In this post, I would like to share the findings on derivation of $\pi$ with Pentadecagon inscribed in unit circle.
Here the side of each chord is $2\sin12^\circ$
(Bisecting the chord by segment which passes through the centre of circle, it is perpendicular to the chord)
Now doubling the sides i.e for 30gon, each side of chord is $2\sin6^\circ$ which can be represented as $\sqrt{2-2\cos12^\circ}$ (by Half angle cosine formula)
For 60gon it is $2\sin3^\circ$ which can be represented as $\sqrt{2-\sqrt{2+2\cos12^\circ}}$
Circumference of 60gon will be $60\cdot\sqrt{2-\sqrt{2+2\cos12^\circ}}$
Approximation for $\pi$ will be $15\cdot2^1\cdot\sqrt{2-\sqrt{2+2\cos12^\circ}}$
When we continue like this
For 120gon it is $2\sin1.5^\circ$ which can be represented as $\sqrt{2-\sqrt{2+\sqrt{2+2\cos12^\circ}}}$
Circumference of 120gon will be $120\cdot\sqrt{2-\sqrt{2+\sqrt{2+2\cos12^\circ}}}$
Approximation for $\pi$ will be $15\cdot2^2\cdot\sqrt{2-\sqrt{2+\sqrt{2+2\cos12^\circ}}}$
We can generalize this to get the more and more accurate value for $\pi$ as $\pi = 15\cdot2^n\cdot\sqrt{2-\sqrt{2+\sqrt{2+...\text{(n times)}+2\cos12^\circ}}}$
$\therefore$ $\pi = \lim_{n \to\infty} 15\cdot2^n\cdot\sqrt{2-\sqrt{2+\sqrt{2+...\text{(n times)}+2\cos12^\circ}}}$
Note:
Infinite expansion happens between right and left extremes of nested radical - (Infinite expansion of balloon nested radical)
Interesting aspects of deriving $\pi$ with Pentadecagon inscribed in unit circle are
$2\cos12^\circ$ as Infinite nested square roots of 2 as $\sqrt{2+{\sqrt{2+\sqrt{2-\sqrt{2-...}}}}}$ here
$2\cos12^\circ$ as a finite nested radical can be represented as $\frac{1}{2}\times\sqrt{9+\sqrt5+\sqrt{(30-6\sqrt5)}}$
Relatively faster convergence than square inscribed in unit circle giving rise to $\pi$
Amazing thing to note is 2 infinities in single nested radical. At the right extreme, cyclic infinite nested square roots of 2 converging to $2\cos\frac{\pi}{15}$ and then another infinite expansion between the extremes of nested radical ultimately converging to $\pi$. It is as though there is strict discipline between 2 different infinities converging to $\pi$ (pun intended). Even though both are infinitely expandable, they are countable in the strict sense, if we observe.
$\pi = \lim_{n \to\infty} 15\cdot2^n\cdot\sqrt{2-\sqrt{2+\sqrt{2+...\text{(n times)}+\sqrt{2+{\sqrt{2+\sqrt{2-\sqrt{2-...}}}}}}}}$
Interestingly infinite product for $\pi$ with perspective of Pentadecagon will be as follows
If $C = 2\cos\frac{\pi}{15}$
$$\pi = 15\cdot\sqrt{2-C}\cdot\frac{2}{\sqrt{2+\sqrt{2+C}}}\cdot\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+C}}}}...$$