# How to prove that 2sin((4pi)/5)cos(pi/5)=sin((2pi)/5)?

May 28, 2018

$\sin \left(\frac{4 \pi}{5}\right) = \sin \left(\pi - \frac{\pi}{5}\right) = \sin \left(\frac{\pi}{5}\right)$

Using $2 \sin x \cos x = \sin \left(2 x\right)$
we obtain

$2 \sin \left(\frac{\pi}{5}\right) \cos \left(\frac{\pi}{5}\right) = \sin \left(\frac{2 \pi}{5}\right)$
and finally

$2 \sin \left(\frac{4 \pi}{5}\right) \cos \left(\frac{\pi}{5}\right) = \sin \left(\frac{2 \pi}{5}\right)$

May 28, 2018

$\text{see explanation}$

#### Explanation:

$\text{using the "color(blue)"product to sum formula}$

â€¢color(white)(x)2sinxcosy=sin(x+y)+sin(x-y)

$\text{here "x=(4pi)/5" and } y = \frac{\pi}{5}$

$\text{consider left side}$

$\sin \left(\frac{4 \pi}{5} + \frac{\pi}{5}\right) + \sin \left(\frac{4 \pi}{5} - \frac{\pi}{5}\right)$

$= \sin \pi + \sin \left(\frac{3 \pi}{5}\right)$

$= 0 + \sin \left(\pi - \frac{3 \pi}{5}\right)$

$= \sin \left(\frac{2 \pi}{5}\right) = \text{ right side"rArr" verified}$
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