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 $\theta$ is one of $0, \dfrac \pi 3, \dfrac \pi 2$.

Lemma A: For any integer $n \ge 1$, there exist a polynomial $F_n \left({x}\right)$ such that:
 * $F_n \left({2 \cos t}\right) = 2 \cos nt$

In addition, $\deg F_n = n$, and $F_n$ is a monic polynomial with integer coefficients.

Proof of lemma A: by induction.

For $n=1$, it is easy to see that $F_1 \left(x\right) = x$ fulfils the propositions.

For $n=2$, $F_2 \left(x \right) = x^2-2$

For $n>2$:
 * $2 \cos (n - 1) t \cos t = \cos n t + \cos (n - 2) t \implies 2 \cos n t = \left({2 \cos (n - 1) t}\right) \left(2\cos t\right)-2\cos (n-2)t = 2\cos t F_{n-1}(2\cos t)- F_{n-2}(2\cos t)$

so:
 * $F_n(x) = xF_{n-1}(x)-F_{n-2}(x) \in \Z[x]$

will fulfil:
 * $F_{n}\left(2\cos t\right) = 2\cos nt$

Because $\deg F_{n-1} = n-1, \deg F_{n-2} = n-2$, we can conclude that:
 * $\deg F_n = \deg \left( xF_{n-1}(x)-F_{n-2}(x)\right) = n$.

In addition, the leading coefficient of $F_n$ is equal to the leading coefficient of $F_{n-1}$, which is $1$.

So we have proved the lemma.

Suppose that $\dfrac \theta \pi$ is rational, meaning $\theta = \frac{2\pi k}{n}$ 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:
 * $F_n(c) = F_n\left(2\cos \frac{2\pi k}{n}\right) = 2\cos \left(2\pi k\right) = 2$

So $c$ is a rational root of $F_n(x)-2$, which is a monic polynomial with integer coefficients, therefore $c$ must be integer.

But, $|c| = |2\cos \theta| \le 2$, so $c=-2,-1,0,1,2$.

Assuming that $0 \le \theta \le \dfrac \pi 2$, we get that $\theta = 0, \dfrac \pi 3, \dfrac \pi 2$.

So we proved that any $\theta$ which in the range $0 \le \theta \le \dfrac \pi 2$ such that both $\dfrac \theta \pi$ and $\cos \theta$ are rational, then $\theta = 0, \dfrac \pi 3, \dfrac \pi 2$.

If, instead of above, we assume that $0 \le \alpha \le \dfrac \pi 2$ and both of $\dfrac \alpha \pi$ and $\sin \alpha$ are rational, then we can denote $\theta = \frac{\pi}{2} - \alpha$ and get that $0 \le \theta \le \dfrac \pi 2$, $\dfrac \theta \pi \in Q$ and $\cos \theta \in \Q$, so $\frac{\pi}{2} - \alpha=\theta = 0, \dfrac \pi 3, \dfrac \pi 2$, therefore $\alpha = 0, \dfrac \pi 6, \dfrac \pi 2$.

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