Hyperbolic Functions: The Other Trigonometry

Learning Objectives

By the end of this section you will be able to:

1. Define $\sinh, \cosh, \tanh$ from the exponential function and explain why those particular combinations matter.
2. Prove the central identity $\cosh^2 x - \sinh^2 x = 1$ and see why it forces the curve onto a hyperbola.
3. Read the graphs of hyperbolic functions — their symmetries, asymptotes, and growth rates.
4. Connect hyperbolic functions to real phenomena: the catenary, special relativity, and the $\tanh$ activation in neural networks.
5. Compute sinh from its Taylor series and use PyTorch's $\tanh$ in a forward pass.

The Story: A Chain, a Sail, a Surprise

Hang a uniform chain between two poles and step back. What shape does it make? Galileo, in 1638, declared it a parabola. He was wrong — but the error was so subtle that the correct answer took another fifty-three years to find. In 1691 Jakob Bernoulli posed the problem as a public challenge; within a year his brother Johann, together with Huygens and Leibniz, produced the answer:

That “cosh” is the hyperbolic cosine. It is not a trigonometric function at all — it is built directly from $e^x$. The same family of functions describes the shape of a sail filled with wind, the velocity of a falling object with air drag, the rapidity of a relativistic particle, and the activation curve of a neuron in a deep network. They show up everywhere there is an underlying balance between exponential growth and decay.

The whole idea in one line: hyperbolic functions are the symmetric and antisymmetric parts of $e^x$. Every fact about them follows from that one decomposition.

Definitions from the Exponential

Any function $f(x)$ can be uniquely split into an even part $\tfrac{1}{2}(f(x) + f(-x))$ and an odd part $\tfrac{1}{2}(f(x) - f(-x))$. Apply this to $f(x) = e^x$. The two pieces are so important that they get their own names:

These are not arbitrary combinations. They are the unique way to decompose $e^x$ into an odd piece (sinh) and an even piece (cosh). Adding them puts the exponential back together:

We also define cosecant, secant, cotangent versions for completeness:

Graphs and Behavior

Drag the slider below and watch the three core curves move together. Hide and show $e^x$ and $e^{-x}$ on the same axes — cosh sits at their midpoint, sinh sits at their half-difference. That single visual is most of the intuition.

Three behaviors to commit to memory:

Two things to notice. First, $\cosh x \geq 1$ everywhere — the chain never sags below its lowest point. Second, $\tanh x$ is bounded between $-1$ and $+1$; it is the squashing function that maps the whole real line into the open interval $(-1, 1)$. That single property is why neural networks use it.

The Fundamental Identity

Every property of sinh and cosh flows from one equation:

It is the hyperbolic mirror of $\cos^2 x + \sin^2 x = 1$. Notice the minus sign — that is the only difference, and it changes the geometry from a circle to a hyperbola. Let's prove it in three lines from the definitions:

Confirm it numerically with the slider in the explorer above. The cell labeled cosh²−sinh² always reads $1.000000$, no matter how you drag $x$.

Why “Hyperbolic”? Circle vs Hyperbola

The name comes from a perfect geometric parallel. Recall:

1. The point $(\cos t, \sin t)$ traces the unit circle $x^2 + y^2 = 1$. And $t$ equals twice the area of the circular sector swept from the positive $x$-axis to that point.
2. The point $(\cosh t, \sinh t)$ traces the unit hyperbola $x^2 - y^2 = 1$. And $t$ equals twice the area of the hyperbolic sector swept from $(1,0)$ to that point.

That is the entire reason for the name. In trigonometry, the parameter is an angle. In hyperbolic trigonometry, the parameter is a hyperbolic angle — a swept area, not an angle in radians.

Analogy clincher. The circle and the hyperbola differ by a single sign: $x^2 \, \textbf{+} \, y^2 = 1$ vs $x^2 \, \textbf{-} \, y^2 = 1$. Every identity, derivative, and series for hyperbolic functions can be obtained from its circular cousin by carefully flipping one sign.

Algebraic Identities

Compare these side-by-side with their circular relatives. Notice how often a sign quietly flips:

These are not coincidence. They drop out of the definitions in one line. For example, expand $\sinh(a+b)$:

The last step is just regrouping $e^a e^b - e^{-a} e^{-b}$ into the two pieces $\sinh a \, \cosh b$ and $\cosh a \, \sinh b$. Try it on paper — it is two lines of algebra.

Inverse Hyperbolic Functions

Because $\sinh$ and $\tanh$ are strictly increasing, they have proper inverses on all of $\mathbb{R}$. $\cosh$ is not one-to-one (it's even), so we restrict to $x \geq 0$ to invert it. The beautiful surprise: all three inverses have closed-form logarithm expressions.

Let's derive $\operatorname{arcsinh}$ to show how clean the algebra is. Set $y = \sinh x = \tfrac{e^x - e^{-x}}{2}$, then solve for $x$. Multiply through by $2 e^x$:

That is the quadratic formula applied to a hidden quadratic in $e^x$. The plus root is taken because $e^x > 0$ rules out the minus root.

The Catenary — Where cosh Really Lives

Now we earn the story we opened with. A flexible chain of uniform mass per unit length, hanging under gravity, satisfies a balance equation between tension and weight. Without doing the derivation (we'll do it carefully in Chapter 22), the resulting shape is:

The constant $a$ is the ratio of horizontal tension to weight per unit length — small $a$ means a tight, heavy chain (sharp dip); large $a$ means a light, taut chain (gentle curve). Drag the sliders below to feel it.

