Infinite ways to represent $\pi$ as product of nested square roots of $2$ and $2^n$ and odd numbers Polygon inscribed in a circle leads to famous infinite product by Viete's formula.here
One way to represent that as infinite nested radical of 2 as follows
$$\pi = \lim_{n\to\infty}2^n \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}$$ being total number of '2's inside the nested square roots is 'n'
After some manipulation in numbers some interesting findings I'm able to observe as follows
$$3\pi = \lim_{n\to\infty}2^n \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...\sqrt{2-\sqrt{2}}}}}}}$$
$$5\pi = \lim_{n\to\infty}2^n \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...\sqrt{2-\sqrt{2-\sqrt{2}}}}}}}}$$
In both these situations total number of '2's inside the nested square roots are 'n'.
As the n grows big each converge to respective multiples of $\pi$ more and more decimals
Python code to verify my findings

from decimal import*

getcontext().prec = 100

# nested sqrts of 2 & $\pi$, $3\pi$,$5\pi$

b = int(input("Enter number of iterations: "))

a = Decimal(2).sqrt()

for i in range(b):
    a = Decimal(2 + a).sqrt()


a = Decimal(2 - a).sqrt()

Pi = 2 ** (b + 2) * a

print("\n\n Pi derived from Square as initial regular Polygon is:\n\n ", Pi)


a = Decimal(2).sqrt()
a = 2 - a
a = Decimal(a).sqrt()

b = int(input("Enter required number of cycles to calculate Pi: "))

for i in range(b):
    a = 2 + a
    a = Decimal(a).sqrt()

a = 2 - a
a = Decimal(a).sqrt()


Pi_3 = 2**(b+3)*a
C = Pi_3/3

print("\n\n",C)

a = Decimal(2).sqrt()
a = 2 - a
a = Decimal(a).sqrt()

a = 2 - a
a = Decimal(a).sqrt()

b = int(input("Enter required number of cycles to calculate Pi: "))


for i in range(b):
    a = 2 + a
    a = Decimal(a).sqrt()

#print("\n\n",a)

a = 2 - a

a = Decimal(a).sqrt()

Pi_5 = 2**(b+4)*a
C = Pi_5/5

print("\n\n",C)
# for practical purposes I have divided by 3 & 5 respectively (next to regular derivation by Viete's method) to get the value of $\pi$

Viete's formula is provable by geometry i.e. square becoming octagon then 16gon and so on ...

Is there any way to prove above 2 formulas by geometric shapes or some other methods? Are there similar infinite ways like those above?(with the help of nested square roots of 2 apart from equilateral triangle, regular Pentagon). Please comment. Thanks in advance

 A: Similarly to formulas, related to Viète's formula, in  your formulas is considered a limit $\lim_{n\to\infty} 2^n\sqrt{2-a_n}$, where $a_n=\sqrt{2+a_{n-1}}$ with the initial conditions
$a_2=-\sqrt{2}$ for the first sequence and $a_3=-\sqrt{2-\sqrt{2}}$ for the second. Assuming $a_{n-1}=2\cos x_{n-1}$ for $x_{n-1}\in [0,\pi]$, we obtain $a_n=2\cos\tfrac {x_{n-1}}{2}$. Putting $x_n=\tfrac {x_{n-1}}2$ for all sufficiently big $n$, we obtain
$$\lim_{n\to\infty} 2^n\sqrt{2-a_n}=\lim_{n\to\infty} 2^n \sqrt{2-2\cos x_n}=\lim_{n\to\infty} 2^{n+1} \sin x_{n+1}.$$
For the first sequence we have $a_2=2\cos\tfrac{3\pi}4$, so $x_2=\tfrac{3\pi}4$ and
$$\lim_{n\to\infty} 2^n\sqrt{2-a_n}=\lim_{n\to\infty} 2^{n+1} \sin \left(\tfrac{3\pi}4\cdot \frac 1{2^{n-1}}\right)=3\pi.$$
For the second sequence we have $a_3=2\cos x_3=-\sqrt{2-\sqrt{2}}<0$. So $x_3\in \left(\tfrac {\pi}2, \pi\right) $ and
$4\cos^2 x_3=2-\sqrt{2}$. Thus $2\cos 2x_3=-\sqrt{2}$ and so $2x_3=\tfrac {5\pi}4$ and
$$\lim_{n\to\infty} 2^n\sqrt{2-a_n}=\lim_{n\to\infty} 2^{n+1} \sin \left(\tfrac{5\pi}8\cdot \frac 1{2^{n-2}}\right)=5\pi.$$
