I wanted to know if the method I used is valid and not just pure luck:
Consider the function: $$\dfrac{\cos(\frac{\pi x}{2})}{(x^2-1)}$$ The roots of the equation are$\quad\pm3,5,7,9,\cdots$
So the function can be represented as $$(1+\frac{x}{-3})(1+\frac{x}{3})(1+\frac{x}{-5})(1+\frac{x}{5})\cdots=(1-\frac{x^2}{3^2})(1-\frac{x^2}{5^2})(1-\frac{x^2}{7^2})\cdots$$
So $$\dfrac{\cos(\frac{\pi x}{2})}{(x^2-1)}=(1-\frac{x^2}{3^2})(1-\frac{x^2}{5^2})(1-\frac{x^2}{7^2})\cdots$$
Substitute $x=1$, the LHS becomes, by L'Hopital's rule, $\dfrac{\pi}{4}$
So $$\dfrac{\pi}{4}=(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})\cdots=\frac{8}{9}\frac{24}{25}\frac{48}{49}$$
This can be rewritten as:$\dfrac{2\cdot4}{3\cdot3}\dfrac{4\cdot6}{5\cdot5}\dfrac{6\cdot8}{7\cdot7}\cdots$
So $$\dfrac{\pi}{2\cdot2}=\dfrac{2\cdot4}{3\cdot3}\dfrac{4\cdot6}{5\cdot5}\dfrac{6\cdot8}{7\cdot7}\cdots$$
So $$\dfrac{\pi}{2}=\dfrac{2\cdot2\cdot4\cdot6\cdot6\cdot8\cdot8\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdot7\cdots}$$