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.
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 $$ b=\sin(2\pi k-\theta)=-\sin\theta,\\ d=\cos(2\pi k-\theta)=\cos\theta. $$