HP 17B Solver - another ARCTAN(Y/X) approximation

[Albert Chan]

(05-31-2021 09:51 PM)Albert Chan Wrote:  \(\displaystyle\arctan(x) = 2\arctan\left( {x \over \sqrt{1+x^2}+1} \right)\)

Another proof, using atanh

Let y = (1+x)/(1-x), we have x = (y-1)/(y+1)

atanh(x) = ln(y)/2       // definition
= ln(√y)
= 2 * atanh((√y-1)/(√y+1))

XCAS> assume(-1 < x < 1)
XCAS> simplify(subst((√y-1)/(√y+1), y = (1+x)/(1-x)))

(-(sqrt(-x^2+1)) + 1) / x

Multiply by conjugate, we have:

atanh(x) = 2 * atanh(x/(√(1-x^2)+1))
atanh(x*i) = 2 * atanh(x/(√(1+x^2)+1)*i)

Divide both side by i, we have atan(x) = 2 * atan(x/(√(1+x^2)+1))

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Just for fun, I try same way, but with ln(y) = 3 * ln(³√y).
Convert back to atan, I got:

atan(x) = 3 * atan(x/(2*cos(atan(x)/3)*√(1+x^2)+1))

Formula is useless, because reduced argument also call atan(x)