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!

Looks right, you can check your answer with Wolfram Alpha:
http://www.wolframalpha.com/input/?i=taylor+series+of+sin%28x%29+about+x%3Dpi
