Given $A \in O(2)$, show that $A$ has one of the following two forms:
$$ A = \left( \begin{matrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{matrix} \right) $$
$$ A = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta) \end{matrix} \right) $$
Suppose $A \in O(2)$, then we have that:
$$ A = \left( \begin{matrix} a & c \\ b & d \end{matrix} \right) $$
Since $A \in O(2)$, it's columns are orthonormal with respect to the inner product $\langle x,y \rangle = x_1y_1 + x_2y_2$ and that $\det(A) = \pm 1$. Thus we arrive at the following system of equations:
$$a^2+b^2 = 1$$
$$c^2+d^2 = 1$$
$$ac+bd = 0$$
$$|ad-bc|=1$$
The first two equations say that $a,b,c,d$ all lie on the unit circle. So we have either $a = \sin(\theta)$ and $b=\cos(\theta)$ or that $a = \cos(\theta)$ and $b = \sin(\theta)$. A similar statement holds for $c$ and $d$.
Anyway, I'm a bit stuck. Am I going about this correctly? Help appreciated. Thanks!
Since $a^2+b^2=1$, there is some $\theta\in\Bbb R$ such that $a=\cos\theta$ and $b=\sin\theta$. And, since $(c,d)$ is orthogonal to $(a,b)$,$$(c,d)=\lambda(-b,a)=(-\lambda\sin\theta,\lambda\cos\theta)$$for some $\lambda\in\Bbb R$. But$$\pm1=\begin{vmatrix}\cos\theta&-\lambda\sin\theta\\\sin\theta&\lambda\cos\theta\end{vmatrix}=\lambda;$$so, $\lambda=\pm1$ and you're done.