Sums of powers of cosines and sines shifted by $2\pi/3$ 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.

This is an answer to question 3.
Let
$$f_n(x):=\cos^n(x)+\cos^n\bigg(x+\frac{2\pi}{3}\bigg)+\cos^n\bigg(x+\frac{4\pi}3\bigg)$$
$$\small g_n(x):=\cos^n(x)\sin^n(x)+\cos^n\bigg(x+\frac{2\pi}3\bigg)\sin^n\bigg(x+\frac{2\pi}3\bigg)+\cos^n\bigg(x+\frac{4\pi}3\bigg)\sin^n\bigg(x+\frac{4\pi}3\bigg) $$
where $n$ is a positive integer.
This answer proves the following two claims :
Claim 1 : $f_n(x)$ is a constant function if and only if $n=1,2,4$.
Claim 2 : $g_n(x)$ is a constant function if and only if $n=1,2,4$.

Claim 1 : $f_n(x)$ is a constant function if and only if $n=1,2,4$
Proof :
You already noticed that $f_1(n),f_2(n)$ and $f_4(n)$ are constant functions.
Now, let us prove that if $f_n(x)$ is a constant function, then $n=1,2,4$ as follows :
$$\begin{align}&\text{$f_n(x)$ is a constant function}
\\\\&\implies f_n(0)=f_n\bigg(\frac{\pi}{6}\bigg)
\\\\&\implies 1+\bigg(-\frac 12\bigg)^n+\bigg(-\frac 12\bigg)^n=\bigg(\frac{\sqrt 3}{2}\bigg)^n+\bigg(-\frac{\sqrt 3}{2}\bigg)^n+0
\\\\&\implies 2^n+2(-1)^n-(\sqrt 3)^n-(-\sqrt 3)^n=0
\\\\&\implies \begin{cases}2^n-2=0&\text{if $n$ is odd}\\\\2(\sqrt 3)^{n-1}\bigg(\bigg(\frac{2}{\sqrt 3}\bigg)^{n-1}-\sqrt 3\bigg)+2=0&\text{if $n$ is even}\end{cases}
\\\\&\implies n=1,2,4\end{align}$$
since for odd $n$ , we have $2^n-2=0\implies n=1$, and for even $n$, letting $h(n):=2(\sqrt 3)^{n-1}\bigg(\bigg(\frac{2}{\sqrt 3}\bigg)^{n-1}-\sqrt 3\bigg)+2$, we see that $h(2)=h(4)=0$ and that $h(n)$ is increasing for $n\ge 6$ with $h(6)=12$.

Claim 2 : $g_n(x)$ is a constant function if and only if $n=1,2,4$.
Proof :
You already noticed that $g_2(n)$ and $g_4(n)$ are constant functions. We have $g_1(n)=0$.
Now, let us prove that if $g_n(x)$ is a constant function, then $n=1,2,4$ as follows :
$$\small\begin{align}&\text{$g_n(x)$ is a constant function}
\\\\&\implies g_n(0)=g_n\bigg(\frac{\pi}{4}\bigg)
\\\\&\implies 0+\bigg(-\frac 12\bigg)^n\bigg(\frac{\sqrt 3}{2}\bigg)^n+\bigg(-\frac 12\bigg)^n\bigg(\frac{-\sqrt 3}{2}\bigg)^n\\&\qquad\qquad =\bigg(\frac{1}{\sqrt 2}\bigg)^n\bigg(\frac{1}{\sqrt 2}\bigg)^n+\bigg(-\frac{1+\sqrt 3}{2\sqrt 2}\bigg)^n\bigg(\frac{\sqrt 3-1}{2\sqrt 2}\bigg)^n+\bigg(\frac{\sqrt 3-1}{2\sqrt 2}\bigg)^n\bigg(-\frac{1+\sqrt 3}{2\sqrt 2}\bigg)^n
\\\\&\implies 2^n+2(-1)^n-(\sqrt 3)^n-(-\sqrt 3)^n=0
\\\\&\implies n=1,2,4\end{align}$$
where the last step is the same as that of the proof for claim 1.
