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:

1. Do the other identities admit similar more or less intuitive interpretations as well?
2. Do such identities have names?
3. 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.

Solution 1:

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 $$ 2 \cos(x+\pi/3) - 2 \cos(x) + 2 \cos(x-\pi/3) = 0 $$

Solution 2:

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.

Solution 3:

$\color{green}{\textbf{Version of 20.07.20.}}$

$\color{brown}{\textbf{Preliminary calculations.}}$

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 $f(t)$ $$\mathcal P_k = \mathcal P_1^k,\\ \mathcal S_3 - 3\mathcal P_3 = \mathcal S_1^3 - 3\mathcal S_1\mathcal Q_1 = 0.$$

Solution 4:

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?