Rational Points on $\sin x$ and $\cos x$ Are there any values for $x$ such that both $\sin x$ and $\cos x$ are rational besides $\displaystyle\frac{n\pi}{2}$ and $n\pi$, where $n$ is an integer? I also only want to include $x$ values that are rational multiples of $\pi$.
If not, how could one prove that there aren't? I was thinking that a proof may use the fact that $\displaystyle\frac{\sin x}{\cos x}=\tan x$, but I don't see how this could be done.
I suspect that no such points exist since this page shows exact values of many angle and none of them are rational. That being said,  I can't prove anything for certain.
 A: Take a Pythagorean triple $(a,b,c)$ so you get an $x$ such that $\sin x=\frac ac$ and $\cos x=\frac bc$.
It is known that in the circle $x^2+y^2=1$ the set of rational points ($(a,b)\in\mathbb Q^2$ such that $a^2+b^2=1$) is dense so there is an infinite set of examples of $x$ with $\sin x$ and $\cos x$ both rational. 
A: First we need to consider right angle triangles with the following edge lengths:
$$
\left(a,\frac{a^2-1}{2},\frac{a^2+1}{2}\right).
$$
It is easy to show that this forms a right angle triangle. 
Now pick $a$ to be prime numbers. Then, you can see that if you pick $a = x_1$ and $a=x_2$, then the triangles formed with these values will not be similar triangles. This is possible only when 
$$
\frac{x_1}{\frac{x_1^2+1}{2}}=\frac{x_2}{\frac{x_2^2+1}{2}}
$$
but this leads to a contradiction as it would require $x_1\cdot x_2 = 1$ (you would need to do the simplification).
Now, notice that for each $a$, the angle corresponding to this edge, say $\theta$ would have
$$
cos(\theta) = \frac{2a}{a^2-1}
$$
and 
$$
\sin(\theta)=\frac{a^2-1}{a^2+1}
$$
both of which are rational numbers. Moreover, for each $a$ these pairs would be unique. Since there are infinitely many prime numbers, you would have infinitely many rational number pairs corresponding to your question. 
This answer covers a subset of values for which your condition is satisfied. I am not sure if these are the only values that satisfy the condition. 
