3 corrected embarrassing typo (sorry for the minimal edit)

Over the reals one has a parametrization of the rotations given by $$\begin{pmatrix} \cos t & - \sin t \\ \sin t & \cos t \end{pmatrix}$$

In other words $$\begin{pmatrix} x & - y \\ y & x \end{pmatrix}$$ with $x^2 + y^2 =1$ with reals $x,y$. Now if one wishes to restrict to the rationals one asks for rational solutions of $x^2 + y^2 =1$ (every such solution will give a rotation and one must not have any irrational entries in the matrix; also cf. André Henriques answer). There is a well-know rational parametrization of these solutions given by $$x(t) = \frac{1-t^2}{1+t^2} \, , \, y(t) = \frac{2t}{1+t^2}$$ with $t$ rational and the one additional solution $(-1, 0)$. (The geometric idea is that the $t$ is the slope of a line through $(-1, 0)$ and the respective solution the other intersection point of this line with the circle.) So one would get all the rotations as $$\begin{pmatrix} \frac{1-t^2}{1+t^2} & - \frac{2t}{1+t^2} \\ \frac{2t}{1+t^2} & \frac{1-t^2}{1+t^2} \end{pmatrix}$$ and $$\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}$$ This is also closey related to parametrization of Phytagorean Pythagorean triples.

Over the reals one has a parametrization of the rotations given by $$\begin{pmatrix} \cos t & - \sin t \\ - \sin t & \cos t \end{pmatrix}$$
In other words $$\begin{pmatrix} x & - y \\ - y & x \end{pmatrix}$$ with $x^2 + y^2 =1$ with reals $x,y$. Now if one wishes to restrict to the rationals one asks for rational solutions of $x^2 + y^2 =1$ (every such solution will give a rotation and one must not have any irrational entries in the matrix; also cf. André Henriques answer). There is a well-know rational parametrization of these solutions given by $$x(t) = \frac{1-t^2}{1+t^2} \, , \, y(t) = \frac{2t}{1+t^2}$$ with $t$ rational and the one additional solution $(-1, 0)$. (The geometric idea is that the $t$ is the slope of a line through $(-1, 0)$ and the respective solution the other intersection point of this line with the circle.) So one would get all the rotations as $$\begin{pmatrix} \frac{1-t^2}{1+t^2} & - \frac{2t}{1+t^2} \\ - \frac{2t}{1+t^2} & \frac{1-t^2}{1+t^2} \end{pmatrix}$$ and $$\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}$$ This is also closey related to parametrization of Phytagorean triples.
Over the reals one has a parametrization of the rotations given by $$\begin{pmatrix} \cos t & \sin t \\ - \sin t & \cos t \end{pmatrix}$$
In other words $$\begin{pmatrix} x & y \\ - y & x \end{pmatrix}$$ with $x^2 + y^2 =1$ with reals $x,y$. Now if one wishes to restrict to the rationals one asks for rational solutions of $x^2 + y^2 =1$ (every such solution will give a rotation and one must not have any irrational entries in the matrix; also cf. André Henriques answer). There is a well-know rational parametrization of these solutions given by $$x(t) = \frac{1-t^2}{1+t^2} \, , \, y(t) = \frac{2t}{1+t^2}$$ with $t$ rational and the one additional solution $(-1, 0)$. (The geometric idea is that the $t$ is the slope of a line through $(-1, 0)$ and the respective solution the other intersection point of this line with the circle.) So one would get all the rotations as $$\begin{pmatrix} \frac{1-t^2}{1+t^2} & \frac{2t}{1+t^2} \\ - \frac{2t}{1+t^2} & \frac{1-t^2}{1+t^2} \end{pmatrix}$$ and $$\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}$$ This is also closey related to parametrization of Phytagorean triples.