Does $\sin(x)=y$ have a solution in $\mathbb{Q}$ beside $x=y=0$ Is there a way to show, that the only solution of 
$$\sin(x)=y$$ 
is $x=y=0$ with $x,y\in \mathbb{Q}$.  
I am seaching a way to prove it with the things you learn in linear algebra and analysis 1+2 (with the knowledge of a second semester student).
 A: One can hit it with a really heavy-duty theorem, the Lindemann-Weierstrass theorem. It is a consequence of this that if $x$ is non-zero algebraic, then $\sin x$ is transcendental.
Ivan Niven, in his book Rational and Irrational numbers, has a nice proof, "elementary" but not entirely simple, of the fact that the sine of a non-zero rational is irrational. That proof does not make any real use of linear algebra. 
A: Sorry for the previous spam. I shall prove for $\cos$, cosine of any rational numbers except for 0 cannot get rational numbers.
By using polynomial argument, we shall only have to prove for integers.
Suppose that $m\in\mathbb{N}$, $\cos(m)\in\mathbb{Q}$. 
For any fixed prime number $p>m$, consider polynomial $x\in(0,m)$
$$f(x) = \frac{(x-m)^{2p}(m^2-(x-m)^2)^{p-1}}{(p-1)!}$$
And 
$$F(x) = \sum_{n = 0}^{2p-1} (-1)^{n+1}f^{2n}(x)$$
Which satisfies
$$(F'(x)\sin(x)-F(x)\cos(x))' = F''(x)\sin(x) +F(x)\sin(x) = f(x)\sin(x)$$
since the other terms cancelled.
$$\int_0^mf(x)\sin(x)dx = F'(m)\sin(m)-F(m)\cos(m)+F(0)$$
Consider $f$ is a polynomial of $(x-m)^2$, thus $F'(m) = 0$, and we can see that 
$$f(m-x) = x^{2p}(m^2-x^2)^{p-1}/(p-1)!$$
By computing, $p|f^{(l)}(m)$ for every $l$. That means $F(m)$ is a multiple of $p$ by definition of $F$, say $pM$.
If $\cos(m) = s/t$, then 
$$t\int_0^m f(x)\sin(x)dx = -spM+tN$$
is an integer,
While
$$f\le \frac{m^{4p-2}}{(p-1)!} $$
thus
$$t\int_0^mf(x)\sin(x)dx\le t\frac{m^{4p-2}}{(p-1)!}\cdot m <1 $$
when $p$ is large enough. Contradiction of it is an integer.
As a result of this, $\sin$ should also satisfy this.
