How find this determinant $\det(\cos^4{(i-j)})_{n\times n}$ Question:
Define the matrix $A_{k}=(a^k_{ij})_{n\times n}\quad$where $a_{ij}=\cos{(i-j)},\quad n\ge 6$
Find the value $$\det(A_{4})=\:?$$

My try:
since 

$$\det(A_{4})=\begin{vmatrix}
1&\cos^4{1}&\cos^4{2}&\cdots&\cos^4{(n-1)}\\
\cos^4{1}&1&\cos^4{1}&\cdots&\cos^4{(n-2)}\\
\vdots&\vdots&\vdots&\ddots&\vdots&\\
\cos^4{(n-1)}&\cos^4{(n-2)}&\cos^4{(n-3)}&\cdots&1
\end{vmatrix}$$

and I know that
$$\det(A_{1})=\det(A_{2})=0$$ 

I suspect the following result: $$\det(A_{4})=0,n\ge 6$$
But I can't prove it.
 A: Hint: A calculation on MATHEMATICA suggests that a vector in the nullspace of this matrix is $$(-1, 1 + 2 \cos(2) + 2 \cos(4), -2 (1 + 2 \cos(2) + \cos(4) + \cos(6)), 2 (1 + 2 \cos(2) + \cos(4) + \cos(6)), -1 - 2 \cos(2) - 2 \cos(4), 1, 0, \cdots, 0, 0)^T$$.
Remark: It would be better to take more care on spelling and grammar in your post.
A: 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.
