Rational rotation matrices are dense in SO(2)

Reduction. $SO(2)\cap M_2(\mathbb{Q})$ corresponds to the points $(c,s)\in\mathbb{Q}^2$ of the unit circle, and the correspondence $(c,s)\mapsto\begin{pmatrix}c&-s\\s&c\end{pmatrix}$ is a homeomorphism from the circle to $SO(2)$. So we must show that the rational points of the unit circle are dense.

Stereographic parametrisation. For $u\in\mathbb{R}$ put
$$\varphi(u)=\left(\frac{1-u^2}{1+u^2},\ \frac{2u}{1+u^2}\right).$$
- $\left(\frac{1-u^2}{1+u^2}\right)^2+\left(\frac{2u}{1+u^2}\right)^2=\frac{(1-u^2)^2+4u^2}{(1+u^2)^2}=1$, so $\varphi(u)$ lies on the circle.
- $\varphi$ is a continuous bijection from $\mathbb{R}$ onto the circle minus the point $(-1,0)$; it is the inverse of the stereographic projection from $(-1,0)$ (the line through $(-1,0)$ with slope $u$ meets the circle again at $\varphi(u)$).
- If $u\in\mathbb{Q}$, then $\varphi(u)\in\mathbb{Q}^2$.

Density. Let $P=\varphi(u_0)$ be a point of the circle other than $(-1,0)$. Rationals $u_k\to u_0$ give rational points $\varphi(u_k)\to P$ by continuity. For the point $(-1,0)$: $\varphi(u)\to(-1,0)$ as $u\to\pm\infty$, and $(-1,0)$ is also approached by $\varphi(k)$, $k\in\mathbb{N}$, integers being rational. So every point of the circle is a limit of rational points.

Example. $u=\frac12$ gives $\left(\frac35,\frac45\right)$, the $3$-$4$-$5$ triple.

Remark. This is the classical parametrisation of Pythagorean triples. In particular $SO(2)$ contains elements of infinite order with rational entries, since $\frac35+\frac45i$ is not a root of unity. Lesson. Rational points on conics with one rational point are dense.