PI Approximation With Arcsine Infinite Series

Question

While Playing with numbers last night I stumbled on an approximation of pi that was quite exciting. I'm sure it has been discovered before, but it was a fun journey nevertheless and I wanted to share it.

I haven't found the approximation on the web yet, but I have seen things that resemble it, which probably means there just needs to be some algebraic adjustments and it will be a well known approximation. Please help me find it.

I was trying find the length of the sides of a triangle with $45^\circ$, $22.5^\circ$, $11.25^\circ$, etc.

$$ \sin(45^\circ) = \sin(\frac{45\pi}{180}) $$

And because I was halving the degree each step I made a function.

$$ f(x)= \sin(\frac{\frac{45\pi}{2^x}}{180}) $$

Which can be simplified to.

$$ f(x)= \sin(\frac{\pi}{2^{x+2}}) $$

At this moment I thought, oh you could use some algebra and trig to get this equation to equal $\pi$.

$$ \pi=2^{x+2}\arcsin(f(x)) $$

We know the answer to f(1) is $\frac{1}{2}\sqrt{2-\sqrt{2}}$.

$$ \pi=2^3\arcsin(\frac{1}{2}\sqrt{2-\sqrt{2}}) $$

We also know we can use the infinite series to remove arcsine.

$$ \pi = 8 \sum_{n=0}^{\infty} (\frac{(2n)!}{ 4^n(n!)^2(2n+1) })(\frac{1}{2}\sqrt{2-\sqrt{2}})^{2n+1} $$

I then did the next x in the f(x), f(2), and it gave me a faster convergence.

$$ \pi = 16 \sum_{n=0}^{\infty} (\frac{(2n)!}{ 4^n(n!)^2(2n+1) })(\frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{2}}})^{2n+1} $$

I then noticed the pattern and this continues to make the convergence on $\pi$ faster as you increase x in f(x).

$$ \pi = 32 \sum_{n=0}^{\infty} (\frac{(2n)!}{ 4^n(n!)^2(2n+1) })(\frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{2 + \sqrt{2}}}})^{2n+1} $$

If any of you know where I can read more about this, I would really enjoy it. I hope you found this as entertaining as I did. Happy Mathing ;)