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. 

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**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 $(n+m)$.