Given that $\cos\left(\dfrac{2\pi m}{n}\right) \in \mathbb{Q}$ prove $\cos\left(\dfrac{2\pi}{n}\right) \in \mathbb{Q}$

HINT:

• For all $\theta\in\mathbb R, m\in\mathbb N$, $\cos(m\theta)$ can be expressed as a polynomial of $\cos(\theta)$ (with rational coefficients), so if $\cos(\theta)$ is rational, so is $\cos(m\theta)$.
• If $m,n\in\mathbb Z$ and $\gcd(m,n)=1$, there exists $k\in\mathbb N$ such that $mk\equiv 1 \bmod n$.
• The cosine function is periodic modulo $2\pi$, that is, $\cos(\theta + 2\pi) = \cos(\theta)$ for all $\theta\in\mathbb R$.

Can you combine these facts to come up with a proof of the desired result?