Find the value of the following determinant of $n \times n$ matrix:
$$\det A=\begin{vmatrix} 1&\cos \theta_{1}&\cos 2\theta_{1}&\dots&\cos (n-1)\theta_{1}\\ 1&\cos \theta_{2}&\cos 2\theta_{2}&\dots&\cos (n-1)\theta_{2}\\ \vdots&\vdots&\vdots&&\vdots\\ 1&\cos \theta_{n}&\cos 2\theta_{n}&\dots&\cos (n-1)\theta_{n}\\ \end{vmatrix}$$
Here's my idea to evaluate it:
Subtracting the $i^{th}$ column by $(i-1)^{th}$ column and get $-2\sin\frac{\theta_{k}}{2}\sin\frac{2i-1}{2}\theta_{k}$ by sum-to-product identities, and there is common factor of $-2$ on each column (expect the first and second columns) can be pulled out.
And the rest I have stucked in it. Any brilliant idea to continue my work? Or getting a new way to go through it?
This is $$\det A=\begin{vmatrix} 1&c_1&T_2(c_1)&\dots&T_{n-1}(c_1)\\ 1&c_2&T_2(c_2)&\dots&T_{n-1}(c_2)\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&c_n&T_2(c_n)&\dots&T_{n-1}(c_n)\\ \end{vmatrix}$$ where $T_r$ is the $r$-th Chebyshev polynomial of the first kind, and $c_j=\cos\theta_j$. By column operations we get $$\det A=\begin{vmatrix} 1&c_1&a_2 c_1^2&\dots&a_{n-1}c_1^{n-1}\\ 1&c_2&a_2 c_2^2&\dots&a_{n-1}c_2^{n-1}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&c_n&a_2 c_n^2&\dots&a_{n-1}c_n^{n-1}\\ \end{vmatrix} =a_2a_3\cdots a_{n-1}\begin{vmatrix} 1&c_1&c_1^2&\dots&c_1^{n-1}\\ 1&c_2&c_2^2&\dots&c_2^{n-1}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&c_n& c_n^2&\dots&c_n^{n-1}\\ \end{vmatrix}$$ where $a_r$ is the leading coefficient of $T_r$ (so $a_r=2^{r-1}$). That's a Vandermonde determinant in the last line.