Proving that real trigonometric polynomials are closed under multiplication 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.
 A: $$2\sin \,A \sin \, B =\cos (A-B) -\cos (A+B),\\ 
2cos \,A \cos \, B =\cos (A-B) +\cos (A+B), \\
2\sin \,A \cos \, B =\sin (A+B)+\sin (A-B),\\
2\cos \,A \sin \, B =\sin (B+A) -\sin  (B-A).$$
A: 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}
$
