Approximation of Pi (π)

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Let's draw a circle with a radius of r, and install a polygon inside it.

As the number of the sides of the polygon increases,

the polygon is approaching a circle, but the polygon will never be a circle.

cutting the polygon into several equal triangles with one of their tips at the center of the circle,

let's call the angle between 2 radii as θ.

as the side of the polygon increases, the angle θ becomes smaller and smaller,

the number of sides also indicates the number of triangles having one of its vertices as the center of the circle we have.

Therefore

θ = 360° / n

where n is the number of sides we have / the number of triangles we have

since we know that the area of a triangle is

Δ = (1/2)ab sinθ

in this case, a = b = r, so

Δ = (1/2) ⋅ r² ⋅ sinθ

since a single triangle has an area like that, then the whole polygon should have n times of that area, so we will have

nΔ = (n/2) ⋅ r² ⋅ sinθ

nΔ = (n/2) ⋅ r² ⋅ sin(360° / n)

as n increases, the polygon will approximately become a circle, and therefore the area would be

<lim,n→∞> nΔ = <lim,n→∞> (n/2) ⋅ r² ⋅ sin(360° / n) = πr²

π = <lim,n→∞> (n/2) ⋅ sin(360° / n)

we can also write the equation with a variable approaching 0 instead of infinity by setting

n = 1/x, when n→∞, then x→0+

therefore

π = <lim,n→∞> (n/2) ⋅ sin(360° / n)

π = <lim,x→0+> (1/2x) ⋅ sin(360° ⋅ x)

π = <lim,x→0+> sin(360° ⋅ x) / 2x