I am able to prove the symmetric functional equation that Riemann gives in his paper, using Poisson Summation and properties of $\theta(x)$.
The functional equation is given like so,
\begin{equation} \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2})\zeta(s) = \pi^{-\frac{1-s}{2}}\Gamma(\frac{1-s}{2}) \zeta(1-s) \end{equation}
Now to get $\zeta(s)$ on its own do I just divide the right hand side by $\pi^{-\frac{s}{2}} \Gamma(\frac{s}{2})$? How do I manipulate that expression to use the result from the product formula for sine? Or better still is there a more efficient way to show the trivial zeroes at the negative even integers?
If you want the sine term you have to make some standard manipulations of the Gamma function. We have $$\zeta\left(s\right)=\pi^{s-1/2}\frac{\Gamma\left(\frac{1}{2}-\frac{s}{2}\right)}{\Gamma\left(\frac{s}{2}\right)}\zeta\left(1-s\right) $$ then by the duplication formula of Gamma and the Euler's reflection, we get $$\zeta\left(s\right)=\pi^{s-1/2}\frac{\Gamma\left(\frac{1}{2}-\frac{s}{2}\right)}{\Gamma\left(\frac{s}{2}\right)}\zeta\left(1-s\right)=\pi^{s}2^{s}\frac{\Gamma\left(1-s\right)}{\Gamma\left(\frac{s}{2}\right)\Gamma\left(1-\frac{s}{2}\right)}\zeta\left(1-s\right)=\pi^{s-1}2^{s}\sin\left(\pi s/2\right)\Gamma\left(1-s\right)\zeta\left(1-s\right). $$