<s>[INST] How Can One Prove $\cos(\pi/7) + \cos(3 \pi/7) + \cos(5 \pi/7) = 1/2$ Reference: http://xkcd.com/1047/
We tried various different trigonometric identities. Still no luck. 
Geometric interpretation would be also welcome.
EDIT: Very good answers, I'm clearly impressed. I followed all the answers and they work! I can only accept one answer, the others got my upvote. 
 [/INST] Hint: start with $e^{i\frac{\pi}{7}} = \cos(\pi/7) + i\sin(\pi/7)$ and the fact that the lhs is a 7th root of -1. 
Let $u = e^{i\frac{\pi}{7}}$, then we want to find $\Re(u + u^3 + u^5)$.
Then we have $u^7 = -1$ so $u^6 - u^5 + u^4 - u^3 + u^2 -u + 1 = 0$.
Re-arranging this we get: $u^6 + u^4 + u^2 + 1 = u^5 + u^3 + u$.
If $a = u + u^3 + u^5$ then this becomes $u a + 1 = a$, and rearranging this gives $a(1 - u) = 1$, or $a = \dfrac{1}{1 - u}$.
So all we have to do is find $\Re\left(\dfrac{1}{1 - u}\right)$.
$\dfrac{1}{1 - u} = \dfrac{1}{1 - \cos(\pi/7) - i \sin(\pi/7)} = \dfrac{1 - \cos(\pi/7) + i \sin(\pi/7)}{2 - 2 \cos(\pi/7)}$
so
$\Re\left(\dfrac{1}{1 - u}\right) = \dfrac{1 - \cos(\pi/7)}{2 - 2\cos(\pi/7)} = \dfrac{1}{2} $
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