[proofplan]
Write an element $A\in SO(2)$ by its action on the standard basis vectors $e_1,e_2\in\mathbb{R}^2$. Orthogonality forces the two column vectors to be unit and perpendicular, and the determinant-one condition selects the positively oriented perpendicular vector. Thus $A$ is determined by a unit vector $(a,c)$, which is parametrized by $(\cos\theta,\sin\theta)$. The uniqueness statement follows by comparing the first columns of $R_\theta$ and $R_\phi$ and using the standard period criterion for sine and cosine.
[/proofplan]
[step:Express an element of $SO(2)$ through its two column vectors]
Let $e_1,e_2\in\mathbb{R}^2$ denote the standard basis vectors. Let $A\in SO(2)$. By the definition of $SO(2)$, $A$ is a real [linear map](/page/Linear%20Map) $A:\mathbb{R}^2\to\mathbb{R}^2$ satisfying $A^\top A=I_2$ and $\det A=1$.
Define [real numbers](/page/Real%20Numbers) $a,b,c,d\in\mathbb{R}$ by
\begin{align*}
Ae_1=a e_1+c e_2
\end{align*}
and
\begin{align*}
Ae_2=b e_1+d e_2.
\end{align*}
Since $A^\top A=I_2$, the columns of $A$ are orthonormal with respect to the Euclidean [inner product](/page/Inner%20Product) on $\mathbb{R}^2$. Therefore
\begin{align*}
a^2+c^2=1,
\end{align*}
\begin{align*}
b^2+d^2=1,
\end{align*}
and
\begin{align*}
ab+cd=0.
\end{align*}
The determinant condition gives
\begin{align*}
ad-bc=1.
\end{align*}
[/step]
[step:Use orientation to identify the second column as the positive perpendicular vector]
We claim that $b=-c$ and $d=a$. Let $v\in\mathbb{R}^2$ be the first column vector $v=a e_1+c e_2$, let $w\in\mathbb{R}^2$ be the second column vector $w=b e_1+d e_2$, and let $u\in\mathbb{R}^2$ be the vector $u=-c e_1+a e_2$.
The relation $ab+cd=0$ says precisely that $\langle v,w\rangle=0$. Also
\begin{align*}
|u|^2=(-c)^2+a^2=1.
\end{align*}
Since $a^2+c^2=1$, the vector $v$ is nonzero. The orthogonal complement of the one-dimensional subspace $\operatorname{span}(v)$ in $\mathbb{R}^2$ is one-dimensional, and $u$ is a unit vector in it. Hence there exists $\lambda\in\mathbb{R}$ such that $w=\lambda u$.
Because $|w|=1$ and $|u|=1$, we get $\lambda^2=1$. In coordinates, $w=\lambda u$ means
\begin{align*}
b=-\lambda c
\end{align*}
and
\begin{align*}
d=\lambda a.
\end{align*}
Substituting these identities into the determinant relation yields
\begin{align*}
ad-bc=a(\lambda a)-(-\lambda c)c=\lambda(a^2+c^2)=\lambda.
\end{align*}
Since $ad-bc=1$, it follows that $\lambda=1$. Therefore $b=-c$ and $d=a$.
[guided]
The orthogonality condition tells us that the second column must be perpendicular to the first column, but in the plane there are two unit choices: the positive perpendicular vector and the negative perpendicular vector. The determinant condition is what chooses between them.
Let
\begin{align*}
v=a e_1+c e_2
\end{align*}
be the first column and
\begin{align*}
w=b e_1+d e_2
\end{align*}
be the second column. The equation $ab+cd=0$ is exactly
\begin{align*}
\langle v,w\rangle=0.
\end{align*}
Now define
\begin{align*}
u=-c e_1+a e_2.
\end{align*}
This is the standard counterclockwise perpendicular vector to $v$. Its norm is
\begin{align*}
|u|^2=(-c)^2+a^2=a^2+c^2=1,
\end{align*}
so $u$ is a unit vector. Also
\begin{align*}
\langle v,u\rangle=a(-c)+ca=0,
\end{align*}
so $u$ lies in the orthogonal complement of $\operatorname{span}(v)$.
Because $a^2+c^2=1$, the vector $v$ is nonzero, and therefore its orthogonal complement in $\mathbb{R}^2$ is one-dimensional. Since both $u$ and $w$ lie in that one-dimensional space, there exists $\lambda\in\mathbb{R}$ such that
\begin{align*}
w=\lambda u.
\end{align*}
Since $w$ is also a unit vector, we have
\begin{align*}
1=|w|^2=|\lambda u|^2=\lambda^2|u|^2=\lambda^2.
\end{align*}
Thus $\lambda=1$ or $\lambda=-1$.
The determinant decides the sign. From $w=\lambda u$, the coordinate identities are
\begin{align*}
b=-\lambda c
\end{align*}
and
\begin{align*}
d=\lambda a.
\end{align*}
Substitute these into the determinant formula:
\begin{align*}
ad-bc=a(\lambda a)-(-\lambda c)c=\lambda(a^2+c^2)=\lambda.
\end{align*}
But $A\in SO(2)$ has determinant $1$, so $ad-bc=1$. Hence $\lambda=1$, and consequently
\begin{align*}
b=-c
\end{align*}
and
\begin{align*}
d=a.
\end{align*}
[/guided]
[/step]
[step:Parametrize the first column by an angle]
From $a^2+c^2=1$, the point $(a,c)\in\mathbb{R}^2$ lies on the unit circle. By the standard trigonometric parametrization of the unit circle, there exists $\theta\in\mathbb{R}$ such that
\begin{align*}
a=\cos(\theta)
\end{align*}
and
\begin{align*}
c=\sin(\theta).
\end{align*}
Using $b=-c$ and $d=a$, we obtain
\begin{align*}
Ae_1=\cos(\theta)e_1+\sin(\theta)e_2
\end{align*}
and
\begin{align*}
Ae_2=-\sin(\theta)e_1+\cos(\theta)e_2.
\end{align*}
By the defining action of $R_\theta$ on the standard basis, this proves $A=R_\theta$.
[/step]
[step:Compare first columns to prove uniqueness modulo $2\pi$]
Let $\theta,\phi\in\mathbb{R}$. Suppose first that $R_\theta=R_\phi$. Applying both maps to $e_1$ gives
\begin{align*}
\cos(\theta)e_1+\sin(\theta)e_2=\cos(\phi)e_1+\sin(\phi)e_2.
\end{align*}
Since coordinates in the standard basis are unique, we have
\begin{align*}
\cos(\theta)=\cos(\phi)
\end{align*}
and
\begin{align*}
\sin(\theta)=\sin(\phi).
\end{align*}
By the standard period criterion for sine and cosine, these two equalities hold simultaneously if and only if $\theta-\phi\in 2\pi\mathbb{Z}$. Therefore $R_\theta=R_\phi$ implies $\theta-\phi\in 2\pi\mathbb{Z}$.
Conversely, suppose $\theta-\phi\in 2\pi\mathbb{Z}$. Then there exists $k\in\mathbb{Z}$ such that $\theta=\phi+2\pi k$. By the $2\pi$-periodicity of sine and cosine,
\begin{align*}
\cos(\theta)=\cos(\phi)
\end{align*}
and
\begin{align*}
\sin(\theta)=\sin(\phi).
\end{align*}
Therefore $R_\theta e_1=R_\phi e_1$ and $R_\theta e_2=R_\phi e_2$. Since two linear maps $\mathbb{R}^2\to\mathbb{R}^2$ agreeing on the standard basis are equal, $R_\theta=R_\phi$. This completes the classification and the uniqueness statement.
[/step]
