<s>[INST] Prove that $\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=3/2$ 
Prove that $$\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=\frac{3}{2}$$

I thought of rewriting $$\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ as $$\cos^2(90^{\circ}+ (\theta +30^{\circ})) + \cos^2(90^{\circ}-(\theta-30^{\circ}))$$ However I don't seem to get anywhere with this.
Unfortunately I don't know how to solve this question. I would be really grateful for any help or suggestions. Many thanks in advance!
 [/INST] As $4\cos^3x-3\cos x=\cos3x$
If  $\cos3x=\cos3A, 3x=2n\pi\pm3A$ where $n$ is any integer
$\implies x=\dfrac{2n\pi}3\pm A$ where $n\equiv0,1,2\pmod3$
So, the roots of $4\cos^3x-3\cos x-\cos3A=0$ are $\cos\dfrac{2n\pi}3\pm A$ where $n\equiv0,1,2\pmod3$
$\implies\sum_{r=0}^2\cos\left(\dfrac{2n\pi}3+A\right)=0$
and $\sum_{r=0}^2\cos\left(\dfrac{2n\pi}3-A\right)=0$
Now use $\cos2A=2\cos^2A-1$
Generalization :  Like Product of cosines: $ \prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right) $ we can prove for positive integer $m$,
$$\cos(mx)=2^{m-1}\cos^mx+0\cdot\cos^{m-1}x+\cdots=0$$
$\implies\sum_{r=0}^{m-1}\cos\left(\dfrac{2n\pi}m-A\right)=\sum_{r=0}^{m-1}\cos\left(\dfrac{2n\pi}m+A\right)=0$
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