]
\[= - \frac{2}{32\sin\frac{\pi}{15}}\left[ \sin\frac{8\pi}{15} \cos\frac{8\pi}{15} \right] \times \frac{\sin\frac{12\pi}{15}}{4\sin\frac{3\pi}{15}}\]
\[ = - \frac{1}{32\sin\frac{\pi}{15}}\left[ \sin\frac{16\pi}{15} \right] \times \frac{\sin\frac{12\pi}{15}}{4\sin\frac{3\pi}{15}}\]
\[ = - \frac{\sin\left( \pi + \frac{\pi}{15} \right)}{128\sin\frac{\pi}{15}} \times \frac{\sin\left( \pi - \frac{3\pi}{15} \right)}{\sin\frac{3\pi}{15}}\]
\[ = - \frac{- \sin\frac{\pi}{15}}{128\sin\frac{\pi}{15}} \times \frac{\sin\frac{3\pi}{15}}{\sin\frac{3\pi}{15}}\]
\[ = \frac{1}{128}\]
\[ = RHS\]
\[\text{ Hence proved} .\]
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