Rational multiples of $\pi/2$ whose sines are also rational

Question

Let $f(x)=\sin(x\frac{\pi}{2})$. Let $R$ the set of $x$ such that :

$0\le x\le 1$
$x \in \mathbb Q$
$f(x) \in \mathbb Q$

Hence, $0\in R$ as $f(0)=0$. $1\in R$ as f(1)=1. And $\frac{1}{3}\in R$ as $f(\frac{1}{3})=\frac{1}{2}$.

Are they any other elements of $R$ ? How to find them all ?

Thanks !

Related: http://math.stackexchange.com/questions/15823/sine-values-being-rational — Hans Lundmark, Jan 10 '13 at 15:25
Is there a proof of Niven's theorem? I can't seem to find it on any of those sites or links... — Adam Rubinson, Jan 10 '13 at 15:58
Actually there is a Proof Sketch in the "Talk" section of the wikipedia article. Thanks — Adam Rubinson, Jan 10 '13 at 16:28

score 0 · Answer 1 · answered Jan 26 '25 at 09:53

Niven's theorem gives us the answer. In radians, one would require that $0 ≤ x ≤ π/2$, that $x/π$ be rational, and that $\sin x$ be rational. The conclusion is then that the only such values are $\sin 0 = 0$, $\sin (π/6) = 1/2$, and $\sin(π/2) = 1$. A proof of this theorem can be found in proofwiki and on MSE itself.

Rational multiples of $\pi/2$ whose sines are also rational

1 Answers1