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 $\cos(x)$?
 A: 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.
A: 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.
A: 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.
A: $\sin(x)$ 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 $\cos(x)$.
A: Yes.  Both are continuous, and so by the intermediate value theorem, take on every value between -1 and 1.
