Question

Cos(π/7).cos(2π/7).cos(4π/7)

Answer 1

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Answer 2

Answer:

$\frac{-1}{8}$

Step-by-step explanation:

According to this question

Given that

$cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})$

Formula, sin 2x = 2 sinx cosx

cosx = $\frac{sin 2x}{2sin x}$

So,

$cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})$ = $\frac{sin \frac{2\pi}{7}\times sin \frac{4\pi}{7}\times sin \frac{8\pi}{7}\times }{2sin \frac{\pi}{7}\times 2sin \frac{2\pi}{7}\times 2sin \frac{4\pi}{7}}$

$cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})$ = $\frac{sin(\pi -\frac{\pi}{8})}{8sin \frac{\pi}{7}}$

$cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})$ = $\frac{sin(-\frac{\pi}{7})}{8sin \frac{\pi}{7}}$

$cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})$ = $\frac{-1}{8}$