Let $\overline{f}$ be a function on the whole real line, such that $\overline{f}$ is continuous and differentiable everywhere, and its derivative $\overline{f}'$ is also continuous everywhere. Now, restrict $\overline{f}$ to a function $f$ defined only on the interval $(0, \pi)$.
Does Fourier sine series of $f$ always converges to $f$ pointwise on $(0, \pi)$? I know that it does not converge uniformly on $(0, \pi)$.
What about Fourier cosine series? Does Fourier cosine series of $f$ always converge to $f$ uniformly on $(0, \pi)$?
My professor hasn't covered much in convergence, so I want to know more about the convergence of functions.
(Edit: extension method is free, first version unnecessarily assumed simple $π$-periodic extension from $(0,π)$ to $\mathbb R$)
For the full Fourier series there is uniform convergence on all closed intervals that contain no jumps and no kinks. See for instance the rather complex result cited in Wikipedia: Convergence of Fourier series.
The sine series obtained as Fourier sine series is an odd $2π$-periodic function. Thus it is obtained as the full Fourier series of the $2π$-periodic odd extension of $\bar f$, that is with $\bar f(x)=-\bar f(-x)=-\bar f(π-x)$. The no jumps condition demands $f(0)=0=f(π)$ from the original function. Differentiability will then be automatic. In general however the jump at the interval ends will produce an oscillation in the series known as "Gibbs phenomenon".
The cosine series obtained as Fourier cosine series is an even $2π$-periodic function. Thus it is obtained as the full Fourier series of the $2π$-periodic even extension of $\bar f$, that is with $\bar f(x)=\bar f(-x)=\bar f(π-x)$. Continuity is now automatic, one reason one uses DCT in JPEG, but differentiability is only given if the derivatives of $f$ at $0$ and $π$ were zero. I think that kinks do not form an obstacle for uniform convergence, as long as left and right limits of the derivative exist in every point.