Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question
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
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

4 Answers 4

up vote 18 down vote accepted

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 $\sin(\pi p/q)\in\{-1,-\frac{1}{2},0,\frac{1}{2},1\}$.

share|cite|improve this answer
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

$\sin(x)$ 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.

share|cite|improve this answer

You might be looking for Niven's theorem:

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

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.