Prove $\cos(2\theta) + \cos\left(2 \left(\frac{\pi}{3} + \theta\right)\right) +\cos\left(2 \left(\frac{2\pi}{3} + \theta\right)\right) = 0$ It's been a while since I've done trig proofs.  I know that
$$\cos(2\theta) + \cos\left(2 \left(\frac{\pi}{3} + \theta\right)\right) + \cos\left(2 \left(\frac{2\pi}{3} + \theta\right)\right) = 0$$
is true.  This can be easily seen by plugging in values, for example where $\theta=0$ we get
$$1 -\frac{1}{2} -\frac{1}{2} = 0$$
and the same can be seen for any $\theta$.  Is there a straight forward way to prove this statement is true.

$$\cos(2\theta) + \cos\left(2 \left(\frac{\pi}{3} + \theta\right)\right) +\cos\left(2 \left(\frac{2\pi}{3} + \theta\right)\right)=\cos{2\theta}+\cos({\frac{2\pi}{3}+2\theta})+\cos({\frac{4\pi}{3}}+2\theta)$$
Expanding this we get $\cos{2\theta}+\cos({\frac{2\pi}{3}+2\theta})+\cos({\frac{4\pi}{3}+2\theta})=\cos{2\theta}+-\frac{1}{2}\cos{2\theta}-\frac{\sqrt{3}}{2}\sin{2\theta}-\frac{1}{2}\cos{2\theta}+\frac{\sqrt{3}}{2}\sin{2\theta}$
which equals $0$
