Let $n\geq3$. How to find the determinant of the following matrix?
$$ \begin{pmatrix} \cos(\alpha_1 -\beta _1) & \cos(\alpha_1 -\beta _2) & \cdots & \cos(\alpha_1 -\beta _n)\\ \cos(\alpha_2 -\beta _1) & \cos(\alpha_2 -\beta _2)& \cdots & \cos(\alpha_2 -\beta _n)\\ \vdots & \vdots& \ddots& \vdots\\ \cos(\alpha_n -\beta _1)& \cos(\alpha_n -\beta _2)& \cdots & \cos(\alpha_n -\beta _n) \end{pmatrix} $$
I tried to use some trigonometric properties but it did not help. Any suggestions? Thanks!
On
Let us call $A$ this matrix.
Using formula $\cos(a-b)=\cos a \cos b + \sin a \sin b$, one can write :
$$A = U^TV \ \ \text{with}$$ $$U:=\begin{pmatrix}\cos\alpha_1&\cos \alpha_2& \cdots&\cos \alpha_n\\ \sin\alpha_1&\sin \alpha_2& \cdots&\sin \alpha_n\end{pmatrix},$$ $$V:=\begin{pmatrix}\cos\beta_1&\cos \beta_2& \cdots&\cos \beta_n\\ \sin\beta_1&\sin \beta_2& \cdots&\sin \beta_n\end{pmatrix}.$$
Knowing that
$$\text{rank}(A)=\text{rank}(U^TV)\le \min(\text{rank}(U^T),\text{rank}(V))\le 2$$
(See there), we can conclude $\det(A)=0$ because it is a $n \times n$ matrix ($n \geq 3$) with rank $\le 2$.
Since the given determinant is equal to: \begin{align*} det\begin{pmatrix} \cos\alpha_1\cos\beta_1 & cos(\alpha_1 -\beta _2) & \cdots & cos(\alpha_1 -\beta _n)\\ \cos\alpha_2\cos\beta_1 & cos(\alpha_2 -\beta _2)& \cdots & cos(\alpha_2 -\beta _n)\\ \vdots & \vdots& \vdots& \vdots\\ cos\alpha_n\cos\beta_1& cos(\alpha_n -\beta _2)& \cdots & cos(\alpha_n -\beta _n) det\end{pmatrix} -det\begin{pmatrix} \sin\alpha_1\sin\beta_1 & cos(\alpha_1 -\beta _2) & \cdots & cos(\alpha_1 -\beta _n)\\ \sin\alpha_2\sin\beta_1 & cos(\alpha_2 -\beta _2)& \cdots & cos(\alpha_2 -\beta _n)\\ \vdots & \vdots& \vdots& \vdots\\ \sin\alpha_n\sin\beta_1& cos(\alpha_n -\beta _2)& \cdots & cos(\alpha_n -\beta _n) \end{pmatrix}\\ = det(\cos\beta_1)\begin{pmatrix} \cos\alpha_1& cos(\alpha_1 -\beta _2) & \cdots & cos(\alpha_1 -\beta _n)\\ \cos\alpha_2& cos(\alpha_2 -\beta _2)& \cdots & cos(\alpha_2 -\beta _n)\\ \vdots & \vdots& \vdots& \vdots\\ cos\alpha_n& cos(\alpha_n -\beta _2)& \cdots & cos(\alpha_n -\beta _n) \end{pmatrix} -det(\sin\beta_1)\begin{pmatrix} \sin\alpha_1 & cos(\alpha_1 -\beta _2) & \cdots & cos(\alpha_1 -\beta _n)\\ \sin\alpha_2& cos(\alpha_2 -\beta _2)& \cdots & cos(\alpha_2 -\beta _n)\\ \vdots & \vdots& \vdots& \vdots\\ \sin\alpha_n& cos(\alpha_n -\beta _2)& \cdots & cos(\alpha_n -\beta _n) \end{pmatrix}\\ = det(\cos\beta_1)\begin{pmatrix} \cos\alpha_1& -\sin\alpha_1\sin\beta _2 & \cdots & -\sin\alpha_1\sin\beta_n\\ \cos\alpha_2& -\sin\alpha_2\sin\beta _2& \cdots & -\sin\alpha_2\sin\beta _n\\ \vdots & \vdots& \vdots& \vdots\\ cos\alpha_n& -\sin\alpha_n\sin\beta _2& \cdots & -\sin\alpha_n\sin\beta_n \end{pmatrix} -det(\sin\beta_1)\begin{pmatrix} \sin\alpha_1 & \cos\alpha_1\cos\beta_2 & \cdots & \cos\alpha_1\cos\beta_n\\ \sin\alpha_2& \cos\alpha_2\cos\beta_2& \cdots & \cos\alpha_2\cos\beta_n\\ \vdots & \vdots& \vdots& \vdots\\ \sin\alpha_n& \cos\alpha_n\cos\beta_2& \cdots & \cos\alpha_n\cos\beta_n \end{pmatrix}\\ \end{align*} the last two determinants are clearly zero. So, the answer is 0.