This, then, is what you are looking for.
Jameson said:
Sorry, I'll clarify. This is the form I'm looking for:
\zeta(1-s)=2^{1-s}{\pi}^{-s}(\sin{\frac{(1-s)\pi}{2}}})(s-1)!\zeta(s)
This one gives the symmetric form of the above, from there you can use your knowledge of the gamma function to obtain the desired form. And by this one, I do mean exercise #21 from pg. 51 of Andrews, G., Askey, R., & Roy, R. (2000). Encyclopedia of mathematics and its applications: special functions . 1st paperback ed. Cambridge, UK: Cambridge University Press., which reads as follows:
Prove that \pi^{-\frac{s}{2}}\Gamma(\frac{s}{2})\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma(\frac{1-s}{2})\zeta(1-s) as follows:
a) Observe that \sum_{n=0}^{\infty}\frac{\sin((2n+1)x)}{2n+1}=(-1)^{m}\frac{\pi}{4} for m\pi<x<(m+1)\pi, m\in\mathbb{N}
b) Multiply the equation by x^{s-1} (0 < s < 1) and integrate over (0,\infty). Show that the left side is \Gamma(s)\sin(\frac{s\pi}{2})(1-2^{-s-1})\zeta(s+1) and that the right represents an analytic function for \Re(s)<1 and is equal to 2(1-2^{s+1}) \zeta(1-s) for \Re(s)<0.
c) Deduce the functional equation for the zeta function. (Hardy

)
(I added the smiley, and note that the text had the sum in part (a) starting at n=1, which my prof. confirmed is a typo).