TITLE: Sums of powers of cosines and sines shifted by $2\pi/3$
QUESTION [12 upvotes]: I have stumbled across these two identities
$$
\begin{split}
\cos^2(x)+\cos^2(x+2\pi/3)+\cos^2(x+4\pi/3) &= 3/2,\\
\cos^4(x)+\cos^4(x+2\pi/3)+\cos^4(x+4\pi/3) &= 9/8.
\end{split}
$$
There is also the more intricate
$$
\begin{split}
\cos^2(x)\sin^2(x)+\cos^2(x+2\pi/3)\sin^2(x+2\pi/3)+\cos^2(x+4\pi/3)\sin^2(x+4\pi/3) &= 3/8,\\
\cos^4(x)\sin^4(x)+\cos^4(x+2\pi/3)\sin^4(x+2\pi/3)+\cos^4(x+4\pi/3)\sin^4(x+4\pi/3) &= 9/128,
\end{split}
$$
and of course the most elementary
$$
\cos(x)+\cos(x+2\pi/3)+\cos(x+4\pi/3)=0.
$$
The last identity admits a rather intuitive interpretation in terms of unitary complex numbers centered about the origin. My questions are:

Do the other identities admit similar more or less intuitive interpretations as well?
Do such identities have names?
Not all powers and combinations produce a constant; What is the general form of the expressions that do?

Context: The first two identities came up while calculating the elastic response of a two-dimensional truss (a planar lattice of nodes connected with springs) that is invariant by rotations of order 3, in which case $x$ describes the orientation of the truss. We know that such trusses must exhibit an isotropic response and that justifies, in a rather convoluted manner, that these expressions must be constants. The other expressions I found by trial and error. I am looking for a satisfying, non-brute-force, non-too-group-theoretic, explanation.

REPLY [2 votes]: $\color{green}{\textbf{Version of 20.07.20.}}$
$\color{brown}{\textbf{Preliminary calculations.}}$
At first,
\begin{cases}
\cos\left(x+\frac{4\pi}3\right) = \cos\left(x-\frac{2\pi}3\right)\\[4pt]
\sin\left(x+\frac{4\pi}3\right) = \sin\left(x-\frac{2\pi}3\right)\tag1
\end{cases}
Denote
\begin{cases}
c= \cos x,\quad s=\sin x,\\[4pt]
\mathcal S_k(f(t),x)=f^k\left(x-\frac{2\pi}3\right) +f^k(x) +f^k\left(x+\frac{2\pi}3\right)\\[4pt]
\mathcal Q_k(f(t),x)=f^k\left(x-\frac{2\pi}3\right)f^k(x) 
+f^k(x)f^k\left(x+\frac{2\pi}3\right)\\[4pt]
\mspace{93mu}+f^k\left(x+\frac{2\pi}3\right)f^k\left(x-\frac{2\pi}3\right)\\[4pt]
\mathcal P_k(f(t),x)=f^k\left(x-\frac{2\pi}3\right) f^k(x) f^k\left(x+\frac{2\pi}3\right).\tag2
\end{cases}
Since
$$\cos\frac{2\pi}3 = -\frac12,\quad \sin\frac{2\pi}3 = \frac{\sqrt3}2,$$
then
$$\cos\left(x\pm\frac{2\pi}3\right) = -\frac c2\mp\frac{s\sqrt3}2,\quad
\sin\left(x\pm\frac{2\pi}3\right) = -\frac s2\pm\frac {c\sqrt3}2.\tag3$$
Applying $(3),$ easily to get
\begin{align}
&\mathcal S_1(\cos t,x) = \cos\left(x-\frac{2\pi}3\right)+\cos x +\cos\left(x+\frac{2\pi}3\right)\\
& = -\frac c2+\frac{s\sqrt3}2\,+c\,-\frac c2-\frac{s\sqrt3}2 = 0,\\[4pt]
&\mathcal Q_1(\cos t,x) = \cos\left(x-\frac{2\pi}3\right)\cos x + \cos x \cos\left(x+\frac{2\pi}3\right)+\cos\left(x+\frac{2\pi}3\right)\cos\left(x-\frac{2\pi}3\right)\\
&= c\left(-\frac c2+\frac{s\sqrt3}2-\frac c2-\frac{s\sqrt3}2\right)
+\left(-\frac c2-\frac{s\sqrt3}2\right)\left(-\frac c2+\frac{s\sqrt3}2\right)\\
&= -\cos^2 x+\frac14\cos^2x-\frac34\sin^2x = -\frac34,\\[4pt]
&\mathcal P_1(\cos t,x) = \cos\left(x-\frac{2\pi}3\right)\cos x \cos\left(x+\frac{2\pi}3\right) = \left(-\frac c2+\frac{s\sqrt3}2\right)\,c\,
\left(-\frac c2-\frac{s\sqrt3}2\right)\\
& = \frac14(\cos^3x-3\cos x\sin^2 x) = \frac14\Re(\cos x + i\sin x)^3 = \frac14\cos 3x.
