Special Orthogonal Group $SO(2)$ The special orthogonal group for $n=2$ is defined as:
$$SO(2)=\big\{A\in O(2):\det A=1\big\}$$
I am trying to prove that if $A\in SO(2)$ then:
$$A=\left(\begin{array}{cc}
\cos\theta& -\sin\theta\\
\sin\theta&\cos\theta
\end{array}\right)$$
My idea is show that $\Phi:S^1\to SO(2)$ defined as:
$$z=e^{\theta i}\mapsto \Phi(z)=\left(\begin{array}{cc}
\cos\theta& -\sin\theta\\
\sin\theta&\cos\theta
\end{array}\right)$$
is an isomorphism of Lie groups. It is easy prove that is an monomorphism of Lie groups. How can I prove that is also surjective?
 A: Let $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)\in \mathrm{SO}(2)$. Then,
$$\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}a^2+c^2&ab+cd\\ab+cd&b^2+d^2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$$
and
$$\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc=1.$$
Thus, $\mathrm{SO}(2)$ is the subset of $\mathbb{R}^4$ satisfying the following four equations:
$$
\begin{align*}
a^2+c^2 &= 1 \\
b^2+d^2 &= 1\\
ad-bc &= 1\\
ab+cd &= 0.
\end{align*}
$$
The first two equations imply that $(a,c)$ and $(b,d)$ lie on a circle, so
$$a=\cos\alpha,\quad c=\sin\alpha,\quad b=\cos\beta,\quad d=\sin\beta$$
for some angles $\alpha,\beta\in\Bbb R$. Inserting in the last two equations, we get
$$
\begin{align*}
\cos\alpha\sin\beta-\cos\beta\sin\alpha &= 1 \\
\cos\alpha\cos\beta+\sin\alpha\sin\beta &= 0. 
\end{align*}
$$
Using the angle sum trigonometric identities, these equations are
$$
\begin{align*}
\sin(\beta-\alpha) &= 1 \\
\cos(\beta-\alpha) &= 0.
\end{align*}
$$
Hence, $\beta-\alpha\in \pi/2+2\pi\Bbb Z$ and we get
$$\begin{pmatrix}a&c\\b&d\end{pmatrix}=\begin{pmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{pmatrix}.$$
A: Hint:
If you want develop your idea, note that the field of complex numbers is isomorphic the the subring of $M(2,\mathbb{R})$ of the matrices of the form
$$
\begin{pmatrix}
a&-b\\
b&a
\end{pmatrix}
$$
( that are commuting matrices, so that they are a field).
So there is a bijection from the complex  numbers of the form $e^{i\theta}=\cos \theta + i \sin \theta$ and the matrix of the form:
$$
$$
\begin{pmatrix}
\cos \theta&-\sin \theta\\
\sin \theta&\cos \theta
\end{pmatrix}
$$
A: Maybe one can argue as follows:
The above first three equations determine the group variety and show it is connected and one dimensional. The suggested matrix realization satisfies them and form a norm torus, which therefore must be the all group.
