Question

prove that


$\sf \: 2cos \frac{\pi}{13}cos \frac{9\pi}{13} + cos \frac{3\pi}{13} + cos \frac{5\pi}{13} = 0$


Note :-

no copy paste answer
don't spam plz ​

Answer 1

$\large\underline{\sf{question \: -}}$

$\rm \: 2cos \frac{\pi}{13}cos \frac{9\pi}{13} + cos \frac{3\pi}{13} + cos \frac{5\pi}{13} = 0$

$\large\underline{\sf{formula \: -}}$

$\rm :\longmapsto\:cosx + cosy = 2cos( \frac{x + y}{2})cos( \frac{x - y}{2})$

$\large\underline{\sf{Solution-}}$

$\sf \: lhs = 2cos \frac{\pi}{13} cos \frac{9\pi}{13} + cos \frac{3\pi}{13} + cos \frac{5x}{13}$

using formula

$\rm :\longmapsto\:cosx + cosy = 2cos( \frac{x + y}{2})cos( \frac{x - y}{2})$

by putting values we get,

$\rm :\longmapsto\: 2cos \frac{\pi}{13} cos \frac{9\pi}{13} + 2cos \binom{ \frac{3\pi + 5\pi}{13 + 13} }{2} cos \binom{ \frac{3\pi - 5\pi}{13 - 13} }{2}$

by further calculation

$\rm :\longmapsto\: 2cos \frac{\pi}{13} cos \frac{9\pi}{13} + 2cos \frac{4\pi}{13} cos( \frac{ - \pi}{13}) \\ \\ \rm :\longmapsto\: 2cos \frac{\pi}{13} cos \frac{9\pi}{13} + 2cos \frac{4\pi}{13} cos \frac{\pi}{13}$

now taking out the common terms

$\rm :\longmapsto\:2cos \frac{\pi}{13} [cos \frac{9\pi}{13} + cos \frac{4\pi}{13}]$

it can be written as

$\rm :\longmapsto\: 2cos \frac{\pi}{13} \big[ 2cos \binom{ \frac{9\pi + 4\pi}{13 + 13} }{2} cos \binom{ \frac{9\pi - 4\pi}{13 - 13} }{2}]$

on further calculation

$\rm :\longmapsto\: 2cos \frac{\pi}{13} [2cos \frac{\pi}{2} cos \frac{5\pi}{26}]$

$\rm :\longmapsto\: 2cos \frac{\pi}{13} \times 2 \times 0 \times cos \frac{5\pi}{26} \\ \\ \rm :\longmapsto\:0$

hence proved