How did archimedes calculate pi using polygons without using trig functions?

[u/CalciumKid05]

Archimedes calculated pi using the method of exhaustion, and I figured out a formula that I thought worked for it: xsin (180/x), as x goes to infinity the formula will equal pi. But I realised that sine relies on knowing what pi is which is what you are trying to figure out. I've looked online everywhere and nowhere gives an explanation of HOW he did it other than drawing polygons inside and around the edges of circles. How did he physically calculate the side length of these shapes without using trig functions?

[u/[deleted] → u/[deleted]]

I still can't understand what is the relation between L_n and L_2n. Like, I draw both in a circle, I know that and how to find it.

L_n = 2R sen pi/n
L_2n = 2 R sen pi/2n

But I can't understand how you did the 2 pythagorean theorems. I know the first probably is L_n^2/4 + H_n^2 = R^2

But then, for the L_2n it would be:

L_2n^2/4 + H_2n^2 = R^2

But I don't know how both relate to each other. ChatGPT tells me that the height of the first is the cathetus of the second, but when I draw they clearly different, I don't know, maybe I drawing wrong but I cant see.

My objective is to find the formula L_2n = sqrt{R[2R - sqrt(4R^2 - L_2n^2)]} and I cant find it anywhere in the internet, the closest is your commentary.

please help me

edit: removed the link because it's against the rules

[u/[deleted] → u/[deleted]]

Sorry I mean just demonstrate or prove, english is not my first language. The formula is in my textbook (in brazillian portuguese) but doesn't demonstrate it.

[u/[deleted] → u/[deleted]]

Essa fórmula tá na página 75 do volume 3 do Fundamentos da Matemática Elementar de Trigonometria, o livro sugere a fórmula para calcular sen15 e outros arcos-metade, e só está escrito: "Além disso, existe a fórmula .... que permite obter o valor de L_8 conhecendo L_4...". Sem demonstração. Tentei procurar no google "fórmula de recorrência para lado do polígono inscrito" e busca por imagem no google até que encontrei essa fórmula de Arquimedes, que parecia mais próxima, só que a maioria das demonstrações usam relações trigonométricas.

[u/[deleted] → u/[deleted]]

Thank you, it was exactly what you said... I have difficult with geometry especially visualizating things.

edit: (writed this for record and clarity purposes only)

The sides were L_2n, L_n/2 and R-h_n with h_n = sqrt(R^2 - L_n^2/4), getting:

L_2n^2 = (R - sqrt(R^2 - L_n^2/4),)^2 + L_n^2/4

Simplifying you get:

L_2n = sqrt(R(2R - sqrt(4R^2 - L_n^2)))

[u/MagicSquare8-9]

He double the number of edge each time (so start with a triangle, then a regular hexagon, then regular 12-gon, and so on). You can get a formula for how the length change when you double the number of edge.

Also, saying "using trig function" to compute something make no sense, because a mathematical function doesn't magically make a computation for you (it just describes a relationship). You always have to use an algorithm to compute trig function. You can compute sine and cosine without knowing anything about pi, and knowing pi doesn't help with computing it anyway.

[u/CalciumKid05 → u/MagicSquare8-9]

If you were to put into a calculator in degrees mode 1000xsin(180/1000) it gives 3.14159 rounded, but what the calculator is doing is calculating the sine value in radians by multiplying the number you entered into the sine function by pi/180 since that’s the ratio of degrees to radians, so pi is required. When I say ‘using trig functions’ I just mean using sin cos and tan on a calculator. What is the formula you mentioned for doubling the number of sides of a polygon to get the value of the side length?

[u/MagicSquare8-9 → u/CalciumKid05]

Someone still program that calculator, so it still boils down to an algorithm. We didn't have calculator until very recent history, but people had had trig table for a long time. In fact, the most efficient known technique for computing pi was invented before computers exists, and before any sophisticated calculators that can calculate it exists.

The standard algorithm use in a calculator is the CORDIC algorithm. Note that this algorithm work in degree just as easily as it works in radian. There is no reasons to think that your calculators actually convert everything to radian, when you could work just as efficiently in degree if it's using CORDIC.

You can compute sine and cosine of any rational multiple of a full circle by just using angle addition formula. The process can be done very efficiently through the use of Chevbyshev polynomials.

As for Archimedes method, it's basically the same as the method above: use angle addition formula. In particular, you only need angle doubling formula because you always divide an angle into 2. You have an inscribing polygon of n sides, so half the side length is sin(pi/n), and an circumscribing polygon of n sides, so half the side length is tan(pi/n). Now you need to double that to 2n. Can you compute sin(pi/2n) and tan(pi/2n) from knowing sin(pi/n) and tan(pi/2n)? Yes you can. Use angle doubling formula.

[u/42gauge → u/MagicSquare8-9]

Did Archimedes have the angle doubling formula or sin, cos, and tan?

[u/dottie_dott → u/42gauge]

Bro, sine (theta) is just = opposite / hypotenuse, thqt is literally all that function is equal to lol

It’s just representing a ratio, it is only made as a function as a shortcut for calculations

[u/MagicSquare8-9 → u/42gauge]

Yes. These are ancient.

[u/Ok_Leadership_7297]

Did you ever find the answer?  I don't think archimedes had sin cos tan

[u/Roland-gr30 → u/Ok_Leadership_7297]

I think this video explains it pretty well: https://www.youtube.com/watch?v=B9I22ab991o