If $\cos\pi\theta$ is algebraic and $\theta$ is irrational, what is the set of possible $\theta$?

Question

I know that $a= \cos \pi \theta$ is an algebraic number ($\theta$ is rational). I want to prove that if $\cos\pi\theta$ is rational, then the possible only possible values of $\theta$ are $0,±1/2,±1$ Thanks for your help.

Answer 1

I am assuming that the question is about Niven's theorem (it is rather poorly phrased) so I give a short proof.

If $ \cos(\pi \theta) = \cos( \pi p / q) $, then we have that $ 2\cos(\pi \theta) = \zeta_{2q}^p + \zeta_{2q}^{-p} $ where $ \zeta_{2q} = e^{\pi i/q} $ is a primitive root of $ X^{2q} - 1 = 0 $, and therefore is an algebraic integer. This implies that $ 2\cos(\pi \theta) $ is an algebraic integer for rational $ \theta $. The only rational algebraic integers are the rational integers, so if $ 2 \cos(\pi \theta) $ is rational it must be an integer. As it is between $ -2 $ and $ 2 $, the result follows immediately.