How do I work out the general form of a $2 \times 2$ special orthogonal matrix?

Question

Also, if I'm not mistaken, the 2 sets of solutions should be the same with the only difference being that the free variable $\theta$ would take different values in each solution set to yield the same result. Also, I think there should be other solution sets (for example, ones in which when b is positive a is always negative and so on). What justifies ignoring all other solution sets and other forms of the first solution set (the one I wrote) and to say that the solution set must take one form and that's that? Can the other solution sets (or at least the other form of the first solution) be interpreted in terms of rotations of vectors?

Answer 1

$a^2 + b^2 = 1$ is the equation of the unit circle in $(a,b)$ space. It can be parametrized by an angle $\theta$ such that $a = \cos(\theta)$ and $b = \sin(\theta)$: this is just the $\theta$ of polar coordinates. Now $a c + b d = 0$ and $c^2 + d^2 = 1$ says $(c,d)$ is also on the unit circle, and orthogonal to $(a,b)$, so either $(\sin(\theta), -\cos(\theta))$ or $(-\sin(\theta), \cos(\theta))$. The other equation, $ad - bc = 1$ tells you which of these it is: the first would give $ad-bc = -1$, the second $1$.