Form of a real orthogonal matrix Q. Let $A$ be a real orthogonal matrix of order $2$ with $|A|=1$. Show that $\exists\theta$ s.t. $$A=\begin{pmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{pmatrix}\tag{4 marks}$$
My answer:
Let $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ where $a,b,c,d\in\Bbb R$. Now, since $A$ is orthogonal, we have,
$$AA^T=I_2\implies \begin{pmatrix}a^2+b^2&ac+bd\\ac+bd&c^2+d^2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\implies\begin{cases}a^2+b^2=c^2+d^2=1\quad(1)\\ ac+bd=0\qquad\qquad\quad~(2)\end{cases}$$
Also, since $|A|=1$, we have $ad-bc=1\tag3$
Using $(1)$ and $(3)$, we get,
$$(a-d)^2+(b+c)^2=0$$
Since $a,b,c,d\in\Bbb R$ and squares of reals are always non-negative, in order to maintain equality, we must have,
$$a-d=0\text{ and }b+c=0\implies d=a~\land~c=-b$$
Putting this back in $(3)$. we see that it's the equation of the unit circle in the $ab$ plane, hence the parametric representation $a=\cos\theta$ and $b=\sin\theta$. The conclusion follows.

My question is how much marks this solution will get (in general) ? Also, it'd be nice to see simpler formal proofs. Thanks.

The solution in the original post is correct.  
Here we present another way forward, which seems fairly efficient.  We note that
$$a^2+b^2=1\implies a=\cos \theta\,\,\text{and}\,\,b=\sin \theta$$
$$c^2+d^2=1\implies c=\cos \phi\,\,\text{and}\,\,d=\sin \phi$$
for some $\theta$ and some $\phi$.
Then, we have
$$ac+bd=0\implies \cos (\theta -\phi)=0\implies \theta -\phi =(2k+1)\pi/2$$
for any integer $k$.  
Eliminating $\phi$, we find that $c=\pm \sin \theta$ and $d=\mp \cos \theta$.  
We can choose the value of $k$ such that $c=-\sin \theta$ and $d=\cos \theta$ and we are done!
