Question
Prove that $\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{\pi}{5}\cos\frac{8\pi}{5}=\frac{-1}{16}$
✓
Answer
$\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5}=\frac{\sin\frac{2^4\pi}{5}}{2^4\sin\frac{\pi}{5}}$ $\Big[\because\cos\text{A}\cos2\text{A}\cos2^2\text{A}\cos2^3\text{A}...\cos2^{\text{n}-1}\text{A}=\frac{\sin2^\text{n}\text{A}}{2^\text{n}\sin\text{A}}\Big]$ $=\frac{\sin\frac{16\pi}{5}}{16\sin\frac{\pi}{5}}$ $=\frac{\sin\Big(3\pi+\frac{\pi}{5}\Big)}{16\sin\frac{\pi}{5}}$ $=\frac{1\big\{3\pi+\frac{\pi}{5}\big\}}{16\sin\frac{\pi}{5}}$ $=\frac{-1}{16}$
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