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!
Hint: use the formula $$ \cos^2x = \frac 12 (1 + \cos(2x)) $$ And use the fact (or prove that) $$ \cos(x) + \cos(x + 240^\circ) + \cos(x - 240^\circ) = 0 $$
$\cos^2(A-120^\circ)+\cos^2(A+120^\circ)=1+\cos^2(A-120^\circ)-\sin^2(A+120^\circ)$
and Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$, $\cos^2(A-120^\circ)-\sin^2(A+120^\circ)=1+\cos(2A)\cos(240^\circ)=?$
$\cos(240^\circ)=\cos(180^\circ+60^\circ)=-\cos60^\circ=?$
Now use $\cos2A=2\cos^2A-1$
I am assuming that $\theta$ is in degrees. Then using the following formulas: \begin{eqnarray} \cos(x+y)&=&\cos x\cos y-\sin x\sin y\\ \cos(x-y)&=&\cos x\cos y+\sin x\sin y\\ (a-b)^2+(a+b)^2&=&2(a^2+b^2)\\ \cos(120)&=&-\frac12\\ \sin(120)&=&\frac{\sqrt3}{2}\\ \cos^2(x)+\sin^2(x)&=&1 \end{eqnarray} we have: \begin{eqnarray} && \cos^2(\theta)+\cos^2(\theta+120)+\cos^2(\theta-120) \\[12pt] &=&\cos^2\theta+[\cos(\theta)\cos(120)-\sin(\theta)\sin(120)]^2\\[8pt] &&+[\cos(\theta)\cos(120)+\sin(\theta)\sin(120)]^2\\[8pt] &=&\cos^2(\theta)+2[\cos^2(\theta)\cos^2(120)+\sin^2(\theta)\sin^2(120)]\\[8pt] &=&\cos^2(\theta)+2\left[\left(-\frac12\right)^2\cos^2(\theta)+\left(\frac{\sqrt3}{2}\right)^2\sin^2(\theta)\right]\\[8pt] &=&\cos^2(\theta)+2\left[\frac14\cos^2(\theta)+\frac34\sin^2(\theta)\right]\\[8pt] &=&\cos^2(\theta)+\frac12\cos^2(\theta)+\frac32\sin^2(\theta)\\[8pt] &=&\left(1+\frac12\right)\cos^2(\theta)+\frac32\sin^2(\theta)\\[8pt] &=&\frac32\cos^2(\theta)+\frac32\sin^2(\theta)\\[8pt] &=&\frac32[\cos^2(\theta)+\sin^2(\theta)]\\[8pt] &=&\frac32\cdot1=\frac32. \end{eqnarray}
We can use static equilibrium of forces, anyhow this application contains trigonometry inside of it. If three equal (can be unit) forces act 120 degree apart on a particle making vector plot as an equilateral triangle, the vector sum is zero.
The given quantity is
$$ \dfrac{1 + \cos 2 \theta}{2} + \dfrac{1 + \cos (2 \theta+240 ^{\circ}) }{2} + \dfrac{1 + \cos (2 \theta-240 ^{\circ}) }{2} $$
$$ =\frac32+ \Sigma _{i=1}^3F(2\pi i /3)$$
$$=\dfrac32 $$
because the three equal forces acting with 120 degree separation vanish in vector addition.
