Is there a generalization of the following equality, for arbitrary fractions of $2\pi$?

Question

In the context of group representations of the Icosahedral group symmetry, I noticed that:

$$ \cos^2\left(\frac{2\pi}{5}\right) + \cos^2\left(\frac{4\pi}{5}\right) = \frac{3}{4}$$

Is it possible to generalize this for other fractions of $2\pi$? In Wikipedia's List of trigonometric identities article, I found Euclid's identity somewhat similar, but I'm not sure it is related, and still I wonder whether a generalization of it exists.

Answer 1

We know $\cos^2 x = \frac 12 (1+\cos(2x))$, so your identity is related to

$$ \frac 12\left(1 + \cos\left(\frac{4\pi}{5}\right) + 1 + \cos\left(\frac{8\pi}{5}\right)\right) = \frac 34 $$ $$ \cos\left(\frac{2\pi}{5}\right)+\cos\left(\frac{4\pi}{5}\right) = -\frac 12 $$

since $\cos \left(\frac{8\pi}{5}\right) = \cos \left(\frac{2\pi}{5}\right)$.

Many trig sums of this type can be proved with the help of complex roots of unity, and a simple application of that technique does help here.

By Euler's formula, $\cos \left(\frac{2k\pi}{5}\right) = \Re (e^{2k\pi i/5})$ where $\Re$ is the real part of a complex number.

The five values $e^{2k\pi i/5}$ with $k \in \{0,1,2,3,4\}$ are the five solutions of the polynomial equation $z^5 - 1 = 0$. The sum of these five roots must be $-\frac{a_4}{a_5}=0$ where $a_4=0$ is the coefficient of $z^4$ and $a_5=1$ is the coefficient of $z^5$.

$$ 1 + e^{2\pi i/5} + e^{4\pi i/5} + e^{6\pi i/5} + e^{8\pi i/5} = 0 $$

Since these are equal as complex numbers, their real parts must also be equal, and $\Re$ is additive:

$$ 1 + \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right) + \cos\left(\frac{8\pi}{5}\right) = 0 $$

And since $\cos\left(\frac{2\pi}{5}\right) = \cos\left(\frac{8\pi}{5}\right)$ and $\cos\left(\frac{4\pi}{5}\right) = \cos\left(\frac{6\pi}{5}\right)$, we do get

$$ \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) = -\frac 12 $$

So to generalize, let $2n+1$ take the place of $5$, with $n$ a positive integer. Take the polynomial $z^{2n+1}-1$; its roots are $e^{2k \pi i/(2n+1)}$ for $k \in \{0,1,\ldots,2n\}$. Since the polynomial's coefficient of $z^{2n}$ is zero, the sum of these roots is zero.

$$ \sum_{k=0}^{2n} e^{2k \pi i/(2n+1)} = 0 $$

The term for $k=0$ is $1$.

$$ \sum_{k=1}^{2n} e^{2k \pi i/(2n+1)} = -1 $$

The real part of this equation is

$$ \sum_{k=1}^{2n} \cos\left(\frac{2k \pi}{2n+1}\right) = -1 $$

Split the sum into the terms where $k=2n+1-2j$ is odd and the terms where $k=2j$ is even:

$$ \left(\sum_{j=1}^n \cos\left(\frac{2(2n+1-2j) \pi}{2n+1}\right)\right) + \left(\sum_{j=1}^n \cos\left(\frac{4j \pi}{2n+1}\right)\right) = -1 $$

But $\cos\left(\frac{2(2n+1-2j) \pi}{2n+1}\right) = \cos\left(\frac{4j \pi}{2n+1}\right)$, so the two sums are equal, and

$$ \sum_{k=1}^n \cos\left(\frac{4k \pi}{2n+1}\right) = -\frac 12 $$

Apply the identity $\cos(2x) = 2\cos^2 x-1$ to get

$$ 2 \left(\sum_{k=1}^n \cos^2\left(\frac{2k \pi}{2n+1}\right)\right) -n = -\frac 12 $$

and finally,

$$ \sum_{k=1}^n \cos^2\left(\frac{2k \pi}{2n+1}\right) = \frac{2n-1}{4} $$

which matches your identity when $n=2$.

Answer 2

This is because $\zeta=\exp(2\pi i/5)$ is a primitive fifth root of $1$.

$$ \begin{align} \cos\left(\frac{2\pi}{5}\right)^2 + \cos\left(\frac{4\pi}{5}\right)^2 &= \left(\frac{\zeta + \bar\zeta}{2}\right)^2 + \left(\frac{\zeta^2 + \bar\zeta^2}{2}\right)^2\\ &=\frac{\zeta^2 + 2\zeta\bar\zeta + \bar\zeta^2 + \zeta^4+2\zeta^2\bar\zeta^2 + \bar\zeta^4}{4}\\ &=\frac{\zeta^2 + 2 + \zeta^{-2} + \zeta^{-1} + 2 + \zeta^{-4}}{4}\\ &=\frac{4+\zeta^2+\zeta^3+\zeta^4+\zeta}{4} = \frac{4-1}{4}=\frac{3}{4} \end{align} $$ Using that $\bar\zeta=\zeta^{-1}$, $\zeta^{k+5l}=\zeta^k$ and $\sum_{k=0}^4 \zeta^k = 0$.