Irrationality of $\cos 2\pi / n$

Question

Let $n$ be a positive integer. Is $\cos (2\pi/n)$ irrational if $n$ is not $1$, $2$, $3$, $4$ or $6$?

Answer 1

This is a proof using Galois theory posted on Wilson Ong's Blog.

We know $\cos(2\pi)=1$ and $\cos(\pi)=-1$. Let $n \geq 3 $ be a positive integer and $\zeta = e^{\frac{2\pi i}{n}}$ be the first root of unity. Assume $\cos(\frac{2\pi}{n})=\frac{\zeta + \zeta^{-1}}{2}$ is a rational number. Consider the Galois group $Gal(\mathbb{Q}(\zeta)/\mathbb{Q}) \cong (\mathbb{Z} / n\mathbb{Z})^{\times}$. If $\cos(\frac{2\pi}{n})$ is rational, any $\sigma \in Gal(\mathbb{Q}(\zeta)/\mathbb{Q})$ must fix it. Since $\zeta$ is a root of the polynomial $t^n -1 \in \mathbb{Q}[t]$, so is $\sigma(\zeta)$. Thus $\sigma =\zeta^{j}$ for some $1 \leq j < n$. Apply this: $$\cos(\frac{2\pi}{n})=\sigma \left( \cos(\frac{2\pi}{n}) \right)=\sigma \left(\frac{\zeta+\zeta^{-1}}{2} \right)=\frac{\zeta^{j} + \zeta^{-j}}{2}=\cos(\frac{2\pi j}{n})$$ It follows that $j \in \{ 1, n-1 \}$. Hence $\sigma \in Gal(\mathbb{Q}(\zeta)/\mathbb{Q}) $ sends $\zeta$ into $\{ \zeta, \zeta^{-1} \}$, which shows $\big| Gal(\mathbb{Q}(\zeta)/\mathbb{Q}) \big| = \big| (\mathbb{Z} / n\mathbb{Z})^{\times} \big| \leq 2$. Thus the value of $ \big| (\mathbb{Z} / n\mathbb{Z})^{\times} \big| = \phi(n)$ is $1$ or $2$. Since the totient function $\phi(n)$ is multiplicative and $\phi(p^k)=p^{k-1}(p-1)$ for prime $p$ and $k \geq 1$, we can conclude that

1. any prime $p \geq 5 $ does not divide $n$
2. $2^k \mid n$ implies $k \leq 2 $
3. $3^k \mid n$ implies $k = 1 $

Thus $n = 3, 4, 6$.

Conversely, we have \begin{align*} &\cos(2\pi)=1 \\ & \cos(2\pi/2) = -1 \\ & \cos(2\pi/3) = -\frac{1}{2} \\ & \cos(2\pi/4) = 0 \\ & \cos(2\pi/6)= \frac{1}{2} \end{align*}

Therefore, for positive integer $n$, $\cos (\frac{2\pi}{n})$ is rational if and only if $n=1,2,3,4,6$.