cos(2pi/7) +cos(4pi/7) + cos(8pi/7) = -1/2
The best way to prove this will be to simplify the left side. To do that we need to be able to try and collect like terms.
There is a trigonometric identity that says:
sin(x)cos(y) = (1/2)[(sin(x+y)+sin(x-y)]
Let's multiply both sides of the equation by sin(pi/7):
sin(pi/7)cos(2pi/7) + sin(pi/7)cos(4pi/7) + sin(pi/7)cos(6pi/7) = -1/2
Now, we apply the identity above:
(1/2)[sin(pi/7+2pi/7) + sin(pi/7-2pi/7)] + (1/2)[sin(pi/7+4pi/7) + sin(pi/7-4pi/7)]
+ (1/2)[sin(pi/7+8pi/7) + sin(pi/7-8pi/7)] = -(1/2)sin(pi/7)
We can multiply both sides by 2 and simplify the terms a bit:
[sin(3pi/7) + sin(-pi/7)] + [sin(5pi/7) + sin(-3pi/7)] + [sin(9pi/7) + sin(-7pi/7)] = -sin(pi/7)
sin(3pi/7) - sin(pi/7) + sin(5pi/7) - sin(3pi/7) + sin(9pi/7) - sin(7pi/7) = -sin(pi/7)
Now, we can collect like terms:
sin(5pi/7) + sin(9pi/7) - sin(7pi/7) = 0
But sin(7pi/7) = 0:
sin(5pi/7) + sin(9pi/7) = 0
At this point you can probably inference whether the identity will be proved, or not.
By the way, was the problem stated/copied correctly? I find myself wondering of the last term of the left half of the equation should be cos(6pi/7), not cos(8pi/7).
Hope this helps!