Famous catenaries in the real world:

• The Gateway Arch in St. Louis is an inverted weighted catenary — designed precisely so that every part of the arch is in pure compression.
• Suspension bridge cables hang very close to a parabola because the deck's weight dominates, not the cable weight. The pure catenary is the answer only when the chain itself is the load.
• The shape of an idealized hanging soap film between two rings is a catenoid, the surface of revolution of a catenary.

Worked Example (Collapsible)

Work through this by hand before peeking. The example threads almost every idea in this section through a single computation.

Plain Python: Building sinh from Scratch

Before reaching for a library, let's build $\sinh$ ourselves from its Taylor series. Recall:

Only the odd powers survive, because $\sinh$ is the odd part of $e^x$ (whose full Taylor series is $\sum x^n / n!$; even terms cancel when we subtract $e^{-x}$). We'll watch the partial sum converge to $\sinh 1$ term by term.

1import math
2
3# Goal: compute sinh(1) by summing the Taylor series term by term.
4# sinh(x) = x + x^3/3! + x^5/5! + x^7/7! + ...
5x = 1.0
6total = 0.0
7
8for n in range(8):
9    power = 2 * n + 1
10    term = x**power / math.factorial(power)
11    total += term
12    print(f"n={n}: term = x^{power}/{power}! = {term:.10f},  running total = {total:.10f}")
13
14print("\nlibrary sinh(1) =", math.sinh(1))
15print("our  approx     =", total)
16print("absolute error  =", abs(total - math.sinh(1)))

PyTorch: tanh as a Neural Activation

In deep learning, the tanh function is a squashing non-linearity: it takes any real number and maps it into the open interval $(-1, 1)$. It is symmetric about the origin (unlike sigmoid, which is centered at $0.5$), which makes gradient flow through deep networks behave better. Here is the smallest meaningful neural-network forward pass that uses it.

1import torch
2
3# A tiny "network": one linear layer followed by tanh.
4# Pre-activation:   z = W x + b
5# Activation:       a = tanh(z)
6torch.manual_seed(0)
7
8x = torch.tensor([1.0, -2.0, 0.5])      # input vector  (3 features)
9W = torch.tensor([[ 0.5, -0.3, 0.2],    # weight matrix (2 outputs, 3 inputs)
10                  [-0.1,  0.4, 0.8]])
11b = torch.tensor([0.1, -0.2])           # bias    (2 outputs)
12
13z = W @ x + b                            # linear pre-activation
14a = torch.tanh(z)                        # squash to (-1, 1)
15
16print("z =", z)
17print("a =", a)
18print("|a| < 1 everywhere ->", torch.all(a.abs() < 1).item())

Where Hyperbolic Functions Show Up

Five places, drawn from very different fields, where you cannot avoid $\sinh$, $\cosh$, or $\tanh$:

Take the falling-object example for a moment. With air drag proportional to $v^2$, Newton's second law $m\dot{v} = m g - k v^2$ has the closed-form solution $v(t) = v_T \tanh(g t / v_T)$. The terminal velocity $v_T = \sqrt{m g / k}$ is the height of the horizontal asymptote you saw in the explorer. Falling raindrops, skydivers, and ping-pong balls all trace out a $\tanh$ curve when plotted velocity vs. time. The same shape governs how a neuron saturates. Same math, different universe.

Summary

1. Definitions. $\sinh x = \tfrac{e^x - e^{-x}}{2}$, $\cosh x = \tfrac{e^x + e^{-x}}{2}$, $\tanh x = \sinh x / \cosh x$. They are the odd and even parts of $e^x$.
2. Identity. $\cosh^2 x - \sinh^2 x = 1$. This is what puts $(\cosh t, \sinh t)$ on the hyperbola $x^2 - y^2 = 1$ for every $t$.
3. Geometry. Trigonometric functions parametrize the unit circle; hyperbolic functions parametrize the unit hyperbola. In both cases the parameter $t$ is twice the swept area, not the arc length.
4. Inverses. Every inverse has a closed-form log expression, e.g. $\operatorname{arctanh} x = \tfrac{1}{2}\ln\!\tfrac{1+x}{1-x}$.
5. Applications. The catenary $y = a\cosh(x/a)$, relativistic rapidity, terminal velocity, and the $\tanh$ neural activation all share the same underlying object.

Exercises

1. Quick definitions. Without a calculator, compute $\sinh(\ln 2)$ and $\cosh(\ln 2)$ in closed form. Hint: $e^{\ln 2} = 2$.
2. Identity hunt. Starting from the definitions, prove the double-angle formula $\cosh(2x) = 2\cosh^2 x - 1$. Then write $\sinh^2 x$ in terms of $\cosh(2x)$.
3. Inverse derivation. Mimic the $\operatorname{arcsinh}$ derivation in this section to derive the closed form for $\operatorname{arctanh} x$. You will end up solving a linear equation in $e^{2x}$, not a quadratic.
4. Catenary fit. A telephone wire hangs between two poles 60 m apart, dipping 5 m below the attachment points. Find $a$ so that $y(x) = a\cosh(x/a)$ matches. Solve numerically — there is no closed form for $a$.
5. Coding exercise. Extend the plain-Python series in this section so it computes $\cosh$ from its Taylor series (only even powers!). Verify your answer against $\cosh(2)$ from the worked example.
6. Neural net check. In the PyTorch snippet, replace $\tanh$ with the sigmoid $\sigma(x) = 1/(1 + e^{-x})$. By hand, predict the new outputs using $\tanh(x) = 2\sigma(2x) - 1$, then run the code to confirm.