Hyperbolic Functions

sinh, cosh and tanh #

The exponential, split into two halves.

The hyperbolic functions are the even and odd halves of the exponential

Average $e^x$eˣ with $e^{-x}$e⁻ˣ and you get cosh; take half their difference and you get sinh.

Added together they give $e^x$eˣ back — that is what makes each of them a half of it.

cosh never drops below 1, with its minimum at x = 0. sinh is increasing and passes through the origin.

tanh is sinh over cosh. It rises from −1 toward 1 and reaches neither.

At zero both exponentials are 1, so cosh starts at 1 and the other two at 0.

The Identity cosh²x − sinh²x = 1 #

A hyperbola where Pythagoras gave a circle.

Squaring the two definitions and subtracting leaves exactly one

Square each definition. The outer terms match; only the middle term differs.

Subtract and the outer terms cancel, leaving 4 over 4. So ${\cosh }^{2}x-{\sinh }^{2}x=1$cosh²x − sinh²x = 1.

Compare: ${\cos }^2\theta +{\sin }^2\theta =1$cos²θ + sin²θ = 1 puts the point $\left(\cos \theta ,\sin \theta \right)$(cos θ, sin θ) on a unit circle.

One sign changes, and (cosh t, sinh t) lies on a hyperbola. That is where the name comes from.

Divide the identity through by ${\cosh }^{2}x$cosh²x and a second identity appears, for tanh.

Differentiating sinh x and cosh x #

Each becomes the other, and no sign is lost.

The hyperbolic pair differentiate into each other with no change of sign

Differentiate the definition of sinh. The minus on $e^{-x}$e⁻ˣ becomes a plus, and cosh appears.

Differentiate cosh the same way. Unlike cosine, no minus sign appears at all.

tanh needs the quotient rule, and the identity turns the numerator into a plain 1.

Read both backwards to get the integrals, and again no minus sign appears.

Inverse Hyperbolic Functions #

Reflect in y = x, once the domain allows it.

A hyperbolic function can be undone only where it takes each value once

sinh always increases and repeats no value, so every output comes from one input.

Reflect it in the line y = x and you have arsinh, defined for every real number.

cosh reaches each height above 1 twice, so reflecting it gives two answers for one input.

Keep only $x\ge 0$x ≥ 0 and the reflection works: arcosh starts at x = 1 and is never negative.

tanh never reaches 1 or −1, so artanh accepts only inputs strictly between them.

Inverse Hyperbolics as Logarithms #

Solve for the input and a logarithm appears.

Solving the exponential definition for its input turns an inverse hyperbolic into a logarithm

Write out what arsinh means: x is sinh of y, an equation in e to the y.

Multiply every term by e to the y and the negative powers disappear.

What is left is a quadratic in e to the y, so the quadratic formula solves it.

One of the two roots is negative, and $e^y$eʸ is never negative, so that root is rejected.

Take logarithms of both sides. The other two inverses are found the same way.

Check it at zero: the formula gives ln 1, which is 0, and sinh 0 is 0 as well.