Finding Pi In A square Plate

 Consider again,

$\theta_0=\cfrac{\pi}{4}$

$Area = r(\cfrac{1}{sin\theta_0}-1).r(\cfrac{1}{sin\theta_0}-1)tan\theta_0=r^2(\cfrac{1}{sin\theta_0}-1)^2tan\theta_0$

For all four quadrant, multiply by $M_0=4$.

Consider the next triangle for which there are two, 

$\theta_1=\cfrac{1}{2}(\pi-(\cfrac{\pi}{2}-\theta_0))=\cfrac{\pi}{4}+\cfrac{\theta_0}{2}=\cfrac{3}{8}\pi$

$Area = r(\cfrac{1}{sin\theta_1}-1).r(\cfrac{1}{sin\theta_1}-1)tan\theta_1=r^2(\cfrac{1}{sin\theta_1}-1)^2tan\theta_1$

$M_1=4*2$

And the next $\theta_2$,

$\theta_2=\cfrac{1}{2}(\pi-(\cfrac{\pi}{2}-\theta_1))=\cfrac{\pi}{4}+\cfrac{\theta_1}{2}=\cfrac{7}{16}\pi$

We have,

$S_{\theta_2}=\cfrac{\pi}{4}+\cfrac{1}{2}\left(\cfrac{\pi}{4}+\cfrac{\theta_0}{2}\right)=\cfrac{\pi}{4}+\cfrac{1}{2}*\cfrac{\pi}{4}+\cfrac{\theta_0}{2^2}$

$S_{\theta_3}=\cfrac{\pi}{4}+\cfrac{1}{2}\left(\cfrac{\pi}{4}+\cfrac{1}{2}\left(\cfrac{\pi}{4}+\cfrac{\theta_0}{2}\right)\right)=\cfrac{\pi}{4}+\cfrac{\pi}{2*4}+\cfrac{\pi}{2^2*4}+\cfrac{\theta_0}{2^3}$

$S_{\theta_4}=\cfrac{\pi}{4}+\cfrac{1}{2}\left(\cfrac{\pi}{4}+\cfrac{\pi}{2*4}+\cfrac{\pi}{2^2*4}+\cfrac{\theta_0}{2^3}\right)=\cfrac{\pi}{4}+\cfrac{\pi}{2*4}+\cfrac{\pi}{2^2*4}+\cfrac{\pi}{2^3*4}+\cfrac{\theta_0}{2^4}$

In general, 

$S_{\theta_i}=\cfrac{\pi}{2^2}+\cfrac{\pi}{2^3}+\cfrac{\pi}{2^4}+...+\cfrac{\pi}{2^{i+1}}+\cfrac{\theta_0}{2^{i}}$

Multiplicative factor due to symmetry, $M_2=4*2^2=16$, in general for $\theta_i$,

$M_i=4*2^{i}=2^{i+2}$

We have a series $E_i$ to estimate $\pi$, with $r=1$, area of square 4,

$\epsilon_i=M_i(\cfrac{1}{sin\theta_i}-1)^2tan\theta_i$

$\theta_0=\cfrac{\pi}{4}$

$S_{\theta_i}=\cfrac{\pi}{2^2}+\cfrac{\pi}{2^3}+...+\cfrac{\pi}{2^{i+1}}+\cfrac{\theta_0}{2^{i}}=\cfrac{\pi}{2^2}+\cfrac{\pi}{2^3}+...+\cfrac{\pi}{2^{i+2}}$

and $M_i=2^{i+2}$

$E_i=4-\epsilon_0-\epsilon_1-\epsilon_2...-\epsilon_i$

$E_0=8(\sqrt{2}-1)=3.313708$

$\epsilon_1=8(\cfrac{1}{sin(\frac{\pi}{4}+\frac{\pi}{8})}-1)^2tan(\frac{\pi}{4}+\frac{\pi}{8})=0.131111$

$E_1=8(\sqrt{2}-1)-0.131111=3.182598$

$\epsilon_2=16(\cfrac{1}{sin(\frac{\pi}{4}+\frac{\pi}{8}+\frac{\pi}{16})}-1)^2tan(\frac{\pi}{4}+\frac{\pi}{8}+\frac{\pi}{16})=0.03087297$

$E_2=8(\sqrt{2}-1)-0.131111-0.03087297=3.151725$

$\epsilon_3=32(\cfrac{1}{sin(\frac{\pi}{4}+\frac{\pi}{8}+\frac{\pi}{16}+\frac{\pi}{32})}-1)^2tan(\frac{\pi}{4}+\frac{\pi}{8}+\frac{\pi}{16}+\frac{\pi}{32})\\=0.0076065$

$E_3=8(\sqrt{2}-1)-0.131111-0.03087297-0.0076065=3.144118$

$\epsilon_4 =64(\cfrac{1}{sin(\frac{\pi}{4}+\frac{\pi}{8}+\frac{\pi}{16}+\frac{\pi}{32}+\frac{\pi}{64})}-1)^2tan(\frac{\pi}{4}+\frac{\pi}{8}+\frac{\pi}{16}+\frac{\pi}{32}+\frac{\pi}{64})  \\=0.001894755$

$E_4=8(\sqrt{2}-1)-0.131111-0.03087297-0.0076065-0.001894755=3.1422236$

$\epsilon_5 =128(\cfrac{1}{sin(\frac{\pi}{4}+\frac{\pi}{8}+\frac{\pi}{16}+\frac{\pi}{32}+\frac{\pi}{64}+\cfrac{\pi}{128})}-1)^2tan(\frac{\pi}{4}+\frac{\pi}{8}+\frac{\pi}{16}+\frac{\pi}{32}+\frac{\pi}{64}+\cfrac{\pi}{128})  \\=0.000473261$

$E_5=8(\sqrt{2}-1)-0.131111-0.03087297-0.0076065-0.001894755-0.000473261\\=3.1417500$

$\epsilon_6 =256(\cfrac{1}{sin(\frac{\pi}{4}+\frac{\pi}{8}+\frac{\pi}{16}+\frac{\pi}{32}+\frac{\pi}{64}+\frac{\pi}{128}+\frac{\pi}{256})}-1)^2tan(\frac{\pi}{4}+\frac{\pi}{8}+\frac{\pi}{16}+\frac{\pi}{32}+\frac{\pi}{64}+\frac{\pi}{128}+\frac{\pi}{256})  \\=0.000118288$

$E_6=8(\sqrt{2}-1)-0.131111-0.03087297-0.0076065-0.001894755\\-0.000473261-0.000118288=3.1416317$

Just for fun.  $\pi$ as a simple series.