Note
that the forces that can act on the particle can be three, four or any other integer in number. If vertical forces are considered instead, then we get the same expression in terms of arbitrary angle $\theta$
$$\sin\, \theta \,+ ... +...$$
in place of the present limited series
$$\cos \,\theta \, + ... +...$$
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 $m$,
$$\cos(mx)=2^{m-1}\cos^mx+0\cdot\cos^{m-1}x+\cdots=0$$
$\implies\sum_{r=0}^{m-1}\cos\left(\dfrac{2n\pi}m-A\right)=\sum_{r=0}^{m-1}\cos\left(\dfrac{2n\pi}m+A\right)=0$
Here's a solution:
$$\cos(\theta + 120^\circ) = \cos\theta \cos 120^\circ-\sin\theta \sin 120^\circ = \frac{\cos \theta}{2} - \frac{\sqrt{3}}{2}\sin \theta$$ $$ \cos(\theta + 120^\circ) = \cos\theta \cos 120^\circ+\sin\theta \sin 120^\circ = \frac{\cos \theta}{2} + \frac{\sqrt{3}}{2}\sin \theta$$
$$ \left(\frac{\cos \theta}{2} - \frac{\sqrt{3}}{2}\sin \theta\right)^2 = \frac{\cos^2\theta}{4} - \frac{\sqrt3}{2} \sin \theta \cos \theta+ \frac{3}{4} \sin^2\theta$$ $$ \left(\frac{\cos \theta}{2} + \frac{\sqrt{3}}{2}\sin \theta\right)^2 = \frac{\cos^2\theta}{4} + \frac{\sqrt3}{2}\sin \theta \cos \theta+ \frac{3}{4} \sin^2\theta$$
$$\cos^2(\theta + 120^\circ) + \cos^2(\theta - 120^\circ) = \frac{1}{2}\cos^2\theta + \frac{3}{2} \sin^2\theta$$
$$\cos^2\theta +\cos^2(\theta + 120^\circ) + \cos^2(\theta - 120^\circ) = \frac{3}{2}(\cos^2\theta + \sin^2\theta) = \frac{3}{2} $$
Note: $\cos^2\theta + \sin^2\theta = 1$
Notice, $$\color{blue}{\sin (A+B)\sin(A-B)=\sin^2A-\sin^2 B}$$ $$\color{blue}{\sin (A+B)+\sin(A-B)=2\sin A\cos B}$$
Now, I will follow OP's steps $$LHS=\cos^2 \theta+\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ $$=\cos^2 \theta+\cos^2(90^{\circ}+ (\theta +30^{\circ})) + \cos^2(90^{\circ}-(\theta-30^{\circ}))$$ $$=\cos^2 \theta+\sin^2(\theta +30^{\circ}) + \sin^2(\theta-30^{\circ})$$ $$=\cos^2 \theta+(\sin(\theta+30^\circ)+\sin(\theta-30^\circ))^2 -2\sin(\theta+30^\circ)\sin(\theta-30^\circ)$$ $$=\cos^2 \theta+(2\sin\theta\cos 30^\circ)^2 -2(\sin^2\theta-\sin^2 30^\circ)$$ $$=\cos^2 \theta+\left(2\sin\theta\frac{\sqrt{3}}{2}\right)^2 -2\left(\sin^2\theta-\frac{1}{4}\right)$$ $$=\cos^2 \theta+3\sin^2\theta -2\sin^2\theta+\frac{1}{2}$$ $$=\cos^2 \theta+\sin^2\theta+\frac{1}{2}$$ $$=1+\frac{1}{2}$$$$=\frac{3}{2}=RHS$$
Since $\cos(2x)=2\cos^2(x)-1$, we have $$\cos^2(x)=\frac {\cos(2x)+1}2$$
Therefore, $$\cos^2(\theta)=\frac {\cos(2\theta)+1}{2}$$ $$\cos^2(\theta+120)=\frac{\cos(2\theta+240^\circ)+1}{2}$$ $$\cos^2(\theta-120)=\frac{\cos(2\theta-240^\circ)+1}{2}$$
So the original equation become: $$\frac{\cos(2\theta)+1+cos(2\theta+240^\circ)+1+\cos(2\theta-240^\circ)+1}2.$$
Using the sum formula, we get $$\frac{3+\cos(2\theta)-\frac12\cos(2\theta)-\sin(2\theta)\sin(240^\circ)-\frac12\cos(2\theta)+\sin(2\theta)\sin(240^\circ)}2$$ $$\Longrightarrow \frac{3+\cos(2\theta)-\cos(2\theta)}2\Longrightarrow\frac32.$$
Hope this is helpful.
