If I plug in 4 for s into:
$$\begin{equation} \zeta(s)=2^s\pi^{s-1}\Gamma(1-s)\sin\left(\frac{\pi s}{2}\right)\zeta(1-s). \end{equation}$$
Doesn't $$\zeta(4) = 0 $$ because of the sine function?
Does this mean that the above functional equation has parameters for where it can be used?