Chebyshev polynomials and rational values of cosine

1. Induction. $P_0=2$ and $P_1=x$ have integer coefficients, and so does $P_{n+1}=xP_n-P_{n-1}$. If $P_{n-1}$ has degree $n-1$ and $P_n$ is monic of degree $n$, then $xP_n$ is monic of degree $n+1$ and $P_{n+1}$ is monic of degree $n+1$.

For the trigonometric identity: $P_0(2\cos\theta)=2=2\cos0$ and $P_1(2\cos\theta)=2\cos\theta$. Assume it holds for $n-1,n$. The sum-to-product formula gives
$$2\cos((n+1)\theta)=2\cos\theta\cdot2\cos n\theta-2\cos((n-1)\theta)=P_{n+1}(2\cos\theta).$$

2. Let $r=\frac pq$ with $q\ge1$ and put $\theta=r\pi$, $x=2\cos\theta$. Then
$$P_{2q}(x)=2\cos(2q\theta)=2\cos(2p\pi)=2,$$
so $x$ is a root of the monic integer polynomial $P_{2q}(t)-2$. If $x$ is rational, the rational root theorem forces $x$ to be an integer. Since $|x|\le2$, $x\in\{-2,-1,0,1,2\}$, i.e. $\cos(r\pi)\in\left\{-1,-\frac12,0,\frac12,1\right\}$.

Remark. This is Niven's theorem. In particular $\cos1^\circ$ is irrational.

Lesson. Express $2\cos n\theta$ as a monic integer polynomial in $2\cos\theta$: it turns trigonometric rationality questions into rational root theorem questions.