Calculating the determinant of a matrix of cosines How to calculate the determinant of the matrix
$$\left|\begin{array}{ccccc}1&1&1&\cdots&1\\
\cos\alpha_1&\cos\alpha_2&\cos\alpha_3&\cdots&\cos\alpha_n\\
\cos2\alpha_1&\cos2\alpha_2&\cos2\alpha_3&\cdots&\cos2\alpha_n\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
\cos(n-1)\alpha_1&\cos(n-1)\alpha_2&\cos(n-1)\alpha_3&\cdots&\cos(n-1)\alpha_n\end{array}\right|$$
My friend told me that De Moivre's formula may be used to estimate it, but I have no idea how to use it. Is this idea right? Or is there another solution to the question? Thank you to your help!
 A: Let $x_i = \cos \alpha_i$ for $i\in\{1,\ldots,n\}.$ The determinant is
$$
\begin{vmatrix}
T_0(x_1)     & T_0(x_2)     & \cdots & T_0(x_n) \\
T_1(x_1)     & T_1(x_2)     & \cdots & T_1(x_n) \\
\vdots       & \vdots       & \ddots & \vdots   \\
T_{n-1}(x_1) & T_{n-1}(x_2) & \cdots & T_{n-1}(x_n)
\end{vmatrix}
$$
with the Chebychev polynomials of the first kind $T_j.$ Each Chebychev polynomial $T_j$ with $j>0$ can be expressed as $2^{j-1}x^j$ plus a linear combination of Chebychev polynomials of smaller degree:
$$
T_j(x) = 2^{j-1}x^j + \sum_{k=0}^{j-1} c_k T_k(x)
$$
This means that I can add multiples of the first $j$ rows of the matrix to row $j+1$ (which does not change the determinant) and get the row $2^{j-1} (x_1^j\;\;x_2^j\;\cdots\;x_n^j)$
When we apply this to all rows of the matrix, we get
$$
\begin{vmatrix}
1     & 1     & 1 & \cdots & 1 \\
x_1     & x_2     & x_3 & \cdots & x_n \\
2x_1^2     & 2x_2^2     & 2x_3^2 & \cdots & 2x_n^2 \\
\vdots       & \vdots       & \vdots & \ddots & \vdots   \\
2^{n-2}x_1^{n-1} & 2^{n-2}x_2^{n-1} & 2^{n-2}x_3^{n-1} & \cdots & 2^{n-2}x_n^{n-1}
\end{vmatrix}
$$
This is the same as
$$
2^0\cdot 2^1 \cdot\ldots \cdot 2^{n-2}\cdot \begin{vmatrix}
1     & 1     & 1 & \cdots & 1 \\
x_1     & x_2     & x_3 & \cdots & x_n \\
x_1^2     & x_2^2     & x_3^2 & \cdots & x_n^2 \\
\vdots       & \vdots     & \vdots  & \ddots & \vdots   \\
x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \cdots & x_n^{n-1}
\end{vmatrix}
$$
which is the same as
$$
2^{0+1+\ldots+(n-2)}\cdot 
\prod_{1\leq i < j\leq n} \left(x_j-x_i\right)
$$
See Vandermonde matrix. So the overall result is
$$
2^{(n-2)(n-1)/2}\cdot \prod_{1\leq i < j\leq n} \left(\cos \alpha_j-\cos\alpha_i\right)
$$
