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.

Yes, any polynomial identity involving $\cos(mx + c)$ and $\sin(mx+c)$ for various constants $c$ and integers $m$ can be written in the form
$R(z) = 0$ where $z = e^{ix}$ and $R$ is a rational function involving the $e^{ic}$.  For this to be true, the numerator of $R(z)$ must simplify to the polynomial $0$.
For example, let's take
$$ \cos^2(x) + \cos^2(x+2\pi/3) + \cos^2(x+4\pi/3)=3/2 $$
Expressed in terms of $z = e^{ix}$, this becomes
$$ \frac{z^2}{4} + \frac{1}{2} + \frac{1}{4z^2} + \frac{z^2}{4} e^{4\pi i/3} + \frac{1}{2} + \frac{1}{4 z^2} e^{-4\pi i/3} + \frac{z^2}{4} e^{8\pi i/3} + \frac{1}{2} + \frac{1}{4 z^2} e^{-8\pi i/3} = \frac{3}{2} $$
which simplifies to
$$ \left(1 + e^{4\pi i/3} + e^{8\pi i/3}\right) \frac{z^2}{4} + 
\left(1 + e^{-4\pi i/3} + e^{-8\pi i/3}\right) \frac{1}{4 z^2} = 0 $$
and that is true, as we verify by showing $$1 + e^{4\pi i/3} + e^{8\pi i/3} = 0$$
and $$ 1 + e^{-4\pi i/3} + e^{-8\pi i/3} = 0$$
Note that if $w = e^{4\pi i/3}$, the first is $1 + w + w^2 = (1-w^3)/(1-w)$, and
$w^3 = e^{4\pi i} = \left(e^{2\pi i}\right)^2 = 1$.  Similarly for the second.
EDIT: For question 3, you basically want to know what polynomial identities are satisfied by the $e^{ic}$.  If there is only one $c$, then $e^{ic}$ must be an algebraic number, and all polynomial identities it satisfies are multiples of its minimal polynomial.  For example, if $c = 2 m \pi/n$
with $m$ and $n$ coprime, then the minimal polynomial is the cyclotomic polynomial $C_n(w)$.  Things can be more complicated if there are several different $c$.
EDIT: For example, the $6$'th cyclotomic polynomial is $C_6(w) = w^2 - w + 1$, and its roots are $e^{2\pi i k/6}$. where $k$ and $6$ are coprime, i.e. $e^{\pi i/3}$ and $e^{- \pi i/3}$.
We might take $$(z+1/z)(w - 1 + 1/w) = z w + \frac{1}{zw} - z - \frac{1}{z} + \frac{z}{w} + \frac{w}{z}$$
which with $w = \exp(i\pi/3)$ and $z = \exp(ix)$ becomes
$$ 2 \cos(x+\pi/3) - 2 \cos(x) + 2 \cos(x-\pi/3) = 0 $$
