Recall that real trigonometric polynomials are functions of the form $$ f(\theta) = \frac{a_0}{2} + \sum_{k=1}^n (a_k \cos(k\theta) + b_k \sin(k\theta)). $$ I want to prove that real trigonometric polynomials are closed under multiplication.
In other words, given two trigonometric polynomials $f$ and $g$, I want to show that the function $h(\theta) = f(\theta)\cdot g(\theta)$ is also a trigonometric polynomial.
Toward this end I've noticed the double-angle formulas $$ \sin(2\theta) = 2\sin(\theta)\,\cos(\theta) \\ \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) $$ might be helpful for the $k=2$ case. But I'm not sure what to do for terms with $k>2$ and I'm not sure how to handle the coefficients $a_k$ and $b_k$.
Any pointers are greatly appreciated.
Wander into the complex domain, write it as a sum
$\begin{array}\\ f(\theta) &= \frac{a_0}{2} + \sum_{k=1}^n (a_k \cos(k\theta) + b_k \sin(k\theta))\\ &= c_0+ \frac12\sum_{k=1}^n (a_k (e^{ik\theta}+e^{-ik\theta}) + b_k e^{ik\theta}-e^{-ik\theta}))\\ &= c_0+ \frac12\sum_{k=1}^n ((a_k+b_k)e^{ik\theta}+(a_k-b_k)e^{-ik\theta})\\ &=\sum_{k=-n}^n c_k e^{ik\theta} \qquad c_k = a_k+b_k, c_{-k} =a_k-b_k \text{for } 1\le k\le n \\ \end{array} $
Multiply them to get a sum $\sum_{k=-m}^m d_ke^{ik\theta} $ and return to real form
$\begin{array}\\ \sum_{k=-m}^m d_ke^{ik\theta} &=\sum_{k=-m}^m d_k(\cos(k\theta)+i\sin(k\theta))\\ &=\sum_{k=-m}^m d_k\cos(k\theta)+i\sum_{k=-m}^m d_k\sin(k\theta)\\ &=d_0+\sum_{k=-m}^{-1} d_k\cos(k\theta) +i\sum_{k=-m}^{-1} d_k\sin(k\theta)\\ &\quad+\sum_{k=1}^m d_k\cos(k\theta)+i\sum_{k=1}^m d_k\sin(k\theta)\\ &=d_0+\sum_{k=1}^{m} d_{-k}\cos(-k\theta)+i\sum_{k=1}^{m} d_{-k}\sin(-k\theta)\\ &\quad+\sum_{k=1}^m d_k\cos(k\theta)+i\sum_{k=1}^m d_k\sin(k\theta)\\ &=d_0+\sum_{k=1}^{m} d_{-k}\cos(k\theta)-i\sum_{k=1}^{m} d_{-k}\sin(k\theta)\\ &\quad+\sum_{k=1}^m d_k\cos(k\theta)+i\sum_{k=1}^m d_k\sin(k\theta)\\ &=d_0+\sum_{k=1}^{m} (d_{-k}+d_k)\cos(k\theta)+i\sum_{k=1}^{m}(d_k- d_{-k})\sin(k\theta)\\ \end{array} $