A trigonometric polynomial was defined as $$f(x) = \frac{a_0}{2} + \sum_{k=1}^{n}(a_k \cos(kx) + b_k \sin(kx))$$
I heard somewhere that trigonometric polynomials have a ring structure, i.e. a product of two polynomials is again a polynomial.
I was wondering how can one see it. A direct attempt to multiply two polynomials seem to result in a mess.
Edit:
Using a suggestion from nicomezi from below I go to the complex representation:
$$f(x) = \frac{a_0}{2} + \frac{1}{2} \sum_{k=1}^{n}(e^{ikx}z_k + e^{-ikx} \bar z_k )$$
We can also "hide" the first term:
$$f(x)=\frac{1}{2} \sum_{k=0}^{n}(e^{ikx}z_k + e^{-ikx} \bar z_k ), z_0 = a_0$$
Now when we multiply two polynomials of degrees $n$ and $m$ we get a sum of terms of the form:
$$\frac{1}{4}(e^{ikx}z_k + e^{-ikx} \bar z_k )(e^{ilx}z_l + e^{-ilx} \bar z_l )= \frac{1}{4}(e^{i(k+l)x}z_kz_l + e^{-i(k+l)x}\overline{z_k z_l}) + \frac{1}{4}(e^{i(k-l)x}z_k\bar z_l + e^{-i(k-l)x} \overline{z_k \bar z_l}), 1 \le k \le n, 1 \le l \le m$$
So each term is a polynomial of degree $(k+l)$, thus the whole product is a polynomial of degree $(n+m)$.
Best Answer
There is no need to multiply two polynomials, that does, indeed, result in a mess. Instead, all you need to show is that:
$\cos(kx)\cos(lx)$ is a trigonometric polynomial,
$\cos(kx)\sin(lx)$ is a trigonometric polynomial, and
$\sin(kx)\sin(lx)$ is a trigonometric polynomial.
The rest comes from linearity.
Let's check $\cos(kx)\sin(lx)$ is a trigonometric polynomial; the rest of the cases are very similar.
Since $e^{ikx}=\cos(kx)+i\sin(kx)$, $\cos(kx)=\frac{1}{2}\left(e^{ikx}+e^{-ikx}\right)$. Similarly, $\sin(lx)=\frac{1}{2}\left(ie^{-ilx}-ie^{ilx}\right)$. Therefore, $$\cos(kx)\sin(lk)=\frac{1}{4}\left(ie^{i(k-l)x}+ie^{-i(k+l)x}-ie^{i(k+l)x}-ie^{i(l-k)x}\right).$$ Since our answer should be real, we are only interested in the real part of this, which is: $$ \cos(kx)\sin(lx)=-\frac{1}{2}\sin((k-l)x)+\frac{1}{2}\sin((k+l)x) $$