Hyperbolic Function Definitions

MATH162

Reference: Stewart 6.7 • Chapter: 6 • Section: 7

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Hyperbolic Function Definitions

Why Study Hyperbolic Functions?

When a heavy cable hangs freely between two supports, it doesn't form a parabola—it forms a shape called a catenary, described by the hyperbolic cosine function. This same function appears in the Gateway Arch in St. Louis, in the cables of suspension bridges, and in the analysis of ocean waves.

Hyperbolic functions arise naturally whenever exponential growth and decay combine in symmetric ways. They're called "hyperbolic" because they relate to the hyperbola $x^2 - y^2 = 1$ the same way trigonometric functions relate to the circle $x^2 + y^2 = 1$.

Before You Start: Quick Self-Check

Can you answer these questions? If not, review the linked prerequisites first.

1. What is $e^{\ln 5}$?

Answer: $e^{\ln 5} = 5$. If this wasn't immediate, review Exponential Functions.

2. Is $\sin(-x) = \sin x$ or $\sin(-x) = -\sin x$?

Answer: $\sin(-x) = -\sin x$ (sine is odd). If this wasn't clear, review Trigonometric Functions.

Prerequisite Map

Quick Reference

Property	Value
Concept	Hyperbolic Functions
Chapter	6.7
Difficulty	Beginner
Time	~15 minutes

Key Concepts

The Basic Definitions

The six hyperbolic functions are defined in terms of exponentials:

$$\boxed{\sinh x = \frac{e^x - e^{-x}}{2} \qquad \cosh x = \frac{e^x + e^{-x}}{2}}$$

$$\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$

The reciprocal functions are:

$$\text{csch } x = \frac{1}{\sinh x} \qquad \text{sech } x = \frac{1}{\cosh x} \qquad \coth x = \frac{\cosh x}{\sinh x}$$

Understanding the Definitions

Think of $\sinh x$ and $\cosh x$ as the "difference" and "sum" of exponential growth and decay:

$\sinh x$ measures how much $e^x$ exceeds $e^{-x}$ (divided by 2)
$\cosh x$ measures the average of $e^x$ and $e^{-x}$

Notice that $\cosh x + \sinh x = e^x$ and $\cosh x - \sinh x = e^{-x}$—these are useful facts for simplification.

Graphs and Behavior

     y                              y
     |     /                        |    \     /
     |    /                         |     \   /
     |   /                          |      \_/ (1,0)
-----+--/-------- x          -------+--------- x
    /  |                            |
   /   |                            |
  /    | y = sinh x                 | y = cosh x

Key observations:

Function	Domain	Range	Even/Odd	Key Feature
$\sinh x$	$\mathbb{R}$	$\mathbb{R}$	Odd	Passes through origin
$\cosh x$	$\mathbb{R}$	$[1, \infty)$	Even	Minimum value of 1 at $x=0$
$\tanh x$	$\mathbb{R}$	$(-1, 1)$	Odd	Horizontal asymptotes at $y = \pm 1$

Why "Hyperbolic"?

For trig functions: if $x = \cos t$ and $y = \sin t$, then $x^2 + y^2 = 1$ (the unit circle).

For hyperbolic functions: if $x = \cosh t$ and $y = \sinh t$, then $x^2 - y^2 = 1$ (the right branch of the unit hyperbola).

This geometric connection gives these functions their name.

Connection to Trig Functions

Trig Function	Hyperbolic Analog
$\sin x$	$\sinh x$
$\cos x$	$\cosh x$
$\tan x$	$\tanh x$
$\csc x$	$\text{csch } x$
$\sec x$	$\text{sech } x$
$\cot x$	$\coth x$

The parallels are striking, but there are important differences in signs (which we'll see in identities and derivatives).

Practice Problems

Level 1 Evaluating Hyperbolic Functions

Find the exact values of $\sinh(\ln 2)$ and $\cosh(\ln 2)$.

Thought Process

Use the definitions with $x = \ln 2$. The key is recognizing that $e^{\ln 2} = 2$ and $e^{-\ln 2} = \frac{1}{e^{\ln 2}} = \frac{1}{2}$.

