# Chebyshev polynomials and algebraic values of the cosine function

Let $p$ be an odd prime and let $T_n(X)$ be the $n$-th Chebyshev polynomial of the first kind. By Lemma 1 on page 5 of this paper, $T_p(X)/X$ is irreducible over $\mathbb Q$. Now

$\displaystyle T_p(\cos\frac{\pi}{2p})=\cos\frac{\pi}{2}=0$,

so the minimal polynomial of $\cos(\pi/2p)$ over $\mathbb Q$ is $T_p(X)/X$. (Note that we are slightly abusing terminology here. By definition, the minimal polynomial is a monic polynomial. But let us drop precision for the sake of brevity.)

By definition (and induction, if you may), $\deg(T_n)=n$. Hence we deduce the following:

Proposition 1. $\cos(\pi/2p)$ is an algebraic number of degree $p-1$.

Since $T_p(X)/X$ consists of only even powers of $X$, $T_p(\cos(\pi/2p))$ can be expressed as a polynomial in $\cos(\pi/p)=2\cos^2(\pi/2p)-1$. Therefore:

Proposition 2. $\cos(\pi/p)$ is an algebraic number of degree $(p-1)/2$.

A nice corollary of this is Niven’s theorem:

Corollary (Niven’s theorem). If $\theta\in (0,\pi/2)$ is a rational multiple of $\pi$ such that $\cos\theta$ is rational, then $\theta=\pi/3$.

Proof. Note that if $\cos\theta$ is rational, then so is $\cos(n\theta)=T_n(\cos\theta)$ for any positive integer $n$. So, (reduction 1) without loss of generality, $\theta=k\pi/p$ for some odd prime $p$ and positive integer $k. (The case $p=2$ is dealt with trivially.)

Further, multiplying by the inverse of $k\pmod p$, (reduction 2) one may assume $k=1$ without any loss of generality.

Now being rational is the same as being algebraic of degree $1$. Hence, by the above, we must have $p=3$ and the conclusion follows immediately. $\square$

Exercise. Find all rational multiples $\theta\in(0,\pi/2)$ of $\pi$ such that $\cos\theta$ is a quadratic irrational.