Consider the following form $$\sum_{1\leq i,j,n}a_{ij}\cos(\theta_i-\theta_j)$$ where $A=\{a_{ij}\}$ is a $n\times n$ matrice.
This form can be written as $$\frac{1}{2}\big<A,I_n-QQ^T\big>$$ where $Q:=[x\ y]\in\mathbb{R}^{n\times 2}$, $x:=[\cos(\theta_1),\cdots,\cos(\theta_n)]^T$, $y:=[\sin(\theta_1),\cdots,\sin(\theta_n)]^T$, and $I_n$ is the identity matrix. Moreover, $\big<U,V\big>$ denotes $\text{trace}(U^TV)$.
My question is:
How to transform $$\sum_{1\leq i,j,n}a_{ij}g(\theta_i-\theta_j)$$ into matrix from when $g$ is an arbitrary function? I realized that the above approach does not work anymore when $g\neq\cos$.
Assume that $g(u)$ is periodic, then $g(u)=\sum_n A_n \cos(nu) + B_n \sin(nu)$, and $$\sum_{ij} a_{ij} g(\theta_{ij})=\sum_{ij}\sum_{n} a_{ij} \big[A_{n} \cos(n\theta_{ij}) + B_{n} \sin(n\theta_{ij})\big]$$ That is $$\sum_n A_{n} \big [\sum_{ij} a_{ij} \cos(n\theta_{ij})\big ] + \sum_n B_n \big [\sum_{ij} a_{ij} \sin(n\theta_{ij})\big ]$$ Each term in the square bracket can be written in terms of your original decomposition.
If $g$ is not periodic, The exercise must be repeated with a Fourier integral.