Show Answer

$$\sinh(\ln 2) = \frac{e^{\ln 2} - e^{-\ln 2}}{2} = \frac{2 - \frac{1}{2}}{2} = \frac{\frac{3}{2}}{2} = \frac{3}{4}$$

$$\cosh(\ln 2) = \frac{e^{\ln 2} + e^{-\ln 2}}{2} = \frac{2 + \frac{1}{2}}{2} = \frac{\frac{5}{2}}{2} = \frac{5}{4}$$

Quick check: $\cosh^2 - \sinh^2 = \frac{25}{16} - \frac{9}{16} = \frac{16}{16} = 1$ ✓

Level 2 Converting to Exponentials

Write $3\sinh x + 4\cosh x$ in terms of $e^x$ and $e^{-x}$.

Thought Process

Replace each hyperbolic function with its exponential definition, then combine like terms. Group the $e^x$ terms together and the $e^{-x}$ terms together.

Show Answer

$$3\sinh x + 4\cosh x = 3 \cdot \frac{e^x - e^{-x}}{2} + 4 \cdot \frac{e^x + e^{-x}}{2}$$

$$= \frac{3e^x - 3e^{-x} + 4e^x + 4e^{-x}}{2}$$

$$= \frac{7e^x + e^{-x}}{2}$$

Level 3 Combining Hyperbolic Values

Compute $\sinh(\ln 4) + \cosh(\ln 4)$ and simplify your answer to a single number.

Thought Process

You could compute each separately using the definitions, but recall the identity $\sinh x + \cosh x = e^x$. This provides a much faster path to the answer.

Show Answer

Quick method using the identity:

$$\sinh(\ln 4) + \cosh(\ln 4) = e^{\ln 4} = 4$$

Verification by direct computation:

$$\sinh(\ln 4) = \frac{4 - \frac{1}{4}}{2} = \frac{15}{8}$$

$$\cosh(\ln 4) = \frac{4 + \frac{1}{4}}{2} = \frac{17}{8}$$

$$\sinh(\ln 4) + \cosh(\ln 4) = \frac{15}{8} + \frac{17}{8} = \frac{32}{8} = 4$$ ✓

Level 4 Hyperbolic Functions of Logarithmic Arguments

Express $\tanh(\ln x)$ as a rational function of $x$ (where $x > 0$).

Thought Process

Write $\tanh(\ln x) = \frac{\sinh(\ln x)}{\cosh(\ln x)}$. Compute each using the exponential definitions with $e^{\ln x} = x$ and $e^{-\ln x} = 1/x$. Then simplify the ratio.

Show Answer

First, compute the numerator and denominator:

$$\sinh(\ln x) = \frac{e^{\ln x} - e^{-\ln x}}{2} = \frac{x - \frac{1}{x}}{2} = \frac{x^2 - 1}{2x}$$

$$\cosh(\ln x) = \frac{e^{\ln x} + e^{-\ln x}}{2} = \frac{x + \frac{1}{x}}{2} = \frac{x^2 + 1}{2x}$$

Therefore:

$$\tanh(\ln x) = \frac{\sinh(\ln x)}{\cosh(\ln x)} = \frac{\frac{x^2-1}{2x}}{\frac{x^2+1}{2x}} = \frac{x^2 - 1}{x^2 + 1}$$

Physical note: As $x \to \infty$, $\tanh(\ln x) \to 1$, consistent with $\tanh$ approaching 1 for large arguments.

Level 5 Rewriting Exponential Combinations

The expression $5e^x + 3e^{-x}$ can be written in the form $\alpha \cosh(x + \beta)$ for appropriate constants $\alpha > 0$ and $\beta$.

(a) Find the exact values of $\alpha$ and $\beta$.

(b) Verify your answer by expanding $\alpha \cosh(x + \beta)$ and showing it equals $5e^x + 3e^{-x}$.

Thought Process

Since both coefficients (5 and 3) are positive, the expression can be written as a shifted $\cosh$. Use the addition formula $\cosh(x + \beta) = \cosh x \cosh \beta + \sinh x \sinh \beta$, then convert to exponentials.

For $\alpha \cosh(x + \beta)$, the coefficient of $e^x$ is $\frac{\alpha e^\beta}{2}$ and the coefficient of $e^{-x}$ is $\frac{\alpha e^{-\beta}}{2}$.

