Consider a series $\sum_{k=2}^\infty a_n \sin(kx)$, where $\{a_n\}$ is a sequence of reals. Is it possible to construct $\{a_n\}$ such that $\sum_{k=2}^\infty a_n \sin(kx)$ converges uniformly to $\sin(x)$ on $[0,\pi]$?
If there were indeed an expression $$ \sin(x) = \sum_{k=2}^\infty a_n \sin(kx) $$ then it implies $$ 0 = \sin(x) - \sum_{k=2}^\infty a_n \sin(kx) $$ saying that $0$ has a Fourier sine series, which is nonsense. But I can't think up a proof to show the problem, so is it really impossible to construct one?
If it is true you get $\int_0^{\pi}\sin^{2}xdx=\sum_k a_k \int_0^{\pi} \sin x \sin (kx)dx=\sum a_k 0=0$ which is a contradiction.