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.

Recall that $\cos(a + b) = \cos a \cos b - \sin a \sin b$
\begin{align}
\cos(\frac{\pi}{3} + \theta) &= \cos \frac{\pi}{3}\cos\theta - \sin \frac{\pi}{3}\sin\theta\\
&= \frac{\sqrt{3}\sin \theta - \cos\theta}{2}\\
\cos(\frac{2\pi}{3} + \theta) &= \cos \frac{2\pi}{3}\cos\theta - \sin \frac{2\pi}{3}\sin\theta\\
&= \frac{\sqrt{3}\sin \theta + \cos\theta}{2}
\end{align}
Remember that $\cos 2\theta = 2\cos^2 \theta- 1$. Now we use the previous results
\begin{align}
\text{L.S} &= \cos 2\theta + \cos(2(\frac{\pi}{3} + \theta)) + \cos(2(\frac{2\pi}{3} + \theta))\\
&= 2\cos^2\theta-1 +2\cos^2(\frac{\pi}{3} + \theta) - 1 + 2\cos^2(\frac{2\pi}{3} + \theta) - 1\\
&= 2\cos^2\theta + 2\bigg(\frac{(\sqrt{3}\sin \theta - \cos\theta)^2}{4} + \frac{(\sqrt{3}\sin \theta + \cos\theta)^2}{4}\bigg) - 3
\end{align}
Recall that $(a + b)^2 + (a - b)^2 = 2a^2 + 2b^2$
\begin{align}
\text{L.S} &= 2\cos^2\theta + 2(\frac{6\sin^2 \theta + 2\cos^2 \theta}{4}) - 3\\
&= 2\cos^2\theta + 3\sin^2\theta + \cos^2\theta - 3\\
&= 3(\sin^2\theta + \cos^2\theta) - 3\\
&= 3 - 3 = 0 = \text{R.S}
\end{align}
Thus concludes the proof. The perfect square pattern saves a lot of work. This method also does not require the sum to product formulas.
