Determinant of matrix whose entries are cosines 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!
 A: 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 $\le 2$.
A: 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. 
