Pi Approximation Day

[Albert Chan]

(07-24-2022 10:16 PM)Thomas Klemm Wrote:  \(
\begin{align}
\tan \frac{\pi}{12} = 2 - \sqrt{3}
\end{align}
\)

Assumed x in first quadrant, to avoid cancellation errors, we can use cotangent

cot(x/2) = cot(x) + √(1 + cot(x)^2)

cot(pi/6) = √3      → cot(pi/12) = √3 + √(1+3) = 2 + √3

3 SQRT 2 +               // cot(pi/12) ≈ 3.732
ENTER X^2 1 + SQRT +     // cot(pi/24) ≈ 7.596
ENTER X^2 1 + SQRT +     // cot(pi/48) ≈ 15.26
ENTER X^2 1 + SQRT +     // cot(pi/96) ≈ 30.55
1/X
11
R/S
96 *

3.141592653589793238462643383279503