This paper extends my previous geometric formula for π 1 by using the full geometry of an inscribed right triangle. My previous formula 1 relied on the small-angle approximation, which uses only the base of the inscribed triangle. This paper adds the tangent to complete the geometry by recovering the height. If I combine both the sine and the tangent in a 2:1 ratio, the cubic terms in the Taylor series cancel exactly. The leading error drops from θ³ to θ⁵. The formula is f(n) = 180 · 10ⁿ · 2sin(θ) + tan(θ)/3, where θ = π/(180 · 10ⁿ). It delivers 4n + 7 correct decimal places with a guaranteed minimum of 4n + 3. I tested it to 2 million digits without encountering an edge case.
Eric Yaw (Thu,) studied this question.