Niven's Theorem

Theorem
Consider the angles $\theta$ in the range $0 \le \theta \le \dfrac \pi 2$.

The only values of $\theta$ such that both $\dfrac \theta \pi$ and $\sin \theta$ are rational are:


 * $\theta = 0: \sin \theta = 0$


 * $\theta = \dfrac \pi 6: \sin \theta = \dfrac 1 2$


 * $\theta = \dfrac \pi 2: \sin \theta = 1$

Proof
We will prove that if both $\dfrac \theta \pi$ and $\cos \theta$ are rational then:
 * $\theta \in \set{ {0, \dfrac \pi 3, \dfrac \pi 2} }$

Lemma
Suppose that $\dfrac \theta \pi$ is rational, meaning:
 * $\theta = \dfrac {2 \pi k} n$

where $k, n \in \Z$ and $n \ge 1$.

Suppose also that $\cos \theta \in \Q$.

Denoting $c = 2 \cos \theta \in \Q$, we get:
 * $\map {F_n} c = \map {F_n} {2 \cos \dfrac {2 \pi k} n} = 2 \map \cos {2 \pi k} = 2$

So $c$ is a rational root of $\map {F_n} x - 2$, which is a monic polynomial with integer coefficients.

By Rational Root Theorem, $c$ must be an integer.

But:
 * $\size c = \size {2 \cos \theta} \le 2$

so:
 * $c \in \set {-2, -1, 0, 1, 2}$

Assuming that $0 \le \theta \le \dfrac \pi 2$, we get that:
 * $\theta \in \set {0, \dfrac \pi 3, \dfrac \pi 2}$

Thus for any $\theta$ in the range $0 \le \theta \le \dfrac \pi 2$ such that both $\dfrac \theta \pi$ and $\cos \theta$ are rational, then:
 * $\theta \in \set {0, \dfrac \pi 3, \dfrac \pi 2}$

Instead of the above, suppose that:
 * $0 \le \alpha \le \dfrac \pi 2$

and both of $\dfrac \alpha \pi$ and $\sin \alpha$ are rational.

Then we can denote $\theta = \dfrac \pi 2 - \alpha$ and get that:
 * $0 \le \theta \le \dfrac \pi 2$
 * $\dfrac \theta \pi \in Q$
 * $\cos \theta \in \Q$

So:
 * $\dfrac \pi 2 - \alpha = \theta \in \set {0, \dfrac \pi 3, \dfrac \pi 2}$

therefore:
 * $\alpha \in \set {0, \dfrac \pi 6, \dfrac \pi 2}$

It is suspected that this result is considerably older, and may date back as far as.