<s>[INST] Compute the determinant of circulant matrix with entries $\cos j\theta$ compute the determinant 
$
\begin{vmatrix}
\cos \theta & \cos 2\theta & \cos 3\theta &\cdots &\cos n\theta \\
\cos n\theta & \cos \theta & \cos 2\theta & \cdots  & \cos (n-1)\theta \\
\cdots \\
\cos 2\theta & \cos 3\theta & \cos 4\theta & \cdots & \cos \theta 
\end{vmatrix}
$
I have tried to apply the formula for circulant matrix $\text{det} =\prod_{i=1}^n f(\omega _i)$ with $\omega _i$ the nth root of unity.This gives an ugly result which I failed to simplify. Should I try another way? Any help will be greatly appreciated on how to obtain a neater result.
By the way, wolfram alpha gives when $n=3$ it's $\cos ^3 \theta + \cos ^3 2\theta + \cos ^3 3\theta -3 \cos \theta \cos 2\theta \cos 3\theta$, and for $n=4$ it also can be factored to a nice form, so I think simplification is possible.
 [/INST] let  $ A = \pmatrix{\cos t & \cos 2t &\ldots & \cos nt\\ 
\cos nt & \cos t & \ldots &\cos(n-1)t\\\ldots & \ldots &\ldots &\ldots\\
\cos 2t & \cos 3t &\ldots & \cos t}
                                 $  and $\omega$ the $n$ the root of unity. that is $\omega^n = 1.$  you can verify that 
$\pmatrix{1 & \omega & \ldots & \omega^{n-1}}^\top$ is an eigenvector corresponding to the eigenvalue $\lambda_\omega = \cos t + \omega \cos 2t + \ldots + \omega^n \cos nt.$  therefore the determinant of $A$ is the product of of the $n$ eignevalues $\{\lambda_\omega : \omega^n = 1\}.$
</s>