How to prove that $cos((2pi)/7)+cos((4pi)/7)+cos((6pi)/7)= -1/2$ ?

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Answer 1 · 2016-03-03T10:00:19

Multiply both sides with $sin(pi/7)$ hence we have that

$sin(π/7)cos(2π/7)+sin(π/7)cos(4π/7)+sin(pi/7)*cos(6pi/7)=-1/2*sin(π/7)$
Then because $sinA*cosB=1/2*[sin(A-B)+sin(A+B)]$ we rewrite this as

$(1/2)[sin(π/7-2π/7) + sin(π/7+2π/7)] + (1/2)[sin(π/7-4π/7) + sin(π/7+4π/7)] + (1/2)[sin(π/7-6π/7) + sin(π/7+6π/7)]= -sin(π/7)/2$

We'll divide by $(1/2)$ both sides:
$-sin(π/7) + sin(3π/7) - sin(3π/7) + sin(5π/7) - sin(5π/7) + sin(7π/7)= -sin(π/7)$

Cancelling the same terms yields to
$-sin(π/7) + sin(π)= -sin(π/7)$

But $sin(π) = 0$ hence

$-sin(π/7) = -sin(π/7)$

Since we have the same results in both sides, the given identity is true.

How to prove that cos((2pi)/7)+cos((4pi)/7)+cos((6pi)/7)= -1/2cos((2pi)/7)+cos((4pi)/7)+cos((6pi)/7)= -1/2?