How to prove that trigonometric functions form a Chebyshev system? How can be proven that $$\{ \operatorname{cos}(kx)\}_{k = 0}^n \text{ and } \{ \operatorname{sin}(kx)\}_{k = 1}^n$$
are Chebyshev systems in the interval $(0, \pi)$? Any ideas will be appreciated.
Thanks in advance! 
 A:   A really simple proof:
  Let $T_n(x) \in \pi_n$ denotes the $n$-th Chebyshev polinomyal of 1-st kind. We have that $T_k(\cos(x))=\cos(kx)$.  Therafore any non-trivial polynomial from $\{\cos(kx)\}_{k=0}^n=\{T_k(\cos(x))\}_{k=0}^n$ is actually a polinomyal of degree at most $n$ for $\cos(x)$ and it has at most $n$ zeroes ($\xi_1, \xi_2, ..., \xi_m$) where $m \leq n$. Since $cos(x)$ is injective in $(0,\pi)$ we have that $ \forall i=1,2,...,m$ $\exists!x_i:  \cos(x_i)=\xi_i$  So now we have that any non-trivial polynomial from $\{\cos(kx)\}_{k=0}^n$ has at most $n$ (particularly distinct) zeroes in $(0,\pi)$.  About $\{\sin(kx)\}_{k=1}^n$ check this out...  Let $\varphi(x)=\lambda_0+\lambda_1\cos(x)+...+\lambda_n\cos(nx)$ is a non-trivial polynomial for $\{\cos(kx)\}_{k=0}^n$ and at least one of $\lambda_1,...,\lambda_n \neq 0$.  It has at most $n$ zeroes in $(0,\pi)$. Now, we have that $\varphi'(x)=g(\cos(x))(-\sin(x))$ where $g(\cos(x))$ is a polynomial from degree $n-1$. Since $\sin(x) \neq 0$ and $\cos(x)$ is injective for $x \in (0,\pi)$ we conclude that $\varphi'(x)=g(\cos(x))(-\sin(x))$ has at most $n-1$ zeroes in $(0,\pi)$.  On the other hand $\varphi '(x)=-\lambda_1 \sin(x)-...-n\lambda_n \sin (nx)$ which is a non-trivial polynomial for $\{\sin(kx)\}_{k=1}^n$. Therefore every non-trivial polynomial over $\{\sin(kx)\}_{k=1}^n$ has at most $n-1$ distinct zeroes...
A: Linear independence follows from $L^2$ orthogonality. That is, the functions are $L^2$-orthogonal (this is a calculus exercise) and orthogonality implies linear independence regardless of the inner product in question. I'll leave that to you; ask in the comments if you need help with it.
Revised:
For zero counting, you can write any linear combination of cosines of angular frequencies $1,2,\dots,n$ as a polynomial of degree at most $n$ in the variable $\cos(x)$ by using the sum-to-product and product-to-sum formulas repeatedly. The actual formula doesn't matter at all, so I don't suggest writing it down; all that matters is that such a polynomial exists.
Having done this transformation, you can now use the fundamental theorem of algebra to get that $\cos(x)$ must be one of at most $n$ values. Then $\cos$ is injective on $(0,\pi)$, so $x$ must be one of at most $n$ values as well. A formal proof would use induction and this general idea.
The case with $\sin$ is slightly more complicated because $\sin$ is not injective on $(0,\pi)$; here I think you need to exploit the odd symmetry in order to get the result.
A: If you use the map $\cos x=\frac{1}{2}(e^{ix}+e^{-ix})$ and $\sin x=\frac{1}{2i}(e^{ix}-e^{-ix})$ in $T(x)=\sum_{i=0}^{n}\cos (ix)$ we get a  polynomial $P(z)=T(x)e^{inx}$ Here $z=e^{ix}$ maps $[a,a+2\pi]$ to unit circle $|z|=1.$ Clearly, if $T$ has more than $2n$ zeros, then $P$ has so. So $P$ must be identically zero. This means $T$ is also identically zero.(G. G. Lorentz PP. 24) 
