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.

If $\cos3y=\cos3x$
$3y=2n\pi\pm3x$ where $n$ is any integer
$y=\dfrac{2n\pi}3+x$ where $n=0,1,2$
Again, $\cos3y=4\cos^3y-3\cos y$
So, the roots of $$4\cos^3y-3\cos y-\cos3x=0$$ are  $p=\cos x,q=\cos\left(\dfrac{2\pi}3+x\right),r=\cos\left(\dfrac{4\pi}3+x\right)$
Using Vieta's formula, $$p+q+r=\dfrac04\ \ \ \  (1)\text{ and }pq+qr+rp=\dfrac{-3}4\ \ \ \  (2)\text{ and }pqr=\dfrac{\cos3x}4\ \ \ \  (3)$$
By $(1),(2)$ $$p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+rp)=?\  \ \  \ (4)$$
By $(1),(3)$ $$p^3+q^3+r^3=3pqr=?\  \ \  \ (5)$$
A little Transformation of equation

*

*Let $c=\cos^2y$
$$(\cos3x)^2=(4\cos^3y-3\cos y)^2$$
$$\implies16c^3-24c^2+9c-\cos^23x=0$$ whose roots are $p^2,q^2,r^2$
Again applying Vieta's formula,  $$p^2+q^2+r^2=\dfrac{24}{16}\  \ \  \ (6)\text{ compare with }(4)$$
$$p^2q^2+q^2r^2+r^2p^2=\dfrac9{16}\  \ \  \ (7)\text{ and } p^2q^2r^2=\dfrac{\cos^23x}{16}\  \ \  \ (8)\text{ compare with }(3)$$
By $(6),(7)$ $$p^4+q^4+r^4=(p^2+q^2+r^2)^2-2(p^2q^2+q^2r^2+r^2p^2)=?\  \ \  \ (9)$$

*

*Let $s=\dfrac1{\cos y}$
$$\dfrac4{s^3}-\dfrac3s-\cos3x=0\iff(\cos3x)s^3+3s^2-4=0$$ whose roots are $\dfrac1p,\dfrac1q,\dfrac1r$
$$\implies\dfrac1p+\dfrac1q+\dfrac1r=-\dfrac3{\cos3x}=-3\sec3x\ \ \ \ (10)$$
Similarly, $$\dfrac1{pq}+\dfrac1{qr}+\dfrac1{rp}=?\ \ \ \ (11)\text{ and }\dfrac1{pqr}=?\ \ \ \ (12)$$
Finally as  $\dfrac1p=\sec x$ etc.,
using $(10,11),$ $$\sec^2x+\sec^2\left(\dfrac{2\pi}3+x\right)+\sec^2\left(\dfrac{4\pi}3+x\right)=\left(\dfrac1p+\dfrac1q+\dfrac1r\right)^2-2\left(\dfrac1{pq}+\dfrac1{qr}+\dfrac1{rp}\right)=?$$
Generalization
$$\cos ny=\cos nx$$ Can this be left as an exercise?
