[proofplan] We classify an orthogonal transformation by its action on the standard basis. The first column is a unit vector, so we write it as $(\cos\theta)e_1+(\sin\theta)e_2$; the second column must be a unit vector orthogonal to it, leaving exactly two possible signs. These two choices are precisely the rotation $R_\theta$ and the reflected rotation $R_\theta R_{e_1}$, and the determinant computation distinguishes the two alternatives. [/proofplan]

[step:Write the first column using an angle] Let $A\in O(2)$. Define $v_1,v_2\in\mathbb{R}^2$ by \begin{align*} v_1:=Ae_1 \end{align*} and \begin{align*} v_2:=Ae_2. \end{align*} By [citetheorem:8283], the vectors $v_1$ and $v_2$ form an [orthonormal basis](/page/Orthonormal%20Basis) of $\mathbb{R}^2$. In particular, $|v_1|=1$. Write \begin{align*} v_1=ae_1+be_2 \end{align*} with $a,b\in\mathbb{R}$. Since $|v_1|=1$, we have \begin{align*} a^2+b^2=1. \end{align*} By the standard parametrization of the unit circle, there exists $\theta\in\mathbb{R}$ such that \begin{align*} a=\cos\theta \end{align*} and \begin{align*} b=\sin\theta. \end{align*} Thus \begin{align*} v_1=(\cos\theta)e_1+(\sin\theta)e_2. \end{align*} [/step]

[step:Determine the two possible second columns]Write \begin{align*} v_2=ce_1+de_2 \end{align*} with $c,d\in\mathbb{R}$. Since $v_1,v_2$ are orthonormal, we have \begin{align*} (\cos\theta)c+(\sin\theta)d=0 \end{align*} and \begin{align*} c^2+d^2=1. \end{align*} Define $u\in\mathbb{R}^2$ by \begin{align*} u:=(-\sin\theta)e_1+(\cos\theta)e_2. \end{align*} Then $|u|=1$ and $\langle v_1,u\rangle=0$. Since the orthogonal complement of the nonzero vector $v_1$ in $\mathbb{R}^2$ is one-dimensional and $u$ is a unit vector in it, the unit vector $v_2$ must satisfy \begin{align*} v_2=u \end{align*} or \begin{align*} v_2=-u. \end{align*} Equivalently, \begin{align*} v_2=(-\sin\theta)e_1+(\cos\theta)e_2 \end{align*} or \begin{align*} v_2=(\sin\theta)e_1-(\cos\theta)e_2. \end{align*}[/step]

[guided]The first column has already fixed a direction \begin{align*} v_1=(\cos\theta)e_1+(\sin\theta)e_2. \end{align*} The second column cannot be arbitrary: because $A$ is orthogonal, [citetheorem:8283] says that the two columns are orthonormal. So if \begin{align*} v_2=ce_1+de_2 \end{align*} with $c,d\in\mathbb{R}$, then orthogonality to $v_1$ gives \begin{align*} (\cos\theta)c+(\sin\theta)d=0, \end{align*} and the unit-length condition gives \begin{align*} c^2+d^2=1. \end{align*} Now define \begin{align*} u:=(-\sin\theta)e_1+(\cos\theta)e_2. \end{align*} This vector is the quarter-turn of $v_1$ compatible with the chosen rotation convention. Direct computation gives \begin{align*} |u|^2=(-\sin\theta)^2+(\cos\theta)^2=1 \end{align*} and \begin{align*} \langle v_1,u\rangle=(\cos\theta)(-\sin\theta)+(\sin\theta)(\cos\theta)=0. \end{align*} Thus $u$ is a unit vector in $v_1^\perp$. The subspace $v_1^\perp\subset\mathbb{R}^2$ has dimension $1$, since $v_1\ne 0$. Therefore every vector in $v_1^\perp$ is a scalar multiple of $u$. Since $v_2\in v_1^\perp$, there is some $\lambda\in\mathbb{R}$ such that \begin{align*} v_2=\lambda u. \end{align*} Taking Euclidean norms and using $|v_2|=|u|=1$, we get \begin{align*} 1=|v_2|=|\lambda u|=|\lambda||u|=|\lambda|. \end{align*} Hence $\lambda=1$ or $\lambda=-1$. Therefore \begin{align*} v_2=(-\sin\theta)e_1+(\cos\theta)e_2 \end{align*} or \begin{align*} v_2=(\sin\theta)e_1-(\cos\theta)e_2. \end{align*}[/guided]

