Consider a regular octagon, inscribed in a circle of radius R

φ = π/4

d² = (x₂ – x₁)² + (y₂ – y₁)² [ ask Mr. Pythagoras ]

d² = (R - Rcos(φ))² + R²sin²(φ)

d² = R² - 2R²cos(φ) + R²cos²(φ) + R²sin²(φ)

d² = R²( 1 + cos²(φ) + sin²(φ)) - 2R²cos(φ)

d² = 2R² - 2R²cos(φ)

d = R SQRT( 2 - 2cos(φ) )

Thus, the octagon’s perimeter is P = 8d, or

P = 8R SQRT( 2 - 2cos(φ) )

Now consider a regular polygon with n8 sides

φ_n = 2π/(8n), where n = 1, 2, 3, 4, …

P_n = 8nR SQRT( 2 - 2cos(2π/(8n)) )

Note: if n >> 1 then 2π/(8n) << 1

Let us look at the cosine function

Consider this approximation for cos(φ) near φ = 0

cos(φ) = 1 − φ²/2! + φ⁴/4! − φ⁶/6! + φ⁸/8! − …

Note: there are no terms like φ, φ ³, φ⁵, …

because cos(-φ) = cos(φ)

Then, with

cos(φ) = 1 − φ²/2! + φ⁴/4! − φ⁶/6! + φ⁸/8! − …

and for φ << 1,

( 2 - 2cos(φ) ) ≈ φ²

Then,

with φ = φ_n = 2π/(8n)

P_n = 8nR SQRT( 2 - 2cos(2π/(8n)) ) becomes

P_n ≈ 8nR(2π/(8n)) ) = 2πR

and if n >> 1,

then P_n ≈ 2πR is the circumference of the circle. Surprise !!