Inverse Hyperbolic Trigonometric Functions
For background on inverse functions, refer to a previous post on the inverse standard trig functions. Looking at the graphs of the hyperbolic trig functions in the previous post, we can see that sinh, tanh, csch, and coth are one-to-one, and therefore have a well-defined inverse. By contrast, cosh and sech are not one-to-one, so we will have to restrict their domain. Since we have exact formulas for the hyperbolic trig functions in terms of ex, we will also get exact formulas for their inverse functions in terms of lnx.
Derivations
Hyperbolic Sine
The standard method for solving for inverse functions is to swap x and y and then isolate for y.
y=sinh(x)=ex−e−x2⟹x=ey−e−y2Now, we are going to isolate ey, which will then let us isolate for y. Multiply both sides by 2ey.
2xey=e2y−1⟹(ey)2−2xey−1=0Use the quadratic formula.
ey=2x±√4x2+42⟹ey=x±√x2+1Since x<√x2+1, the root x−√x2+1 will result in a negative number and thus no solution for y.
ey=x+√x2+1⟹y=arcsinh(x)=ln(x+√x2+1)Hyperbolic Cosine
The steps are almost identical to arcsinh. I will skip a few steps to save space.
y=cosh(x)=ex+e−x2x=ey+e−y2(ey)2−2xey+1=0ey=x±√x2−1ey=x+√x2−1y=arccosh(x)=ln(x+√x2−1)One thing to note is that the reason we neglect the root ey=x−√x2−1 is different than before. x>√x2−1, therefore the root will be positive and produce a solution for y. Our choice of using the root ey=x+√x2−1 is due to our choice of domain restriction.
Hyperbolic Tangent
The methodology is very similar to the above. Again, I will skip some steps to save space.
y=tanh(x)=ex−e−xex+e−xx=ey−e−yey+e−y=e2y−1e2y+1e2y=1+x1−xy=arctanh(x)=12ln(1+x1−x)Hyperbolic Secant
This is almost identical to the derivation of arccosh.
y=sech(x)=2ex+e−xx=2ex+e−x(ey)2−2xey+1=0ey=1x±√1x2−1ey=1x+√1x2−1y=arcsech(x)=ln(1x+√1x2−1 )Hyperbolic Cosecant
This is almost identical to the derivation of arcsinh.
y=csch(x)=2ex−e−xx=2ex−e−x(ey)2+2xey+1=0ey=1x±√1x2+1ey=1x+√1x2+1y=arccsch(x)=ln(1x+√1x2+1 )Hyperbolic Cotangent
This is almost identical to the derivation of arctanh.
y=coth(x)=ex+e−xex−e−xx=ey+e−yey−e−y=e2y+1e2y−1e2y=x+1x−1y=arccoth(x)=12ln(x+1x−1)