Computing π from Scratch: A Formula Hiding Between Sine and Tangent

Archimedes showed that an inscribed polygon and a circumscribed polygon bracket π from below and above. The inscribed polygon undershoots the arc while the circumscribed polygon overshoots it. A Taylor series expansion shows these errors are in ratio 1:2 with opposite signs. If I give sine twice the weight of tangent, the errors cancel. The resulting formula: P(N) = N · [2 sin(π/N) + tan(π/N)] / 3 converges to π with fourth order accuracy O(1/N⁴) compared to O(1/N²) for Archimedes' bounds. I show the 2:1 weighting is the only one that works. I compute π from √3 using half angle formulas, then bracket π from both sides, and finally, I generate higher order formulas by repeating the same cancellation.