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Question
Prove that: \[\cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{1}{16}\]
Solution
\[ = \frac{1}{2 \times 2\sin\frac{2\pi}{15}} \times \left( 2\sin\frac{4\pi}{15} \times \cos\frac{4\pi}{15} \right) \times \cos\frac{8\pi}{15} \times \cos\frac{16\pi}{15} \]
\[ = \frac{1}{2 \times 4\sin\frac{2\pi}{15}}\left( 2\sin\frac{8\pi}{15} \times \cos\frac{8\pi}{15} \right) \times \cos\frac{16\pi}{15}\]
\[ = \frac{1}{2 \times 8\sin\frac{2\pi}{15}}\left( 2\sin\frac{16\pi}{15} \times \cos\frac{16\pi}{15} \right)\]
\[ = \frac{1}{16\sin\frac{2\pi}{15}}\left( \sin\frac{32\pi}{15} \right)\]
\[= - \frac{1}{16\sin\frac{2\pi}{15}}\left( \sin2\pi - \frac{32\pi}{15} \right) \left[ \because \sin\left( 2\pi - \theta \right) = - sin\theta \right]\]
\[ = - \frac{1}{16\sin\frac{2\pi}{15}}\sin\left( - \frac{2\pi}{15} \right)\]
\[ = \frac{1}{16} = RHS\]
\[\text{ Hence proved } .\]
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