Suppose you are given this $2 \times 2$ matrix of trig functions:
\begin{vmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{vmatrix}
The zeros of which give the identity matrix:
\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}
Noticing the original matrix is an example of a wronskian, extending it to the $3 \times 3$ case:
\begin{vmatrix} \cos\theta & \sin\theta & -\cos\theta \\ -\sin\theta & \cos\theta & \sin\theta \\ -\cos\theta & -\sin\theta & \cos\theta \end{vmatrix}
Evaluating at zero yields:
\begin{vmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}
which is symmetric about the diagonal, leading me to believe that all the odd powers have determinants equal to zero. In the next even case e.g. $4 \times 4$:
\begin{vmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 1 \end{vmatrix}
which is also symmetric about the diagonal, leading me to believe the even powers greater than two equal zero.
Originally when I attempted the case where n equals five, I wrote the determinant as the sum of products in terms of trig functions without evaluating at zero. I noticed that they grow as a function of n at the same rate as the elements of the symmetric group.
So, I have two questions:
was I right? is the determinant equal to 0 for n greater than 2 based on the symmetry of the matrix about the diagonal?
since there is an isomorphism between the sum of products representation of the determinant and $S_n$, is there a group theoretic proof to be had?