How find this determinant $\det(\cos^4{(i-j)})_{n\times n}$

Since $\cos^2(x)=\frac{\cos(2x)+1}{2}$, we have

$$ \cos^4(x)=\bigg(\frac{\cos(2x)+1}{2}\bigg)^2= \frac{\cos^2(2x)+2\cos(2x)+1}{4}= \frac{\cos(4x)+4\cos(2x)+3}{8} \tag{1} $$

So by induction on $n$, for every $n>0$ we have $$ \cos^4(x+n)=a_n\cos(4x)+b_n\sin(4x)+c_n\cos(2x)+d_n\sin(2x)+e_n \tag{2} $$

where the sequences $a_n,b_n,c_n,d_n,e_n$ are defined by $a_0=\frac{1}{8},b_0=0,c_0=\frac{1}{2},d_0=0,e_0=\frac{3}{8}$ and the recurrence relation

$$ \left(\begin{matrix} a_{n+1} \\ b_{n+1} \\ c_{n+1} \\ d_{n+1} \\ e_{n+1} \end{matrix}\right)= B \times \left(\begin{matrix} a_{n} \\ b_{n} \\ c_{n} \\ d_{n} \\ e_n \end{matrix}\right), \text{ with }B= \left(\begin{matrix} \cos(4) & \sin(4) & 0 & 0 & 0 \\ -\sin(4) & \cos(4) & 0 & 0 & 0 \\ 0 & 0 & \cos(2) & \sin(2) & 0 \\ 0 & 0 & -\sin(2) & \cos(2) & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{matrix}\right) \tag{3} $$

The characteristic polynomial of $B$ is

$$ \begin{array}{lcl} \chi_B &=& (X^2-2\cos(2)X+1)(X^2-2\cos(4)X+1)(X-1) \\ &=& X^5-(1+2\cos(2)+2\cos(4))X^4 +(2+4\cos(2)+2\cos(4)+2\cos(6))X^3\\ & & -(4\cos(4)+2)X^2+(1+2\cos(2)+2\cos(4))X-1 \end{array} \tag{4} $$

By the Cayley Hamilton-theorem, we have for any $x$,

$$ \begin{array}{lcl} \cos^4(x+5)&=& (1+2\cos(2)+2\cos(4))\cos^4(x+4) \\ & & -(2+4\cos(2)+2\cos(4)+2\cos(6))\cos^4(x+3) \\ & & +(4\cos(4)+2)\cos^4(x+2) \\ & & +(1+2\cos(2)+2\cos(4))\cos^4(x+1)-\cos^4(x) \end{array} \tag{5} $$

So for any $n$, matrix $A_n$ has rank at most $5$. In particular, ${\sf det}(A_n)=0$ for $n\geq 6$.

Remark : the rank of $A_n$ is exactly $5$ for $n\geq 6$, because the $\cos(kx) (k\geq 0)$ are linearly independent.

Hint: A calculation on MATHEMATICA suggests that a vector in the nullspace of this matrix is $$(-1, 1 + 2 \cos(2) + 2 \cos(4), -2 (1 + 2 \cos(2) + \cos(4) + \cos(6)), 2 (1 + 2 \cos(2) + \cos(4) + \cos(6)), -1 - 2 \cos(2) - 2 \cos(4), 1, 0, \cdots, 0, 0)^T$$.

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