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

Question

29221 views around the world

You can reuse this answer
Creative Commons License

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.