第(i, j)项为cos( (j-1) * x_i )的n阶行列式的求解

待求解的行列式

$$\left|A_{n\times n}\right|=\left|\begin{array}{ccccc}1& \cos {\theta }_1& \cos 2{\theta }_1& \cdots & \cos \left(n-1\right){\theta }_1\\ 1& \cos {\theta }_2& \cos 2{\theta }_2& \cdots & \cos \left(n-1\right){\theta }_2\\ 1& \cos {\theta }_3& \cos 2{\theta }_3& \cdots & \cos \left(n-1\right){\theta }_3\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 1& \cos {\theta }_n& \cos 2{\theta }_n& \cdots & \cos \left(n-1\right){\theta }_n\end{array}\right|$$

求解过程

直接在$\left|A\right|$上运用Laplace定理展开计算是不现实的，为了方便计算，我们寻求$A$中不同行或者不同列之间是否存在某种线性关系，以便对原行列式进行化简。
由欧拉公式$e^{i\theta }=\cos \theta +i\sin \theta$可推导出De Moivre公式
$$\cos k\theta +i\sin k\theta =e^{i\left(k\theta \right)}={\left(e^{i\theta }\right)}^k={\left(\cos \theta +i\sin \theta \right)}^k,k=1,2,\cdots .\text{\hspace{1em}(1)}$$
对式（1）的右边进行二项式展开，得到
$$\begin{array}{rl}& {\left(\cos \theta +i\sin \theta \right)}^k\\ & =\sum _{p=0}^k{C}_k^p{\cos }^{k-p}\theta \cdot {\left(i\sin \theta \right)}^p\\ & =\left(\sum _{0\le 4q\le k}{C}_k^{4q}{\cos }^{k-4q}\theta \cdot {\sin }^{4q}\theta -\sum _{0\le 4q+2\le k}{C}_k^{4q+2}{\cos }^{k-\left(4q+2\right)}\theta \cdot {\sin }^{4q+2}\theta \right)\\ & +i\left(\sum _{0\le 4q+1\le k}{C}_k^{4q+1}{\cos }^{k-\left(4q+1\right)}\theta \cdot {\sin }^{4q+1}\theta -\sum _{0\le 4q+3\le k}{C}_k^{4q+3}{\cos }^{k-\left(4q+3\right)}\theta \cdot {\sin }^{4q+3}\theta \right).\end{array}\text{\hspace{1em}(2)}$$
对比式（1）等号左边和式（2）等号右边的实部和虚部，可得
cos ⁡ k θ = ∑ 0 ≤ 4 q ≤ k C k 4 q cos ⁡ k − 4 q θ ⋅ sin ⁡ 4 q θ − ∑ 0 ≤ 4 q + 2 ≤ k C k 4 q + 2 cos ⁡ k − ( 4 q + 2 ) θ ⋅ sin ⁡ 4 q + 2 θ = ∑ 0 ≤ 4 q ≤ k C k 4 q cos ⁡ k − 4 q θ ⋅ ( 1 − cos ⁡ 2 θ ) 2 q − ∑ 0 ≤ 4 q + 2 ≤ k C k 4 q + 2 cos ⁡ k − ( 4 q + 2 ) θ ⋅ ( 1 − cos ⁡ 2 θ ) 2 q + 1 = ∑ p = 0 k d k , p cos ⁡ p θ , (3) \begin{aligned} & \cos k\theta =\sum\limits_{0\le 4q\le k}^{ {}}{C_{k}^{4q}{ {\cos }^{k-4q}}\theta \centerdot { {\sin }^{4q}}\theta }-\sum\limits_{0\le 4q+2\le k}^{ {}}{C_{k}^{4q+2}{ {\cos }^{k-\left( 4q+2 \right)}}\theta \centerdot { {\sin }^{4q+2}}\theta } \\ & =\sum\limits_{0\le 4q\le k}^{ {}}{C_{k}^{4q}{ {\cos }^{k-4q}}\theta \centerdot { {\left( 1-{ {\cos }^{2}}\theta \right)}^{2q}}}-\sum\limits_{0\le 4q+2\le k}^{ {}}{C_{k}^{4q+2}{ {\cos }^{k-\left( 4q+2 \right)}}\theta \centerdot { {\left( 1-{ {\cos }^{2}}\theta \right)}^{2q+1}}} \\ & =\sum\limits_{p=0}^{k}{ { {d}_{k,p}}{ {\cos }^{p}}\theta }, \\ \tag{3} \end{aligned}