Let $\phi: \mathbb R \to \mathbb R$ be $2\pi$-periodical function with $\phi|_{[-\pi,\pi[}= 1$
Prove $\forall N \in \mathbb N$ with $|x|<2\pi$ the fourier polynomial $F_{N}\phi$ satisfies:
$$(F_{N}\phi)(x+\pi)=x-\int^{x}_{0}\frac{\sin{[(N+1/2)t]}}{\sin{(t/2)}}dt$$
Steps so far:
First looking at the fourier coefficients $c_{k}$ for $\phi$, we've got:
By definition $$c_{k}=\frac{1}{2\pi}\int^{\pi}_{-\pi}f(x)e^{-ikx}dx$$ and through partial integration we get:
$$c_{k}=\frac{\cos(k\pi)}{-ik}=\frac{(-1)^{k}}{-ik}$$
Now, looking at the Fourier polynomial, defined as:
$$(F_{N}\phi)(x)=\sum^{N}_{k=-N}c_{k}e^{-ikx}$$
That means:
$$(F_{N}\phi)(x)=\sum^{N}_{k=-N}(-1)^{k}\frac{e^{-ikx}}{-ik}$$
and since $c_{0}=0$;
$$(F_{N}\phi)(x)=\sum^{N}_{k=1}(-1)^{k}\frac{e^{-ikx}}{-ik}+\sum^{-1}_{k=-N}(-1)^{k}\frac{e^{-ikx}}{-ik}$$
It therefore follows that:
\begin{align} (F_{N}\phi)(x+\pi)&=\sum^{N}_{k=1}(-1)^{k}\frac{e^{-ik(x+\pi)}}{-ik}+\sum^{-1}_{k=-N}(-1)^{k}\frac{e^{-ik(x+\pi)}}{-ik}\\ &=\sum^{N}_{k=1}(-1)^{2k}\frac{e^{-ikx}}{-ik}+\sum^{-1}_{k=-N}(-1)^{2k}\frac{e^{-ikx}}{-ik}\\ &=\sum^{N}_{k=1}\frac{e^{-ikx}}{-ik}+\sum^{-1}_{k=-N}\frac{e^{-ikx}}{-ik} \end{align}
This is where I stop and I now have no idea how to get it back into integrals, let alone proving anything. Does the $\forall N \in \mathbb N$ suggest induction as a means of proving?
Any help is greatly appreciated.