Problem in trigonometry solved with more advanced topics - $\displaystyle \cos(q\pi) \in \mathbb{Q} \to \cos(q\pi) \in \{0, \pm \frac{1}{2}, \pm 1\}$

Question

Prove the following affirmation: If $q$ is a rational number and $\cos(q\pi)$ is also a rational number, prove that $\cos(q\pi)$ must be one of the elements of the set $\{0, \pm \frac{1}{2}, \pm 1\}$.

Context: I have encountered this exercise while studying abstract algebra - polynomials, algebraic numbers and algebraic integers. More precisely, I have studied in the chapter: Fundamental Theorem of Algebra, Symmetric Polynomials, Fundamental Theorem of Symmetric Polynomials, Kronecker's Theorem on roots of unity, algebraic numbers and algebraic integers - definitions, elementary/minimum polynomial, closure of $\overline{\mathbb{Z}}$ and $\overline{\mathbb{Q}}$ to $+$ and $\cdot$.

However, I can't relay to neither of this topic when studying trigonometric functions. I have observed, obviously, that those particular values are specific to rational 'multiples' of transcendental number $\pi$, but I can't obtain any more information.

Answer 1

Step 1. There is a unique polynomial $F_n(x) \in \mathbf{Z}[x]$ of degree $n$ for $n > 0$ with leading term $x^n$ such that, for any real number $\theta$,

$$F_n(2 \cos(\theta)) = 2 \cos(n \theta).$$

For example $F_0(x) = 2$, $F_1(x) = x$, and $F_2(x) = x^2-2$ because

$$4 \cos^2(\theta) - 2 = 2(2 \cos^2(\theta) - 1) = 2 \cos(2 \theta)$$

by the double angle formula. More generally, by the addition and subtraction formula one has:

$$\cos((n+1) \theta) = \cos(n \theta) \cos(\theta) - \sin(n \theta) \sin(\theta),$$ $$\cos((n-1) \theta) = \cos(n \theta) \cos(\theta) + \sin(n \theta) \sin(\theta),$$

and thus

$$\cos((n+1) \theta) + \cos((n-1) \theta) = 2 \cos(\theta) \cos(n \theta),$$

or, after multiplying both sides by $2$,

$$F_{n+1}(x) + F_{n-1}(x) = x F_{n}(x).$$

Here $F_n(x)$ is (up to scaling) known as a Chebyshev polynomial of the first kind. (https://en.wikipedia.org/wiki/Chebyshev_polynomials)

Step 2. If $q = a/b$ is a rational number then there is an integer $n$ such that $(n-1)q$ is an integer and is a multiple of $2$, for example, $n = 1+2b$. But then if $\theta = q \pi$, it follows that $n \theta - \theta$ is a multiple of $2 \pi $, and thus $\cos(n \theta) = \cos(\theta)$. But that means that if $x = 2 \cos(\theta)$, then

$$T_n(x) = 2 \cos(n \theta) = 2 \cos(\theta) = x,$$

in particular, for $n > 1$, $x$ is a root of the polynomial $T_n(x) - x = 0$ which has integral coefficients and is monic.

Step 3. Gauss' Lemma: if $x$ is a rational root of a monic polynomial with integer coefficients, then $x$ is an integer. S ee https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomials), or more specifically the special case of the rational root theorem (https://en.wikipedia.org/wiki/Rational_root_theorem).

Step 4. We have deduced that if $q$ is rational and $\cos(q \pi)$ is rational, then $2 \cos(q \pi)$ is an integer. But sin ce cosine takes values between $[-2,2]$, the only integers it can equal are $\{-2,1,0,1,2\}$, or equivalently the only way $\cos(q \pi)$ can be rational for $q$ rational is if it takes values in $\{-1,-1/2,0,1/2,1\}$.

Answer 2

Suppose $\cos(\frac{2\pi m}n)=\frac12(e^{2\pi im/n}+e^{-2\pi im/n})\in\mathbb Q(e^{2\pi i/n})$ is rational, with $m$ and $n$ coprime. Then, the Galois group $Gal(\mathbb Q(e^{2\pi i/n})/\mathbb Q)$ acts trivially on it, so for any $d$ coprime to $n$, we have: $$\cos(\frac{2\pi m}n)=\cos(\frac{2d\pi m}n),$$ since the Galois action sends $e^{2\pi i/n}\mapsto e^{2\pi id/n}$. This condition is only met for very few possibilities of $n$.