What is $Cos^2(pi/7)+cos^2(2pi/7)+cos^2(4pi/7)$ is equal to?

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Answer 1 · 2016-04-30T13:43:25

$5/4$

LHS $=cos^2(pi/7)+cos^2(2pi/7)+cos^2(4pi/7)$
using formula $cos^2theta=1/2(1+cos2theta)$

$=1/2(1+cos(2pi/7))+1/2(1+cos(4pi/7))+ 1/2(1+cos(8pi/7))$
$=3/2+1/2(cos(2pi/7)+cos(4pi/7)+cos(8pi/7))$

$=3/2+1/(4sin(pi/7)) (2sin(pi/7)cos(2pi/7)+2sin(pi/7)cos(4pi/7)+2sin(pi/7)cos(8pi/7))$

$=3/2+1/(4sin(pi/7)) (cancel(sin(3pi/7))-sin(pi/7)+sin(5pi/7)-cancel(sin(3pi/7))+sin(9pi/7)-sin(7pi/7))$

$=3/2+1/(4sin(pi/7)) (-sin(pi/7)+sin(pi-2pi/7)+sin(pi+2pi/7)-0)$

$=3/2+1/(4sin(pi/7)) (-sin(pi/7)+cancel(sin(2pi/7))-cancel(sin(2pi/7)))$

$=3/2+1/(4cancel(sin(pi/7))) (-cancel(sin(pi/7)))$

$=3/2-1/4=(6-1)/4=5/4$

What is Cos^2(pi/7)+cos^2(2pi/7)+cos^2(4pi/7)Cos^2(pi/7)+cos^2(2pi/7)+cos^2(4pi/7) is equal to?