Evaluate $$\begin{vmatrix} \sin x_1 & \sin x_2 & \dots & \sin x_n \\ \sin 2x_1 & \sin 2x_2 & \dots & \sin 2x_n \\ \vdots & \vdots & & \vdots \\ \sin nx_1 & \sin nx_2 & \dots & \sin nx_n \end{vmatrix}$$ where $x_1,x_2,\dots,x_n \in \mathbb{R}$.

I tried to somehow expand $\sin kx_i$ in terms of $\sin x_i$ and $\cos x_i$, but things got really complicated and I couldn't go further. Also, I tried substracting the first column from the others, but nothing came out of it.

  • $\begingroup$ Think about Chebyshev polynomials. $\endgroup$ – Lord Shark the Unknown Jan 24 '18 at 18:54

Let $y_k=\cos x_k$. We have:

$$\begin{vmatrix} \sin x_1 & \sin x_2 & \dots & \sin x_n \\ \sin 2x_1 & \sin 2x_2 & \dots & \sin 2x_n \\ \vdots & \vdots & & \vdots \\ \sin nx_1 & \sin nx_2 & \dots & \sin nx_n \end{vmatrix}=\prod_{k=1}^{n}(\sin x_k)\begin{vmatrix} U_0(y_1) & U_0(y_2)& \dots & U_0(y_n) \\ U_1(y_1)& U_1(y_2)& \dots & U_1(y_n) \\ \vdots & \vdots & & \vdots \\ U_{n-1}(y_1)& U_{n-1}(y_2)& \dots & U_{n-1}(y_n) \end{vmatrix}$$ and the last determinant is a constant multiple of $\prod_{1\leq j < k\leq n}(y_j-y_k)$, since it is a polynomial in the variables $y_1,\ldots,y_n$ with degree $\binom{n}{2}$ which vanishes if $y_j=y_k$ for some $j\neq k$. You just have to find what is the multiplicative constant, for instance by computing the LHS at $x_k=\frac{\pi k}{n+1}$. It is worth recalling that $\prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)=\frac{2n}{2^n}$.

  • 1
    $\begingroup$ Very clean answer. +1 $\endgroup$ – Fimpellizieri Jan 24 '18 at 19:30

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