\end{align}
At the same time,
$$\sin x = \cos\left(x-\frac\pi2\right),\tag4$$
and then
\begin{align}
&\mathcal S_k(\sin t,x) = \sin^k\left(x-\frac{2\pi}3\right)+\sin^k x +\sin^k\left(x+\frac{2\pi}3\right) = \mathcal S_k\left(\cos t,x-\frac\pi2\right) ,\\[4pt]
&\mathcal Q_k(\sin t,x) = \sin^k\left(x-\frac{2\pi}3\right)\sin^k x + \sin^k x \sin^k\left(x+\frac{2\pi}3\right)\\[4pt]
&+\sin^k\left(x+\frac{2\pi}3\right)\sin^k\left(x-\frac{2\pi}3\right)
=\mathcal Q_k\left(\cos t,x-\frac\pi2\right),\\
&\mathcal P_k(\sin t,x) = \sin^k\left(x-\frac{2\pi}3\right)\sin^k x \sin^k\left(x+\frac{2\pi}3\right)
= \mathcal P_k\left(\cos t,x-\frac\pi2\right).\\[4pt]
\end{align}
On the other hand,
\begin{cases}
\cos x \sin x = \frac12\sin(2x)\\[4pt]
\cos\left(x-\frac{2\pi}3\right) \sin \left(x-\frac{2\pi}3\right) 
= \frac12\sin\left(2x-\frac{4\pi}3\right)
= \frac12\sin\left(2x+\frac{2\pi}3\right)\\[4pt] 
\cos\left(x+\frac{2\pi}3\right) \sin \left(x+\frac{2\pi}3\right) 
= \frac12\sin\left(2x+\frac{4\pi}3\right)
= \frac12\sin\left(2x-\frac{2\pi}3\right).\tag5
\end{cases}
then
\begin{align}
&\mathcal S_k(\cos t \sin t, x) = \cos^k\left(x-\frac{2\pi}3\right)\, \sin^k\left(x-\frac{2\pi}3\right)+ \cos^k x\, \sin^k x\\
& + \cos^k\left(x+\frac{2\pi}3\right)\, \sin^k\left(x+\frac{2\pi}3\right)\\ 
&=\frac1{2^k}\left(\sin^k\left(2x+\frac{2\pi}3\right)+ \sin^k 2x 
+ \sin^k\left(2x-\frac{2\pi}3\right)\right) = \frac1{2^k}\mathcal S_k(\sin t,2x),\\[4pt]
&\mathcal Q_k(\cos t \sin t,x) = \cos^k\left(x-\frac{2\pi}3\right)\, \sin^k\left(x-\frac{2\pi}3\right) \cos^k x\, \sin^k x\\
& + \cos^k x\, \sin^k x \cos^k\left(x+\frac{2\pi}3\right)\, \sin^k\left(x+\frac{2\pi}3\right)\\
& + \cos^k\left(x+\frac{2\pi}3\right)\, \sin^k\left(x+\frac{2\pi}3\right)\cos^k\left(x-\frac{2\pi}3\right)\, \sin^k\left(x-\frac{2\pi}3\right)\\
&=\frac1{4^k}\left(\sin^k\left(2x+\frac{2\pi}3\right)\sin^k 2x + \sin^k 2x 
\sin^k\left(2x-\frac{2\pi}3\right)\right.\\
&\left.+ \sin^k\left(2x-\frac{2\pi}3\right)\sin^k\left(2x+\frac{2\pi}3\right)\right)\\
& = \frac1{4^k}\mathcal Q_{k}(\sin t,2x),\\[4pt]
&\mathcal P_{k}(\cos t \sin t,x) = \cos^k\left(x-\frac{2\pi}3\right)\, \sin^k\left(x-\frac{2\pi}3\right) \cos^k x\, \sin^k x\\ &\times\cos^k\left(x+\frac{2\pi}3\right)\, \sin^k\left(x+\frac{2\pi}3\right)\\
&=\frac1{8^k}\sin^k\left(2x+\frac{2\pi}3\right)\sin^k 2x \sin^k\left(2x-\frac{2\pi}3\right) = \frac1{8^k}\mathcal P_{k}(\sin t, 2x).
