Are there infinitely many rational outputs for sin(x) and cos(x)?

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 $\cos(x)$?

Solution 1:

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.

Solution 2:

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

Solution 3:

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