Expand $sin(z)$ about $\pi$ It seems like all you have to do is use the fact that:
$f(z)=\sum_{n=0}^\infty\frac{f^n(z_o)}{n!}(z-z_o)^n$
In this scenario, $z_o=\pi$.
The solution I got is:
$\frac{sin(\pi)}{0!}(z-z_o)^0 + \frac{cos(\pi)}{1!}(z-z_o)^1 + \frac{-sin(\pi)}{2!}(z-z_o)^2 + \frac{-cos(\pi)}{3!}(z-z_o)^3 + \frac{sin(\pi)}{4!}(z-z_o)^4 + ...$ 
This simplifies to:
$(z-\pi) - \frac{(z-\pi)^3}{6} + \frac{(z-\pi)^5}{120} - ...$
I do not know if that is right, however.  Any help would be appreciated.  Thanks a lot!

It might be of help if you proceed as follows: Note that
$$\sin(z) = \sin(z-\pi + \pi) = \sin(z-\pi) \cos(\pi) + \cos(z-\pi) \sin(\pi) = - \sin(z-\pi)$$
Setting $z-\pi = w$, we now have
$$\sin(z) = - \sin(w) = \sum_{k=0}^{\infty}(-1)^{k+1} \dfrac{w^{2k+1}}{(2k+1)!} = \sum_{k=0}^{\infty}(-1)^{k+1} \dfrac{(z-\pi)^{2k+1}}{(2k+1)!}$$
