Rational and Irrational Values of cosθ, sinθ, and tanθ

# Rational and Irrational Values of cosθ, sinθ, and tanθ

 Lemma 1: If $r \in \mathbb{Q}$ and $\theta = r \pi$ where $\cos \theta \neq \pm 1$ then for each $n \in \mathbb{N}$ there is a monic polynomial $P_n$ with degree $n$ such that $2 \cos n \theta = P_n(2 \cos \theta)$.
• Proof: In the case when $n = 1$ let $P_1(x) = x$. Then $P_1$ is monic and:
(1)
\begin{align} \quad P_1(2 \cos \theta) = 2 \cos \theta = 2 \cos 1 \theta \end{align}
• In the case when $n = 2$, let $P_2(x) = x^2 - 2$. Then $P_2$ is monic and:
(2)
\begin{align} \quad P_2(2 \cos \theta) = (2 \cos \theta)^2 - 2 = 4 \cos^2 \theta - 2 = 2(2 \cos^2 \theta - 1) = 2 \cos 2 \theta \end{align}
• For the inductive step, suppose that $P_n$ and $P_{n-1}$ are monic polynomials of degree $n$ and $n-1$ respectively for which $P_n(2 \cos \theta) = 2 \cos n \theta$ and $P_{n-1}(2 \cos \theta) = 2 \cos (n-1)\theta$. Let:
(3)
\begin{align} \quad P_{n+1}(x) = xP_n(x) - P_{n-1}(x) \end{align}
• Then $P_{n+1}$ is a monic polynomial of degree $n + 1$, and so by trigonometric identities:
(4)
\begin{align} P_{n+1}(2 \cos \theta) &= [2 \cos \theta]P_n(2 \cos \theta) - P_{n-1}(2 \cos \theta) \\ &= [2 \cos \theta][2 \cos n \theta] - [2 \cos ((n-1) \theta)] \\ &= 4[\cos \theta][\cos n \theta] - [2 \cos ((n-1)\theta)] \\ &= 4 \left [ \frac{1}{2} \cdot \cos ((n - 1)\theta) + \cos ((n+1) \theta) \right ] - [2 \cos ((n-1)\theta)] \\ &= 2 \cos ((n+1) \theta) \quad \blacksquare \end{align}
 Theorem 1: If $r \in \mathbb{Q}$ and $\theta = r \pi$ then: a) $\cos \theta$ is equal to $0$, $\pm 1$, $\pm \frac{1}{2}$, or is irrational. b) $\sin \theta$ is equal to $0$, $\pm 1$, $\pm \frac{1}{2}$, or is irrational. c) $\tan \theta$ is equal to $0$, $\pm 1$, undefined, or is irrational.
• Proof: Let $r = \frac{a}{b} \in \mathbb{Q}$, $\theta = \frac{a}{b} \pi$ and let $P_b$ be the monic polynomial of degree $b$ for which:
(5)
\begin{align} \quad P_b(2 \cos \theta) = 2 \cos b \theta \end{align}
• Which exists by Lemma 1. Then $P_b(x) \pm 2$ is a polynomial which has $2 \cos \theta$ as a root since:
(6)
\begin{align} \quad P_b(2 \cos \theta) \pm 2 = 2 \cos a \pi \pm 2 = 0 \end{align}
• By The Rational Roots Theorem, any solutions must be irrational or $2 \cos \theta$ is an integer and since $P_b$ is monic, the rational possibilities are $0$, $\pm 2$, or $\pm 1$. So $\cos \theta$ is either irrational or the rational possibilities are $0$, $\pm 1$, or $\pm \frac{1}{2}$. $\blacksquare$