\end{align}
Therefore,
\begin{cases}
\mathcal S_1(\cos t, x)=0\\
\mathcal Q_1(\cos t, x) = -\frac34\\
\mathcal P_1(\cos t, x) = \frac14\cos3x\\
\mathcal S_k(\sin t, x) = \mathcal S_k\left(\cos t,x-\frac\pi2\right)\\[4pt]
\mathcal Q_k(\sin t, x) = \mathcal Q_k\left(\cos t,x-\frac\pi2\right)\\[4pt]
\mathcal P_k(\sin t, x) = \mathcal P_k\left(\cos t,x-\frac\pi2\right)\\[4pt]
\mathcal S_k(\cos t \sin t, x) = \frac1{2^k}\mathcal S_k(\sin t,2x)\\[4pt]
\mathcal Q_k(\cos t \sin t, x) = \frac1{4^k}\mathcal Q_k(\sin t,2x)\\[4pt]
\mathcal P_k(\cos t \sin t, x) = \frac1{8^k}\mathcal P_k(\sin t,2x).\tag6
\end{cases}
$\color{brown}{\textbf{Calculation of sums.}}$
$\mathcal S_1(\cos t, x),\ \mathcal Q_1(\cos t, x),\ \mathcal P_1(\cos t, x)\ $ can be considered as the elementary symmetric polynomials. In this way,
If $f(t) = \cos t,$ then
$$\begin{align}
&\mathcal S_2 = \mathcal S_1^2 - 2\mathcal Q_1 = \frac32,\\
&\mathcal Q_2 = \mathcal Q_1^2 - 2\mathcal S_1 \mathcal P_1 = \frac9{16},\\
&\mathcal S_4 = \mathcal S_2^2 - 2\mathcal Q_2 = \frac98.
\end{align}\tag7$$
Formulas $(6)-(7)$ allow to fill the table $(8).$
Besides,
$$\begin{align}
&\mathcal P_2(\cos t, x) + \mathcal P_2(\sin t, x)\\ 
&= \cos^2\left(x-\frac{2\pi}3\right)\cos^2 x \cos^2\left(x+\frac{2\pi}3\right)+\sin^2\left(x-\frac{2\pi}3\right)\sin^2 x \sin^2\left(x+\frac{2\pi}3\right) =\frac1{16}.
\end{align}$$
\begin{vmatrix}
k & f(t) & \mathcal S_k(f(t),x) & \mathcal Q_k(f(t),x) & \mathcal P_k(f(t),x)\\
1 & \cos t       & 0         & -\dfrac34    & \dfrac14\cos 3x\\
1 & \sin t       & 0         & -\dfrac34    & -\dfrac14\sin 3x\\
1 & \cos t\sin t & 0         & -\dfrac3{16} & -\dfrac1{32}\sin 6x\\
2 & \cos t       & \dfrac 32 & \dfrac9{16}  & \dfrac1{16}\cos^2 3x\\
2 & \sin t       & \dfrac 32 & \dfrac9{16}  & \dfrac1{16}\sin^2 3x\\
2 & \cos t\sin t & \dfrac 38 & \dfrac9{256} & \dfrac1{1024}\sin^2 6x\\
4 & \cos t       & \dfrac98  & & \\
4 & \sin t       & \dfrac98  & & \\
4 & \cos t\sin t & \dfrac9{128} & & \tag8
\end{vmatrix}
Any function which depends only from the considered constant polynomials, should be a constant too.
In particular, for the considered functions $f(t)$
$$\mathcal P_k = \mathcal P_1^k,\\
\mathcal S_3 - 3\mathcal P_3 = \mathcal S_1^3 - 3\mathcal S_1\mathcal Q_1 = 0.$$