[step:Identify the determinant-one case as a rotation] Assume first that \begin{align*} v_2=(-\sin\theta)e_1+(\cos\theta)e_2. \end{align*} Then $A$ and $R_\theta$ agree on the basis vectors $e_1,e_2$: \begin{align*} Ae_1=R_\theta e_1 \end{align*} and \begin{align*} Ae_2=R_\theta e_2. \end{align*} Since two linear maps $\mathbb{R}^2\to\mathbb{R}^2$ agreeing on a basis are equal, $A=R_\theta$. Using the basis $e_1,e_2$, the determinant of $R_\theta$ is \begin{align*} \det R_\theta=(\cos\theta)(\cos\theta)-(-\sin\theta)(\sin\theta)=\cos^2\theta+\sin^2\theta=1. \end{align*} Thus this case has determinant $1$. [/step]

[step:Identify the determinant-minus-one case as a reflected rotation] Assume now that \begin{align*} v_2=(\sin\theta)e_1-(\cos\theta)e_2. \end{align*} By the definition of $R_{e_1}$, \begin{align*} R_{e_1}e_1=-e_1 \end{align*} and \begin{align*} R_{e_1}e_2=e_2. \end{align*} Therefore \begin{align*} R_\theta R_{e_1}e_1=R_\theta(-e_1)=-(\cos\theta)e_1-(\sin\theta)e_2. \end{align*} This differs from $Ae_1=v_1$, so the stated product convention corresponds to the reflection across the line perpendicular to $e_1$ followed by the rotation with angle shifted by $\pi$. Indeed, since \begin{align*} R_{\theta+\pi}e_1=-(\cos\theta)e_1-(\sin\theta)e_2 \end{align*} and \begin{align*} R_{\theta+\pi}e_2=(\sin\theta)e_1-(\cos\theta)e_2, \end{align*} we obtain \begin{align*} R_{\theta+\pi}R_{e_1}e_1=(\cos\theta)e_1+(\sin\theta)e_2 \end{align*} and \begin{align*} R_{\theta+\pi}R_{e_1}e_2=(\sin\theta)e_1-(\cos\theta)e_2. \end{align*} Thus $A$ and $R_{\theta+\pi}R_{e_1}$ agree on the basis $e_1,e_2$, and hence \begin{align*} A=R_{\theta+\pi}R_{e_1}. \end{align*} Since $\theta+\pi$ is again a real angle, this is of the required form $R_\phi R_{e_1}$ with $\phi:=\theta+\pi$. The determinant is multiplicative, so \begin{align*} \det(R_\phi R_{e_1})=(\det R_\phi)(\det R_{e_1}). \end{align*} The previous determinant computation gives $\det R_\phi=1$. Since $R_{e_1}$ sends $e_1$ to $-e_1$ and fixes $e_2$, its determinant in the basis $e_1,e_2$ is $-1$. Hence \begin{align*} \det(R_\phi R_{e_1})=1\cdot(-1)=-1. \end{align*} [/step]

[step:Separate the two alternatives by determinant] The preceding steps show that every $A\in O(2)$ is either $R_\theta$ for some $\theta\in\mathbb{R}$ or $R_\phi R_{e_1}$ for some $\phi\in\mathbb{R}$. The determinant calculation gives determinant $1$ in the first case and determinant $-1$ in the second case. Since a matrix cannot have both determinant $1$ and determinant $-1$, the two alternatives are disjoint. This proves the classification and the determinant assertion. [/step]