Arc SOHCAHTOA method

[Albert Chan]

For fun, I try the same idea to arc hyperbolics.

asinhq(x) = asinh(√x) = ln(√(x) + √(x+1))
acoshq(x) = acosh(√x) = ln(√(x) + √(x-1))
atanhq(x) = atanh(√x) = 1/2 * ln((1+√x) / (1-√x))

→ asinhq(x) = acoshq(x+1)      // O = x, A = 1+x, H = 1

This suggested for hyperbolics, A = O + H
TOA suggested asinhq(x) = atanhq(x/(1+x))

Lets proof this by getting tanh(y = asinh(sqrt(x)))
To simplify away the exponential, let z = e^y

tanh(y) = (z-1/z) / (z+1/z) = (z^2-1) / (z^2+1)

z = √(x) + √(x+1)      // Goal, remove all square roots.
z - √(x) = √(x+1)
z^2 - 2*√(x)*z + x = x + 1
(z^2 - 1) = 2*√(x)*z
(z^2-1)^2 = 4*x*z^2

If atanhq() expression is correct, we should have (z^2-1)/(z^2+1) = √(x/(1+x))
Or, (z^2-1)^2 * (1+x) = (z^2+1)^2 * x

(z^2-1)^2 * (1+x) - (z^2+1)^2 * x
= 1 - 2*z^2 + z^4 - 4*x*z^2
= (z^2-1)^2 - 4*x*z^2
= 0      QED

Relative to atanhq(x), we have O = x, A = 1, H = A-O = 1-x

                        SOH                        CAH

atanhq(x) = asinhq(x/(1-x)) = acoshq(1/(1-x))


Update: I was being stupid, and should prove with more familiar sinh/cosh
Or, simply work with sin/cos, then flip sign using Osborn's rule

y = asinhq(x)  → sinh(y)² = x  → cosh(y)² = 1+x  → tanh(y)² = x/(1+x)  → y = atanhq(x/(1+x))