Question

Prove that cos(π/7)*cos(2π/7)*cos(4π/7)=(-1/8) !

Answer 1

\[\sin2x=2 \sin x \cos x\]\[\cos x = \frac{\sin 2x}{2 \sin x}\]using above,\[\cos \frac{\pi}{7}\cos\frac{2\pi}{7}\cos \frac{4\pi}{7}=\frac{\sin \frac{2\pi}{7}\sin \frac{4\pi}{7}\sin \frac{8\pi}{7}}{8\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}}\]we know that,\[\sin(\pi + \alpha)=-\sin \alpha\]\[\sin \frac{8\pi}{7}=\sin (\pi + \frac{\pi}{7})=-\sin\frac{\pi}{7}\]=\[\frac{\sin \frac{2\pi}{7}\sin \frac{4\pi}{7}\sin\frac{8\pi}{7}}{8 \sin \frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}}=-\frac{\cancel{\sin\frac{2\pi}{7}}\cancel{\sin\frac{4\pi}{7}}\cancel{\sin\frac{\pi}{7}}}{8\cancel{\sin\frac{\pi}{7}}\cancel{\sin\frac{2\pi}{7}}\cancel{\sin\frac{4\pi}{7}}}=-\frac{1}{8}\]

Answer 2

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