Prove that the number of free parameters in an $n\times n$ orthogonal transformation matrix is equal to $\frac{n(n-1)}{2}$.
For example parametrization of $2 \times 2$ orthogonal matrix requires only one parameter, ie $\theta$.
And the parametric form is
$$ M_2 = \pm\left(\begin{array}{cc}
\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta
\end{array}\right) .$$
Best Answer
There are $n^2$ free parameters in an $n\times n$ matrix. Orthogonality imposes $n(n+1)/2$ constraints, namely $n$ normalization constraints on the columns and $\binom n2=n(n-1)/2$ orthogonality constraints, one for each pair of columns. That leaves $n^2-n-n(n-1)/2=n(n-1)/2$ degrees of freedom.