Prove cos (5π/7) + cos (3π/7) + cos (π/7) = 1/2?

Consider the roots of x^7 + 1 = 0. They are e^iπ/7, e^i3π/7, ... , e^i13π/7 and must have sum zero since there is no x^6 term. Hence, in particular, their real parts sum to zero. But cos7π/7 = - 1 and the others are equal in pairs, because cos(2π - x) = cos x. So we get cos π/7 + cos 3π/7 + cos 5π/7 = 1/2. Finally since cos(π - x) = - cos x, cos 5π/7 = - cos 2π/7.

Let me tell you what I'm thinking, and perhaps you can construct a proof out of it:

If you visualize a regular heptagon (7-sided polygon) inscribed in the unit circle with one of vertices at (0,Pi), then this polygon has vertices at each of the points you're intrested in: Pi/7, 3Pi/7, 5Pi/7, as well as -Pi/7, -3Pi/7, and -5Pi/7.

Well, if you take the sum of the cosine of each vertex in this heptagon, you *should* get zero. This is sticky in my mind right now, because I'm not 100% sure how to show this, but perhaps you have a good idea for it. If this is the case, then the proof becomes simple, because cos(Pi/7)+cos(3Pi/7)+cos(5Pi/7)= cos(-Pi/7)+cos(-3Pi/7)+cos(-5Pi/7),

and cos(Pi) = -1. So, the two sums above must both equal 1/2.

maybe this helps? Good Luck!