# When is $\sin(x)$ rational?

Obviously, there are some points (e.g. $\pi$, $30^\circ$) but I am unsure if there are more. How can it be proved that there are no more points, or, if there are, what those points will be?

EDIT: I largely meant to ask what is supposed in the first comment. Put another way, are there numbers such that $\sin x$ and $\arcsin x$ are both rational?

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If you look at the graph of $\sin(x)$ it goes through every rational from -1 to 1. I doubt there is a simple characterization when it is rational. – Aleks Vlasev Dec 2 '11 at 15:40
almost nowhere! – Vinicius M. Dec 2 '11 at 15:53
– Martin Sleziak Dec 2 '11 at 15:54
like pi,30 ? what? 30 degrees? that's not rational, only algebraic – leonbloy Dec 2 '11 at 15:54
@leonbloy I presume what was meant was that $\sin 30^\circ$ is rational. – Michael Hardy Dec 2 '11 at 16:49

I assume you mean to ask: when $x$ is in whole degrees, when is $\sin(x)$ rational?

If $x$ is in whole degrees, then $x^\circ=\pi x/180\text{ radians}=\pi p/q\text{ radians}$, so we wish to 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 $\left(e^{\pm i\pi p/q}\right)^q-(-1)^p=0$. Thus, $2\sin(\pi p/q)=-i\left(e^{i\pi p/q}-e^{-i\pi p/q}\right)$ 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 $\sin(\pi p/q)\in\{-1,-\frac{1}{2},0,\frac{1}{2},1\}$.

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But what we can say about rationality of sin(pi x) when x is not rational? – kp9r4d Aug 10 at 12:41
@kp9r4d: This is an entirely different question. Just knowing that $x\not\in\mathbb{Q}$, tells us nothing about the rationality of $\sin(x)$. – robjohn Aug 10 at 14:12

You might be looking for Niven's theorem:

If $\sin(r \pi) = q$ where $r,q$ are rationals, then $q$ is $0$, $\pm 1/2$, $\pm 1$.

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There are infinitely many points where both the sine and the cosine are rational, namely $$\left( \frac{n^2-m^2}{n^2+m^2}, \frac{2mn}{n^2+m^2} \right).$$ You can see that if $(x,y)=\text{that pair}$ then $x^2+y^2=1$, so it's on the unit circle.

Maybe a more interesting question is when do you have an angle that is a rational number of degrees (or, equivalently, a rational multiple of $\pi$ radians) for which the sine is rational. I think in that case you get only the obvious ones: $\sin 0^\circ=0$, $\sin 30^\circ=1/2$, $\sin 90^\circ=1$, and the counterparts in the other quadrants. I'm not sure right now how to prove that.

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I attempted to show why in my answer. I learned this trick from Robert Israel back on sci.math. – robjohn Dec 2 '11 at 19:17
ahhh.. but then what type do m and n have to be? naturals? integers? rationals? roots? complex? transcendentals? – don bright May 25 '14 at 15:41
Integers will give you as many points as rationals. If $n,m$ are rational then one could multiply both the numerator and denominator by a suitable integer to get an expression with the same value with $n,m$ integers. In which cases you get a rational point when either $n$ or $m$ is not rational doesn't really matter here. One already gets all rational points on the circle just from integer values of $n$ and $m$. ${}\qquad{}$ – Michael Hardy May 25 '14 at 15:47

$\sin(x)$ is locally a bijection (with the exception of maxima/minima), so it's rational whenever evaluated on $\arcsin(r)$ for rational $r$ unless this is maximum/minimum.

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