How to calculate the infinite series of $\sin x $ without using $\pi$ while solving for the Maclaurin series for arcsin(x)
$$ 
\arcsin(x) = \sum_{0}^{\infty}\frac{(2n-1)!!x^{2n+1}}{(2n)!!(2n+1)}
$$
Taking $$\arcsin(\sin\frac{\pi}{6})$$
I got
$$
\pi = 6\sum_{0}^{\infty}\frac{(2n-1)!!(\frac{1}{2})^{2n+1}}{(2n)!!(2n+1)}
$$
This gives around 16 digits of pi per 22 iterations.
Trying to improve the convergence of the function I took
$$
\arcsin(\sin\frac{\pi}{12})
$$
Considering,
$$
\sin\frac{\pi}{12}=\sqrt{2-\sqrt{3}}
$$
We get,
$$
\pi = 12\sum_{0}^{\infty}\frac{(2n-1)!!{\sqrt{2-\sqrt{3}}}^{2n+1}}{(2n)!!(2n+1)}
$$
Which yielded a faster convergence of pi than the first equation. I have not calculated the number of digits of pi per iteration since I focused on on the below equation.
$$
\arcsin(\sin\frac{\pi}{24})
$$
Considering,
$$
\sin\frac{\pi}{24}=\sqrt{2-\sqrt{2+\sqrt{3}}}
$$
We get,
$$
\pi = 24\sum_{0}^{\infty}\frac{(2n-1)!!{\sqrt{2-\sqrt{2+\sqrt{3}}}}^{2n+1}}{(2n)!!(2n+1)}
$$
Which gives about 18 digits of pi per 10 iterartions, which is an improvement over the first equation.
I've tried using identities involving e
i.e
$$
\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}
$$
I do not know how to calculate sin(x) using the above equation
since complexity of the equation seems to increase with n increasing with
$$
x=\frac{\pi}{n}
$$
I've also tried using the golden ratio using the identity
$$
\phi=1+2\sin\frac{\pi}{10}
$$
We get,
$$
\sin\frac{\pi}{10}=\frac{\phi-1}{2}
$$
$$
\sin\frac{\pi}{10}=\frac{\sqrt{5}-1}{4}
$$
Taking,
$$
\arcsin(\sin\frac{\pi}{10})
$$
We get,
$$
\pi = 10\sum_{0}^{\infty}\frac{(2n-1)!!(\frac{\sqrt{5}-1}{4})^{2n+1}}{(2n)!!(2n+1)}
$$
With my limited observation, seemed to converge at a similar rate to  $\arcsin(\sin\frac{\pi}{24})$
Is there any way to find  $\sin\frac{\pi}{n}$
Without using $\pi$ and isn't computationally very tasking?
Thanks.
 A: For a general value of $n$ I have good news and bad news. The good news is that $\sin(\pi/n)$ is always available as radicals. The bad news is ... well, this is the result for $n=9$:
$$
\frac{1}{4}\,\sqrt {-\sqrt [3]{4+4\,i\sqrt {3}}-4\,{\frac {1}{\sqrt [3]{4+4\,i
\sqrt {3}}}}+8+8\,i\sqrt {3} \left( \frac{1}{8}\sqrt [3]{4+4\,i\sqrt {3}}-\frac{1}{2}\,{\frac {1}{\sqrt [3]{4+4\,i\sqrt {3}}}} \right) }
$$
Oh yeah, that is one of the simpler cases.  In general, you can render $\sin(\pi/n)$ with real radicals only, and thus reasonably economical for computation, only if all the radicals are square roots. This is equivalent to the regular $n$-gon being constructible from rational and square-root operations, thus a Euclidean construction.
Threading the needle
If you can construct a regular $n$-gon, then you can easily construct a regular $2n$-gon.  This suggests that you can construct $\sin[\pi/(2n)]$ from $\sin(\pi/n)$. Let's explore how that might be accomplished algebraically.
Start from the double-angle identity
$\sin(\pi/n)=2\sin[\pi/(2n)]\cos[\pi/(2n)]$
Square both sides and substitute $1-\sin^2$ for $\cos^2$:
$\sin^2(\pi/n)=4\sin^2[\pi/(2n)][1-\sin^2[\pi/(2n)]]$
And then
$4\sin^4[\pi/(2n)]-\sin^2[\pi/(2n)]+\sin^2(\pi/n)=0$
This can be solved for $\sin^2[(\pi/(2n)]$ by using the Quadratic Formula, but the operation becomes ill-conditioned with the form we usually use.  Instead use this alternative solution for $ax^2+bx+c=0$:
$x=\dfrac{2c}{-b\pm\sqrt{b^2-4ac}}$
Plugging this into our equation and taking the smaller root leads to
$\sin^2[\pi/(2n)]=\dfrac{\sin^2(\pi/n)}{2[1+\sqrt{1-\sin^2(\pi/n)}]}$
Thus we get the iterative relationship
$\sin[\pi/(2n)]=\dfrac{\sin(\pi/n)}{\sqrt{2[1+\sqrt{1-\sin^2(\pi/n)}]}}$
so that given $\sin(\pi/n)$ we can get $\sin[\pi/(2n)]$ fairly readily, and upon iteration get the denominator large enough for an accurate rendering of $\pi$ (if you get the square roots to required accuracy, which can be done with Newton's Method) even with just one term in the series!
