Why can you replace $1/\sin(\phi/2)$ with $2/\phi$ in an integral?

Question

I am walking myself through a proof of convergence of Fourier series.

For the partial sum $S_Nf(\theta)$, and any constant $S$, we have that

$$ S_Nf(\theta) - S = \frac{1}{2 \pi} \int_0^\pi ( f(\theta + \phi) + f(\theta - \phi) - 2S ) \frac{\sin(N + 1/2) (\phi)}{\sin(\phi/2)} \ d \phi $$

Here, the author (Pinsky) says that we can replace $1/\sin(\phi/2)$ with $2/\phi$ and an error that tends to zero, uniformly in $\theta$, but I do not understand how you are allowed to do this.

Answer 1

Using the first term, the Taylor series expansion of $\frac 1{\sin(\phi/2)}=\csc(\phi/2) \approx 2/\phi$ http://www.efunda.com/math/taylor_series/trig.cfm