How to write a square of a trigonometric polynomial cosine?

Question

How to write a square of a polynomial of the form $$\left(1 + 2\sum_{k=1}^n a_k \cos k \theta\right)^2$$ with an explicit formula for just the coefficient of $$\cos k\theta$$ in terms of $k$ and the sequence $$a_1,\cdots , a_n$$ for some given $k$.

Answer 1

Upon squaring, this becomes $$4\left(\sum_{k=1}^n a_k\cos(k\theta)\right)^2+4\sum_{k=1}^n a_k\cos(k\theta)+1$$ So let's look at the coefficient of $\cos(k\theta)$ from each specific term. We obviously can't get $\cos(k\theta)$ from the last term. The coefficient of $\cos(k\theta)$ in the second term is just $4a_k$. The first term is the tricky one. Let's forget about the $4$ for a moment: $$(a_1\cos(1\theta)+a_2\cos(2\theta)+\cdots+a_n\cos(n\theta))(a_1\cos(1\theta)+a_2\cos(2\theta)+\cdots+a_n\cos(n\theta))$$ How do we get $\cos(k\theta)$? Well, if we choose some $a_j\cos(j\theta), a_t\cos(t\theta)$, they must multiply to $\cos(k\theta)$. Hence $$\cos(j\theta)\cos(t\theta)=\cos(k\theta)$$ So the coefficient of $\cos(k\theta)$ $$4\left(\sum_{1\le j,t\le n, \cos(j\theta)\cos(t\theta)=\cos(k\theta)}a_ja_t\right)+4a_k$$ Can you simplify the sum?