Why $\cos(t-s)$ is a "positive function"?

Question

Problem: Let $c: \mathbb R\longrightarrow \mathbb R$ be a finite support function, i.e. the set $\{t: c(t)\neq 0\}$ is finite. Prove that \begin{align}\label{eq1} \sum_{t, s\in \mathbb R} c(t)c(s) \cos(t-s)\geq 0. \end{align}

Reason: The reason I ask this question is because $\cos(s-t)$ can be considered as a covariance function (see the answer here). And we know that every covariance function is positive (in the sense that above inequality holds for all finite support functions). However, I don't know how to prove above inequality independently. Could anyone give me a hint, thank you!

Answer 1

\begin{align} \sum_{s,t} c(s)c(t)\cos(s-t) &= \sum_{s,t}\big(c(s)c(t)\cos(s)\cos(t)+c(s)c(t)\sin(s)\sin(t)\big) \\ &= \left(\sum_s c(s)\cos(s)\right)^2 + \left(\sum_s c(s)\sin(s)\right)^2 \ge 0. \end{align}