# When is sin(x) rational?

Niven’s theorem states that if $x/\pi$ and $\sin{x}$ are both rational, then the sine takes values 0, $\pm\frac{1}{2}$ , and $\pm 1$.

\begin{aligned} 2\cos2\alpha &= (2\cos\alpha)^2 - 2\\ &= (\frac{q}{p})^2 - 2\\ &= \frac{q^2-2p^2}{p^2} \end{aligned}

Find all rational multiples of $\pi$ so that $\sin(\pi p/q)$ is rational.

If $p/q\in\mathbb{Q}$, then $e^{\pm i\pi p/q}$ is an algbraic integer since $(e^{\pm i\pi p/q})^q−(−1)^p=0.$

Thus, $$2\sin(\pi p/q)=−i(e^{i\pi p/q}−e^{−i\pi p/q})$$ is the difference and product of algebraic integers, and therefore an algebraic integer.

However, the only rational algebraic integers are normal integers. Thus, the only values of $\sin(\pi p/q)$ which could be rational, are those for which $2\sin(\pi p/q)$ is an integer, that is $$\sin(\pi p/q)\in \{−1,−1/2,0,1/2,1\}.$$