Why and how are pi and nested square roots of 2 related? How

I think both stem from iterating sin- and cos- relations to each other.

>>9151174
the last squareroot is unclear to me. it says sqrt(2 + ... + sqrt(2)). what happens at the "..."?

Pi is the arc length of the curve y = sqrt(x^2 + 1), so you take an arc length integral to get this interesting thing as an intermediate step if you work the integral in a certain verbose way.

>>9151209
I understant it as :
[math] \pi = \lim 2^k \cdot U_k [/math]
Where [math] U_1 = \sqrt 2 \\ U_{n+1} = \sqrt {U_n + 2} [/math]

Actually, there's yet another relation between them.

The cosine of pi/(2^k) can be expressed with sqrt 2s and you can sum them up to build the area of a quarter circle from the inside with triangles.

>>9151339
edit: you can sum their reciprogs up

Can someone provide a step by step (at least 5 steps) guide on how to calculate the hypotenuses gotten from dividing the circle into 2^n parts and connecting the dots?
I can't calculate past the third? step which yields sqrt(2-sqrt(2))
I'm so retarded

>>9151570
And this only because I know sin(pi/4) yields sqrt(2)/2.
How should I "know" that sin(pi/8) is sqrt(2-sqrt(2))/2 and so forth?

>>9151339
>and you can sum them up to build the area of a quarter circle from the inside with triangles
Is it beneficial to calculate the areas of the triangles when their hypotenuses will yield pi already, with the 2^n multiplier.

File: 450px-Viète_nested_polygons.svg.png (44KB, 450x450px) Image search: [Google]

44KB, 450x450px

Just what is it with these circles and square roots of two? It's so pretty.

>>9151238
Oh, didn't notice this post. Would you have the time to work out the integral in a certain verbose way? Or someone else? This is fundamental and interesting after all.

>>9151300
Doesn't that actually lead to 2^(k+1)

Note, sqrt(2+sqrt(2+sqrt(2+...)))=2.

/sci/ why aren't you more interested, these are really interesting limits for pi.
Please be more interested and do that wonky arc length integral in a verbose way for the fun of it. And explain how it leads to the result in OP image.

http://mathhelpboards.com/math-notes-49/sines-cosines-infinitely-nested-radicals-8199.html

Very thorough post. Interesting stuff.

Wtf are you mathfags even saying

>>9151576
You start with the double-angle formula.

[eqn] \sin(2x) = 2 \sin(x) \cos(x) \\
\sin^2(2x) = 4 \sin^2(x) (1 - \sin^2(x)) \\
4 \sin^4(x) - 4 \sin^2(x) + \sin^2(2x) = 0 \\
\sin^2(x) = \frac{1}{2} \left(1 - \sqrt{1 - \sin^2(2x)} \right)
[/eqn]
If you let x=pi/8 you get
[eqn] \sin^2(\frac{\pi}{8}) = \frac{1}{2} \left(1 - \sqrt{1 - \sin^2(\frac{\pi}{4})}\right)
= \frac{1}{2} \left(1 - \sqrt{1 - \frac{1}{2}}\right) \\
= \frac{1}{4} \left( 2 - \sqrt{2} \right) [/eqn]

Now you can plug in x = pi/16 to find sin(pi/16) and so on.

>>9151300
There is a minus there somewhere.

As long as you can establish that:
[eqn]\sin{\frac{\pi}{2^{k+1}}}=\frac{1}{2}\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots\sqrt{2}}}}}}}_{k\text{ square roots}}[/eqn]
(As demonstrated in >>9151911)
Then the result follows from the fact that:
[eqn]\lim_{x\rightarrow0}{\frac{\sin{x}}{x}}=1[/eqn]

>>9152234
Check out an analysis book newfriend.

>>9152247
I came up with a way to do that with just complex numbers

ITT:brainlets
https://en.wikipedia.org/wiki/Indiana_Pi_Bill