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. 

Let $y = \cos\theta + i \cdot \sin\theta$,  where $\theta$ has either of the values $$\frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \pi, \frac{9\pi}{7}, \frac{11\pi}{7} \ \text{and} \ \frac{13\pi}{7}$$ Then $y^{7} = \cos{7\theta} + i \:\sin{7\theta} =-1$. Then you have $$y^{7}+1=0 \Rightarrow (y+1)(y^{6}-y^{5}+y^{4}-y^{3}+y^{2}-y +1)=0$$ Now the root $y=-1$ corresponds to the value $\theta=\pi$. The roots of the equation $$y^{6}-y^{5}+y^{4}-y^{3}+y^{2}-y +1 =0 \qquad \cdots (1)$$ are therefore $\cos\theta + i\: \sin\theta$ where $\theta$ takes other values than $\pi$. Put $2x = y + \frac{1}{y} = 2 \cos\theta$ and note that:


*

*$\displaystyle y^{2}+\frac{1}{y^2} = 4x^{2}-2$

*$\displaystyle y^{3} + \frac{1}{y^3} = 8x^{3}-6x$. 
Now divide $(1)$ by $y^{3}$ and get the answer.
Another method to look at this:
$7 \theta = \text{an odd multiple of} \: \pi$, therefore $\cos{4\theta} = -\cos{3\theta}$. Take $c=\cos{\theta}$. From this you have
\begin{align*}
2\cdot (2c^{2}-1)^{2} -1 &= -(4c^{2}-3c) \\\ 8c^{4} + 4c^{2}-8c^{2} -3c + 1 &=0 \\\ (c+1) \cdot (8c^{3} - 4c^{2} -4c+1) &=0
\end{align*}
The following page also may be helpful: http://www.isibang.ac.in/~sury/luckyoct10.pdf
