Take the 2-minute tour ×
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.

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 $\cos(x)$?

share|improve this question
    
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
2  
@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
2  
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

5 Answers 5

up vote 32 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.

share|improve this answer
    
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

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.

share|improve this answer

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.

share|improve this answer

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

share|improve this answer

$\sin(x)$ 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 $\cos(x)$.

share|improve this answer
    
And for infinitely many $x$ between $0$ and $\pi/2$, both $\sin x$ and $ \cos x$ are rational. –  Andres Caicedo Jul 13 '13 at 4:18

Your Answer

 
discard

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.