Set these equal to 5 and 3 respectively, then solve the system.

Show Answer

(a) Finding $\alpha$ and $\beta$:

From $\alpha \cosh(x + \beta) = \frac{\alpha}{2}[e^{\beta}e^x + e^{-\beta}e^{-x}]$, we need:

$$\frac{\alpha e^\beta}{2} = 5 \quad \text{and} \quad \frac{\alpha e^{-\beta}}{2} = 3$$

Multiplying these equations: $\frac{\alpha^2}{4} = 15$, so $\alpha = 2\sqrt{15}$.

Dividing: $e^{2\beta} = \frac{5}{3}$, so $\beta = \frac{1}{2}\ln\left(\frac{5}{3}\right)$.

(b) Verification:

$$\alpha \cosh(x + \beta) = 2\sqrt{15} \cdot \frac{e^{x+\beta} + e^{-(x+\beta)}}{2}$$

$$= \sqrt{15}[e^x \cdot e^\beta + e^{-x} \cdot e^{-\beta}]$$

With $e^\beta = \sqrt{5/3}$ and $e^{-\beta} = \sqrt{3/5}$:

$$= \sqrt{15} \cdot e^x \cdot \sqrt{\frac{5}{3}} + \sqrt{15} \cdot e^{-x} \cdot \sqrt{\frac{3}{5}}$$

$$= e^x \sqrt{15 \cdot \frac{5}{3}} + e^{-x}\sqrt{15 \cdot \frac{3}{5}} = e^x \sqrt{25} + e^{-x}\sqrt{9}$$

$$= 5e^x + 3e^{-x}$$ ✓

CCI-Style Conceptual Questions

Question 1: Which statement correctly describes the relationship between $\sinh x$ and $\cosh x$?

(A) $\sinh x + \cosh x = 1$ (B) $\sinh x + \cosh x = e^x$ (C) $\sinh x \cdot \cosh x = 1$ (D) $\sinh x / \cosh x = 1$

Answer

(B) Adding the definitions: $\frac{e^x - e^{-x}}{2} + \frac{e^x + e^{-x}}{2} = \frac{2e^x}{2} = e^x$

Question 2: The minimum value of $\cosh x$ is:

(A) 0 (B) 1 (C) $e$ (D) There is no minimum

Answer

(B) The function $\cosh x$ achieves its minimum at $x = 0$, where $\cosh 0 = 1$. Since $\cosh x$ is even and increases as $\vert x\vert $ increases, 1 is the global minimum.

Question 3 (Explain Why): A student claims that $\sinh x$ can never equal $\cosh x$. Is this correct? Explain your reasoning.

Answer

Incorrect. We have $\sinh x = \cosh x$ when $\frac{e^x - e^{-x}}{2} = \frac{e^x + e^{-x}}{2}$, which simplifies to $e^{-x} = 0$. Since $e^{-x} > 0$ for all real $x$, there is no solution for finite $x$.

However, as $x \to +\infty$, both functions approach $\frac{e^x}{2}$, so they become arbitrarily close. The student's claim is technically correct for finite values, but the reasoning should reference that $e^{-x} \neq 0$ rather than making an unfounded assertion.

Mastery Checklist

[ ] I can write $\sinh x$ and $\cosh x$ in terms of exponentials
[ ] I can evaluate hyperbolic functions at specific values (including $0$ and $\ln a$)
[ ] I can convert expressions between exponential and hyperbolic form
[ ] I understand why $\cosh x \geq 1$ for all $x$
[ ] I can identify which hyperbolic functions are even vs. odd
[ ] I understand the connection between hyperbolic functions and the hyperbola $x^2 - y^2 = 1$

Mental Model

The Sum-Difference Pair:

Think of $\cosh x$ and $\sinh x$ as the "average" and "half-difference" of exponential growth ($e^x$) and decay ($e^{-x}$). Just as the average of two numbers is always at least as large as their difference, $\cosh x \geq \vert \sinh x\vert $ always.

When $x$ is large and positive, $e^{-x} \approx 0$, so both functions behave like $\frac{1}{2}e^x$.

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Last updated: 2026-01-22