TITLE: Orthogonal Matrix with Determinant 1 is a Rotation Matrix
QUESTION [7 upvotes]: I am confused with how to show that an orthogonal matrix with determinant 1 must always be a rotation matrix.
My approach to proving this was to take a general matrix $\begin{bmatrix}a&b \\c&d\end{bmatrix}$ and using the definition of a matrix being orthogonal, work out some restrictions on $a,b,c,d$ such that the matrix must be a rotation matrix.
Doing this; I end up with $\begin{bmatrix}a&b \\c&d\end{bmatrix}^T\begin{bmatrix}a&b \\c&d\end{bmatrix}=\begin{bmatrix}a^2+c^2&ab+cd \\ab+cd&b^2+d^2\end{bmatrix}=\begin{bmatrix}1&0 \\0&1\end{bmatrix}$
We also know $ad-bc=1$ from the determinant restriction.
I thought we could say since $a^2+c^2=1$ then we could say $a=\cos(\theta)$ and $c=\sin(\theta)$. From here we could chose the second column of the matrix to a vector such that $b=-\cos(\pi/2-\theta)$ and $d=\sin(\pi/2-\theta)$ which gives $b=-\sin(\theta)$ and d=$\cos(\theta)$.
This would give $\begin{bmatrix}\cos(\theta)&-\sin(\theta) \\\sin(\theta)&\cos(\theta)\end{bmatrix}$ which is the rotation matrix I need to show.
Is this correct and is there a better approach to help me prove that any orthogonal matrix with determinant 1 must be a rotation matrix?
EDIT:
We can say $a=\cos(\theta)$ and $c=\sin(\theta)$ as we know $a^2+c^2=1$ and so $a$ and $c$ must lie on the unit circle, hence we can parameterise the variables in the matrix as such.

REPLY [3 votes]: The sentence "From here we could chose the second column of the matrix..." is ambiguous. You cannot choose $b$ and $d$, they are given. You need to prove that they are $\sin$ and $\cos$. 
For example, since
$$
a^2+c^2=1,\\
d^2+b^2=1,
$$
the points $(a,c)$ and $(d,b)$ are on the unit circle and can be parameterized as
$$
(a,c)=(\cos\theta,\sin\theta),\\
(d,b)=(\cos\phi,\sin\phi).
$$
Now another two equations that you know are
$$
ab+cd=0 \quad\Leftrightarrow\quad\cos\theta\sin\phi+\sin\theta\cos\phi=\sin(\theta+\phi)=0,\\
ad-bc=1\quad\Leftrightarrow\quad\cos\theta\cos\phi-\sin\theta\sin\phi=\cos(\theta+\phi)=1.
$$
It gives $\theta+\phi=2\pi k$, which makes
$$
b=\sin(2\pi k-\theta)=-\sin\theta,\\
d=\cos(2\pi k-\theta)=\cos\theta.
$$