How to calculate the infinite series of $\sin x $ without using $\pi$

Question

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.

Answer 1

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!