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Question
Prove that: \[\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15} = \frac{1}{16}\]
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Solution
\[= \frac{2\sin\frac{\pi}{15}\cos\frac{\pi}{15}}{2\sin\frac{\pi}{15}}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}\]
\[ \left[ \text{ On dividing and multiplying by } 2\sin\frac{\pi}{15} \right]\]
\[= \frac{2\sin\frac{2\pi}{15} \times \cos\frac{2\pi}{15}}{2 \times 2\sin\frac{\pi}{15}}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}\]
\[ = \frac{2\sin\frac{4\pi}{15} \times \cos\frac{4\pi}{15}}{2 \times 2 \times 2\sin\frac{\pi}{15}}\cos\frac{7\pi}{15}\]
\[ = \frac{\sin\frac{8\pi}{15}}{2 \times 2 \times 2\sin\frac{\pi}{15}}\cos\frac{7\pi}{15}\]
\[ = \frac{2\sin\frac{8\pi}{15}\cos\frac{7\pi}{15}}{16\sin\frac{\pi}{15}}\]
\[ = \frac{\sin\left( \frac{8\pi}{15} + \frac{7\pi}{15} \right) + \sin\left( \frac{8\pi}{15} - \frac{7\pi}{15} \right)}{16\sin\frac{\pi}{15}} \left[ \because 2\text{ sin } A\text{ cos } B = \sin\left( A + B \right) + \sin\left( A - B \right) \right]\]
\[ = \frac{0 + \sin\frac{\pi}{15}}{16\sin\frac{\pi}{15}}\]
\[ = \frac{\sin\frac{\pi}{15}}{16\sin\frac{\pi}{15}}\]
\[ = \frac{1}{16}\]
\[ = RHS\]
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