(HP-67) Barkers's Equation

[Albert Chan]

(12-06-2019 06:39 PM)Albert Chan Wrote:  y = ³√(W + sign(W) √(W²+1)) = ³√(cot(acot(W)/2))

Prove: let θ = acot(W), c = cot(θ/2)

Half angle formula: cot(θ) = W = (c²-1) / (2c)

c² - 2W c - 1 = 0   → c = W ± √(W²+1)

Since |θ| < pi/2, c has same sign of θ, which has same sign of W (assumed sign(0)=1)
However, 2 roots of c have opposite sign. Matching c and W signs, we have:

c = cot(acot(W)/2) = W + sign(W) √(W²+1)