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<p/2. (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.

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Filed under Algebra, Number theory

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