<s>[INST] 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.
 [/INST] We know that $$\cos 6\theta = 4\cos^{3}2\theta - 3\cos 2\theta$$ and hence $x = \cos 2\theta$ is a root of the cubic equation $$4x^{3} - 3x -\cos 6\theta = 0$$ and clearly the sum of the roots of this equation is $0$. One of the roots is $x= \cos 2\theta$ and other two roots are $\cos((6\theta + 2\pi)/3)$ and $\cos((6\theta + 4\pi)/3)$ and the sum of these $3$ roots is $0$.
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