Show `cos (2pi/7)+cos (4pi/7) + cos (6pi/7) = -1/2`

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It has to be shown that `cos (2pi/7)+cos (4pi/7) + cos (6pi/7) = -1/2`

`cos (2pi/7)+cos (4pi/7) + cos (6pi/7)`

Multiply and divide by `sin(pi/7)`

=> `(sin(pi/7)cos(2pi /7)+sin(pi /7)cos(4pi /7)+sin(pi /7)cos(6pi /7))/sin(pi/7)`

Convert from product to sum

=>`(sin(pi/7-2pi/7)+sin(pi/7+2pi/7)+sin(pi/7-4pi/7)+ sin(pi/7+4pi /7)+sin(pi/7-6pi /7)+ sin(pi/7+6pi/7))/(2*sin(pi/7))`

=> `(sin(-pi/7)+sin(3pi/7)+sin(-3pi/7)+ sin(5pi/7)+sin(-5pi/7)+ sin(7pi/7))/(2*sin(pi/7))`

=> `(sin(-pi/7) + sin pi)/(2*sin(pi/7))`

=> `(sin(-pi/7) + 0)/(2*sin(pi/7))`

=> `-1/2`

This proves that `cos (2pi/7)+cos (4pi/7) + cos (6pi/7) = -1/2`

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