Question

Prove that $cos\frac{2\pi}{7} . cos\frac{4\pi}{7} . cos\frac{8\pi}{7} = \frac{1}{8}$.

Answer 1

Hi there!
Here's the answer:.

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¶ POINTS TO REMEMBER :

• $Sin(2A) = 2 Sin A. CosA$

• $Sin(2\pi+x) = sin x$

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¶¶¶ SOLUTION:

Let
$P = cos(\frac{2\pi}{7}) . cos(\frac{4\pi}{7}) . cos(\frac{8\pi}{7})$

Multiply and Divide by $8 sin(\frac{2\pi}{7})$and group the terms

=>$P = [\frac{1}{8sin(\frac{2\pi}{7})}] × [ 2 sin(\frac{2\pi}{7}) . cos(\frac{2\pi}{7})] ×[4 cos (\frac{4\pi}{7}) . cos(\frac{8\pi}{7})]$

Apply 2 Sin A Cos A= Sin 2A

=>$P = [\frac{1}{8sin(\frac{2\pi}{7})}] × [ sin(\frac{4\pi}{7})] ×[4 cos (\frac{4\pi}{7}) . cos(\frac{8\pi}{7})]$

=>$P = [\frac{1}{8sin(\frac{2\pi}{7})}] × [2 sin(\frac{4\pi}{7}) . cos(\frac{4\pi}{7})] × [2 cos(\frac{8\pi}{7})]$

=>$P = [\frac{1}{8sin(\frac{2\pi}{7})}] × [2 sin(\frac{4\pi}{7}) . cos(\frac{4\pi}{7})] × [2 cos(\frac{8\pi}{7})]$

=>$P = [\frac{1}{8sin(\frac{2\pi}{7})}] × [ sin(\frac{8\pi}{7})] × [2 cos(\frac{8\pi}{7})]$

=>$P = [\frac{1}{8sin(\frac{2\pi}{7})}] × [2 sin(\frac{8\pi}{7}). cos(\frac{8\pi}{7})]$

=>$P = [\frac{1}{8sin(\frac{2\pi}{7})}] × [sin(\frac{16\pi}{7})]$

Here,

$sin(\frac{16\pi}{7}) = sin(2\pi + \frac{2\pi}{7})$

=> $sin(\frac{16\pi}{7})= sin(\frac{2\pi}{7})$

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$P = \frac{sin(\frac{2\pi}{7})}{8×sin(\frac{2\pi}{7})}$

•°• $P = \frac{1}{8}$

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