Determinant of a matrix consisting of sines
From page 166 of Li Jiong Sheng's Linear Algebra (ISBN:9787312022982),
$$\begin{vmatrix} \sin(\theta_1) & \sin(2\theta_1) & \cdots & \sin(n\theta_1) \\ \sin(\theta_2) & \sin(2\theta_2) & \cdots & \sin(n\theta_2) \\ \vdots & \vdots & \ddots & \vdots \\ \sin(\theta_n) & \sin(2\theta_n) & \cdots & \sin(n\theta_n) \\ \end{vmatrix}$$
Using Matlab, I got a complex result, like a cyclic expansion of each term. I wonder how to get this cyclic result.
Solution 1:
I thank Rodrigo de Azevedo for providing the link for
the solution to a similar question.
Let $U_n(x)$ be the $n$-th degree Chebychev polynomial
of the second kind.
- The leading term of $U_n(x)$ is $2^nx^n.$
-
$U_n(\cos\theta))\sin\theta=\sin(n+1)\theta.$
Observe that row $k$ of the given matrix is $$\sin\theta_k(1, U_1(\cos\theta_k),\dots,U_{n-1}(\cos\theta_k))\tag1$$ After applying suitable determinant-preserving column operations to the matrix, row $k$ becomes $$\sin\theta_k(1, 2x_k,\dots,2^{n-1}x^{n-1}_k),\tag2$$ where $x_k=\cos\theta_k.$ If we remove suitable factors along rows and columns, we get a Vandermonde matrix. Thus, the original determinant equals $$2^{n(n-1)\over2}\sin\theta_1\cdots\sin\theta_n\prod_{1\le j\lt k\le n}(\cos\theta_k-\cos\theta_j)$$