Isometries of $\mathbb{R}^2$ 
Show that if $A:\mathbb{R}^2\to \mathbb{R}^2$ is a proper rotation,
  then it may be represented by a matrix of the form $$\pmatrix{
 \cos(\theta)& -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \\}.$$ Further, any
  improper rotation is given by $$\pmatrix{ 1 & 0 \\ 0 & -1 \\} \dot\
 \pmatrix{ \cos(\theta)& -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \\}.$$
  Conclude then that any isometry of $\mathbb{R}^2$ is a composition of
  a translation, a proper rotation and possibly a reflection with
  respect to the y-axis.

I do not know how to do this problem. Any help with be greatly appreciated. 
Note: If $f:\mathbb{R}^n\to \mathbb{R}^n$ is an isometry, then $$f(p)=f(o)+A(p),$$ where $o$ is the origin of $\mathbb{R}^n$ and $A$ is an orthogonal transformation. So if $f:\mathbb{R}^n\to \mathbb{R}^n$ is an isometry with $f(o)=o$ we say that it is a rotation, and if $A=f-f(o)$ is identity we say that $f$ is a translation.
 A: 
I think I have come up with an answer (from some hints from my
  professor), but I am having difficulty finishing it.

Since a proper rotation is a type of rigid motion, we can use an orthogonal transformation to show this. 
Let $A$ be an orthogonal transformation of $\mathbb{R}^2$ (and suppose $A$ is the identity matrix):
$$\pmatrix{
 a& b \\ c & d \\}.$$
I know:


*

*$\text{det}A =\text{ad - bc}=1$  

*$A^TA=\pmatrix{
 a& b \\ c & d \\}\pmatrix{
 a& c \\ b & d \\} = \pmatrix{
 a^2+b^2& ac+bd \\ ac+bd & c^2+d^2\\}.$


Thus: 
$\begin{cases} a^2+b^2=1 \\ c^2+d^2 = 1 \\ ac+bd=0 \\ \text{ad - bc}=1 \end{cases}$
I can note that $a^2+b^2=1$ and $c^2+d^2 = 1$ are just points on the unit circle which can be expressed as $\cos\theta$ and $\sin\theta$. Also, since $ac+bd=0 $ then $\pmatrix{
 a \\b}\dot\ \pmatrix{
 c \\d}=0$ which means they are orthogonal.
But this is where I am having difficulty. Why is $a=\cos\theta$, $b=-\sin\theta$, $c=\sin\theta$, and $d=\cos\theta$? Also, how does this matrix rotate by an angle theta? 
