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I know this may be a dumb question but I know that it is possible for $\sin(x)$ to take on rational values like $0$, $1$, and $\frac {1}{2}$ and so forth, but can it equal any other rational values? What about $\cos(x)$?

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I had a similar question once, which was: are there infinitely many values $q\pi$ for $q\in \mathbb{Q}$ such that $\sin(q\pi)$ is algebraic? Galois theory cleared that up a bit for me, but I still wonder... – Eric Auld Jul 13 '13 at 4:27
@EricAuld What is it you still wonder about? $\sin(q\pi)$ is always algebraic, being the average of two roots of unity. – Erick Wong Jul 13 '13 at 6:54
Incidentally, a consequence of the Lindemann–Weierstrass theorem is that if both $\sin x$ and $\cos x$ are rational and $x\neq 0$, then $x$ is irrational. – Jonas Meyer Jul 13 '13 at 7:32
@ErickWong Oh, I didn't know that! Thanks for the interesting information. – Eric Auld Jul 13 '13 at 13:15
up vote 34 down vote accepted

There are infinitely many primitive Pythagorean triples, that is, triples $(a,b,c)$ of positive integers such that $a$, $b$, and $c$ are positive integers $\gt 1$ such that $a^2+b^2=c^2$.

Any such triple determines a right triangle. The sines and cosines of the two non-right angles are the rationals $\frac{a}{c}$ and $\frac{b}{c}$. So there are infinitely many angles between $0$ and $\frac{\pi}{2}$ such that $\sin x$ and $\cos x$ are both rational.

You are undoubtedly familiar with the triples $(3,4,5)$ and $(5,12,13)$. There are infinitely many more. The Wikipedia article linked to above gives a detailed description.

We can already get infinitely many examples by letting $n$ be any integer $\gt 1$, and setting $a=n^2-1$, $b=2n$, and $c=n^2+1$. Make the right-triangle $ABC$, with the right angle at $C$, and $a,b,c$ as above.

Note that $\triangle ABC$ really is a right-triangle, since $(n^2-1)^2+(2n)^2=(n^2+1)^2$.

Let $x=\angle A$. Then $\sin x=\frac{n^2-1}{n^2+1}$ and $\cos x=\frac{2n}{n^2+1}$. Thus both are rational.

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Do you mean $(5, 12, 13)$? – Ben Jul 13 '13 at 5:50
Thanks! Yes, ridiculous no? At that stage I was wondering about also mentioning $(8,15,17)$. So I can't even call it a typo. – André Nicolas Jul 13 '13 at 5:52

$\sin(x)$ takes on every value between $-1$ and $1$, so it can take on any rational value between $-1$ and $1$ inclusive. The same is true of $\cos(x)$.

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And for infinitely many $x$ between $0$ and $\pi/2$, both $\sin x$ and $ \cos x$ are rational. – Andrés E. Caicedo Jul 13 '13 at 4:18

Yes. Both are continuous, and so by the intermediate value theorem, take on every value between -1 and 1.

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Both $\sin x$ and $\cos x$ take on all rational values within the range $-1$ to $1$. That doesn't mean you can find the $x$ such that $\sin x=\frac {25}{149}$ for example. But there is such an $x$ (in fact many), which you can approximate as closely as you want.

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You may be interested in Niven's theorem, which is that the only rational values of $\sin x$ when $x$ is a rational multiple of $\pi$ (i.e. a rational number of degrees) are $0$, $\pm\frac12$ and $\pm1$. Of course $\sin x$ takes many other rational values but these do require $x/\pi$ to be irrational.

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