For $j\in\mathbb{N}_0$ let $\sin_j(x)=\sin(jx)$ and $\cos_j(x)=\cos(jx)$ for all $x\in\mathbb{R}$. How do I prove that the finite series of functions $$ (1,\cos_1,\sin_1, \cos_2, \sin_2,...,\cos_m,\sin_n) $$ is a $\mathbb{R}$-Chebyshev-system for $n\in\mathbb{N}_0$ on $\mathbb{R}/2\pi \cong (-\pi,\pi]$?
Thankfull for every answer and your help!
EDIT: Definition of a Chebyshev-System
"A system of linearly independent functions $S=(ϕ_i)_{i=0}^n$ in a space $C(Q)$ with the property that no non-trivial polynomial in this system has more than n−1 distinct zeros."$
Let $n \in \mathbb{N}_{\geq 1}$ be a fixed order. Based, for example, on the books
we know that \begin{align} &~\det \left( \begin{array}{cccccccc} 1 & \cos\left(x_0\right) & \sin\left(x_0\right) & \cdots & \cos\left(nx_0\right) & \sin\left(nx_0\right)\\ 1 & \cos\left(x_1\right) & \sin\left(x_1\right) & \cdots & \cos\left(nx_1\right) & \sin\left(nx_1\right)\\ \vdots & \vdots & \vdots & & \vdots & \vdots\\ 1 & \cos\left(x_{2n}\right) & \sin\left(x_{2n}\right) & \cdots & \cos\left(nx_{2n}\right) & \sin\left(nx_{2n}\right)\\ \end{array} \right) \\ \\~ =& ~2^{2n^2} \prod_{0\leq j < i \leq 2n} \sin\left(\frac{x_i - x_j}{2}\right) > 0\end{align} whenever $$ x_0 < x_1 < \ldots < x_{2n} < x_0 + 2\pi. $$
If I am not mistaken, your question is equivalent to the inequality cited above. (Considering the exponent of the factor $2^{2n^2}$ (appearing before the product operator), it seems, that equation (7.19) of [Karlin, 1968] contains a typo.)
You may also check the answer given to a very similar question.