Arc SOHCAHTOA method

[Albert Chan]

(04-02-2022 12:10 AM)Albert Chan Wrote:  TOA suggested asinhq(x) = atanhq(x/(1+x))

Lets proof this by getting tanh(y = asinh(sqrt(x))) ...

Here is another way, by comparing slopes.

XCas> s1 := factor(diff(asinh(sqrt(x))))                  → 1/(sqrt(x)*sqrt(1+x)*2)
XCas> s2 := factor(diff(atanh(sqrt(x)/sqrt(1+x))))  → 1/(sqrt(x)*sqrt(1+x)*2)

Slope matches. Now check domain.
This is important, see blackpenredpen video, THE CONFUSING DERIVATIVES

Domain: for x = 0 to inf, we have x/(1+x) = 1 - 1/(1+x) = 0 to 1. Domain OK

asinhq(x = 0) = asinh(0) = 0
atanhq(0/(1+0)) = atanh(0) = 0

2 sides slopes have same anti-derivative, and matching constant of integration.      QED

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Even simpler, use arc-trig identity, asinq(x) = atanq(x/(1-x))

asinhq(x) = asinq(-x) / i = atanq(-x/(1+x)) / i = atanhq(x/(1+x))      QED

Update:

Just realized OP arc-trig identity was stated without proof, but its is trivial.
With a right triangle, O² + A² = H² --> O=√x, H=1, A=√(1-x)

θ = asin(√x) = acos(√(1-x)) = atan(√(x/(1-x)))      QED

Trivia: acosq ↔ acoshq formulas, argument have same pattern.

acosq(x) = asinq(1-x) = asinhq(x-1)/i = acoshq(x)/i

--> acos(z) = acosh(z)/i = pi/2 - asin(z)