Hyperbolic Functions

sinh, cosh and tanh #

The exponential, split into two halves.

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

Read the full lesson →

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

Read the full lesson →

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

Read the full lesson →

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

Read the full lesson →

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

Read the full lesson →

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.