Showing that the set of $2 \times 2$ real orthogonal matrices has a particular parameterization 
Theorem Every orthogonal matrix in $\mathbb{R}^{2, 2}$ is in the form
  \begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}
  or
  \begin{bmatrix}
\cos\theta & \sin\theta \\
\sin\theta & -\cos\theta
\end{bmatrix}

I am able to prove that
$$\mathbf{M} = \begin{bmatrix}
     a & \pm b \\
     b & \pm a
    \end{bmatrix},
    |\operatorname{det}(\mathbf{M})|
    = |a^2 \pm b^2|
    = 1,$$
but I don't know how to continue.
 A: First, note that the general form of $\bf M$ is
$$\pmatrix{a & \mp b \\ b & \pm a},$$
that is, e.g., if one takes $-b$ for the $(1, 2)$ entry, one must take $+a$ for the $(2, 2)$ entry. (One can show this for example by writing a general matrix $A$ and determining the algebraic conditions on the entries determined by the defining condition $A^T A = I$.) The determinant of this matrix is
$$\det {\bf M} = a(\pm a) - (\mp b) b = \pm (a^2 + b^2),$$ and so the determinant condition is simply
$$a^2 + b^2 = 1.$$
Geometrically, this equation defines the unit circle in the $ab$-plane, and so by the usual unit circle definitions of $\sin$ and $\cos$, for any point $(a, b)$ such that $a^2 + b^2$ there is an angle $\theta$ such that $$a = \cos \theta, \qquad b = \sin \theta,$$ namely, the anticlockwise angle from the positive $x$-axis to the point $(a, b)$. This angle is determined uniquely up to multiples of $2 \pi$.
A: Do you mean:
$$\mathbf{M} = \begin{bmatrix}
     a & \pm b \\
     b & \mp a
    \end{bmatrix},
    |\operatorname{det}(\mathbf{M})
    = |a^2 + b^2|
    = 1$$
?
To complete the proof, note that $|a^2 + b^2|= 1$ imply that $a \in [-1,1] $, so there exists such a $\theta \in [0,2\pi)$, that $\cos(\theta)=a$. Next, by the same equality, you have:
$$a^2+b^2=1=\cos^2(\theta)+\sin^2(\theta)$$
Because of $a^2=\cos^2(\theta)$ you have:
$$b^2=\sin^2(\theta)$$
So:
$$b=\sin(\theta)$$
or
$$b=-\sin(\theta)$$
