it is well known that
$$\frac{2}{\pi}=\sqrt{\frac12}{\sqrt{\frac12+\frac12\sqrt{\frac12}}{\sqrt{\frac12+\frac12\sqrt{\frac12+\frac12\sqrt{\frac12}}}{\sqrt{\frac12+\frac12\sqrt{\frac12\cdots}}}}}$$
My Idea is to know what about above product if it is a power as shown below : $A=\sqrt{\frac12}^{\sqrt{\frac12+\frac12\sqrt{\frac12}}^{\sqrt{\frac12+\frac12\sqrt{\frac12+\frac12\sqrt{\frac12}}}^{\sqrt{\frac12+\frac12\sqrt{\frac12\cdots}}}}}$ ?
First, let us try to find out what $\sqrt{\frac12+\frac12\sqrt{...}}$ can evaluate to.
By a recursive trick, we can set up an equation as below
$$x = \sqrt{\frac12+\frac12x}\to x^2-\frac12x-\frac12=0$$
So, $x=-0.5,1$. Now, we note that the radical could not converge to $-0.5$ as the number of recursive units grows, so if it even converges, it would approach $1$.
That's not terribly helpful, since $x^1=x$ for all $x$. But, what could be useful is seeing how fast it converges to $1$, to see how many terms we need to carefully look at in the exponentiation.
We note that if $f(a) =\sqrt{\frac12+\frac12a}$ and $b=\sqrt{\frac12}$ (Inspired by commenter @Mason), then $$f(f(f(f(f(b)))))\sim0.9996988$$
So, 4 levels of recursion should be very close to the final limit value of this exponentiation, which gives us an approximation of $0.725657309196$. I will add more to this answer if I find a more satisfying approach.