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.

By a formula that is well known,
$$cos(A)+cos(B)=2cos((A+B)/2)cos((A-B)/2)$$
Let $A = 2\theta $, $B=4\pi/3 + 2\theta$. Then by the above,
$$
cos(2\theta)+cos(4\pi/3 + 2\theta)=2cos(2\theta+2\pi/3)cos(2\pi/3) = -cos(2\theta +2\pi/3)
$$ 
The first formula used above must have been rote learnt or at least encountered in high school. It's use gives a 2 line proof as I have shown.
