Which trigonometric polynomials are identically zero? There are a number of elementary polynomial relations between functions of the form $\cos(nθ)$ and $\sin(nθ)$. For instance, $\cos^{2}(nθ) + \sin^{2}(nθ) - 1 = 0$, or $\sin(2nθ)-2\sin(nθ)\cos(nθ) = 0$. Such identities can be thought of as polynomials in $\cos(θ), \sin(θ), \cos(2θ), \sin(2θ), ...$
Is there a neat enumeration of all (or many) such polynomials? In other words, which trigonometric polynomials are identically zero?
 A: I'll answer the same question for the complex exponential functions $e^{in\theta}$ ($n\in\mathbf{Z}$) for notational reasons, then one can take real and imaginary parts and solve the problem for trigonometric functions.
The set of polynomials $p(e^{-in\theta},\ldots,e^{in\theta})\in\mathbf{R}[1,e^{i{\theta}}, e^{-i{\theta}},e^{2i{\theta}},\ldots]$ that are identically zero is an ideal.
One can write
$$p(e^{-in\theta},\ldots,e^{in\theta})= \sum_{N}\sum_{n_1+\ldots+n_k=N}\alpha_{n_1,\ldots,n_k}e^{iN\theta}=0$$
Since the latter is an equality between holomorphic functions valid for real $\theta$, it holds for every $\theta$. If there was an $N$ with $\sum_{n_1+\ldots+n_k=N}\alpha_{n_1,\ldots,n_k}\ne 0$, for lange imaginary $\theta$ you get a contradiction.
So $p$ is a combination of polynomials of the form
$$\sum_{n_1+\ldots+n_k=N}\alpha_{n_1,\ldots,n_k}\prod_{k=1}^je^{in_k\theta}$$
with $\sum_{n_1+\ldots+n_k=N}\alpha_{n_1,\ldots,n_k}=0$
A polynomial of such form is a linear combination of polynomials of the form
$$\prod_{k=1}^je^{in_k\theta}-
e^{iN\theta}$$
with $\sum n_k =N$.
