Hyperbolic Functions (Extended)

flowchart TD
    A[Hyperbolic Functions] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

This topic explores fundamental concepts that shape our understanding of the world.

This document provides a rigorous treatment of hyperbolic functions, their identities, inverses, and Calculus.

• \sinh 0 = 0$$\cosh 0 = 1$$\tanh 0 = 0.
• $\sinh x$ is odd: $\sinh(-x) = -\sinh x$.
• $\cosh x$ is even: $\cosh(-x) = \cosh x$.
• $\tanh x$ is odd.
• As $x \to +\infty$: \sinh x \to +\infty$$\cosh x \to +\infty$$\tanh x \to 1.
• As $x \to -\infty$: \sinh x \to -\infty$$\cosh x \to +\infty$$\tanh x \to -1.

Proof. $\cos(ix) = \dfrac{e^{i(ix)} + e^{-i(ix)}}{2} = \dfrac{e^{-x} + e^{x}}{2} = \cosh x$.

$-i\sin(ix) = -i \cdot \dfrac{e^{i(ix)} - e^{-i(ix)}}{2i} = \dfrac{e^{-x} - e^{x}}{-2} = \dfrac{e^x - e^{-x}}{2} = \sinh x$. $\blacksquare$

• $\sinh x$: passes through the origin, increasing, resembles $y = x/2$ near the origin and $y = e^x/2$ for large positive $x$.
• $\cosh x$: minimum at $(0, 1)$ Symmetric about the $y$-axis, resembles $y = 1 + x^2/2$ near the origin and $y = e^x/2$ for large positive $x$.
• $\tanh x$: S-shaped curve with horizontal asymptotes at $y = \pm 1$ Passing through the origin with gradient 1.

Proof.

Proof. Divide the fundamental identity by $\cosh^2 x$:

Every trigonometric identity has a corresponding hyperbolic identity, obtained by:

• Replacing $\cos$ with $\cosh$.
• Replacing $\sin$ with $\sinh$.
• Negating every product (or power of 2 or more) of $\sinh$.

Example: $\cos^2 x + \sin^2 x = 1$ becomes $\cosh^2 x - \sinh^2 x = 1$ (the $\sinh^2$ term is Negated).

Example: $\sin 2x = 2\sin x\cos x$ becomes $\sinh 2x = 2\sinh x\cosh x$ (the product $\sinh x \cosh x$ is negated since it contains $\sinh$ But $\sinh 2x$ on the left is not a product, So the overall sign changes on both sides cancel).

Proof of $\cosh 2x = 2\cosh^2 x - 1$:

Proof of the addition formula for $\sinh$:

Problem. Given $\sinh x = 3$ Find $\cosh x$ and $\tanh x$ without finding $x$.

From $\cosh^2 x - \sinh^2 x = 1$:

(We take the positive root since $\cosh x \geq 1$ for all $x$.)

Let $y = \operatorname{arsinh}\,x$ So $x = \sinh y = \dfrac{e^y - e^{-y}}{2}$.

This is a quadratic in $e^y$:

Since $e^y \gt 0$ and $\sqrt{x^2 + 1} \gt |x|$ We take the positive root:

Let $y = \operatorname{arcosh}\,x$ So $x = \cosh y = \dfrac{e^y + e^{-y}}{2}$.

Since $e^y \geq 1$ and $x \geq 1$ We need $e^y \geq 1$. Both roots are positive when $x \geq 1$. The convention is to take $e^y = x + \sqrt{x^2 - 1}$ (which gives $y \geq 0$):

Let $y = \operatorname{artanh}\,x$ So $x = \tanh y$.

Problem. Evaluate $\operatorname{arsinh}\,2$ and $\operatorname{artanh}\,\dfrac{1}{3}$ in exact Logarithmic form.

Proof of $\dfrac{d}{dx}\sinh x = \cosh x$:

Proof for $\operatorname{arsinh}\,x$. Let $y = \operatorname{arsinh}\,x$ So $x = \sinh y$.

Proof for $\operatorname{artanh}\,x$. Let $y = \operatorname{artanh}\,x$ So $x = \tanh y$.

The differentiation results give standard integrals:

Proof of the first formula. Let $u = x/a$:

Problem. Differentiate $f(x) = \sinh(3x^2)$.

Problem. Evaluate $\displaystyle\int \frac{1}{\sqrt{4x^2 + 9}}\,dx$.

Problem. Evaluate $\displaystyle\int_0^1 \frac{dx}{\sqrt{x^2 + 1}}$.

Problem. Evaluate $\displaystyle\int \sqrt{x^2 + 4}\,dx$.

Use the substitution $x = 2\sinh u$, $dx = 2\cosh u\,du$:

Using $\cosh^2 u = \dfrac{1 + \cosh 2u}{2}$:

Since $x = 2\sinh u$: $\sinh u = \dfrac{x}{2}$ $\cosh u = \sqrt{1 + \dfrac{x^2}{4}} = \dfrac{\sqrt{x^2 + 4}}{2}$ $u = \operatorname{arsinh}\!\left(\dfrac{x}{2}\right)$.

Solve the equation $4\sinh x - 3\cosh x = 0$.

Prove that $\sinh 3x = 3\sinh x + 4\sinh^3 x$.

Evaluate $\displaystyle\int_0^2 \frac{dx}{\sqrt{x^2 + 16}}$.

Differentiate $f(x) = x\,\operatorname{arcosh}\,x$ for $x \gt 1$.

Proof. Let $u = x/a$ So $dx = a\,du$:

Now let $u = \cosh t$ So $du = \sinh t\,dt$:

Proof. Let $u = x/a$ So $dx = a\,du$:

Theorem. A uniform chain hanging under gravity takes the shape $y = a\cosh\!\left(\dfrac{x}{a}\right) + c$.

Proof (sketch). Consider a small element of the chain between horizontal positions $x$ and $x + \delta x$. Let the tension at position $x$ be $T$ Making angle $\theta$ with the horizontal.

Horizontal equilibrium: $T\cos\theta = T_0$ (constant).

Vertical equilibrium: $\dfrac{d}{dx}(T\sin\theta) = w$ where $w$ is the weight per unit length.

Since $T = T_0\sec\theta$ and $T\sin\theta = T_0\tan\theta$:

Let $y' = \tan\theta$ So $\dfrac{dy'}{dx} = \sec^2\theta\,\dfrac{d\theta}{dx} = \dfrac{w}{T_0}$.

Integrating: $y' = \dfrac{w}{T_0}\,x + C_1$. Taking $C_1 = 0$ by symmetry:

Integrating again: $y = a\cosh\!\left(\dfrac{x}{a}\right) + C$. $\blacksquare$

(a) Solve the equation $\cosh x = 2.5$ Giving your answer in exact logarithmic form.

(b) Hence solve $\cosh 2x = 2.5$.

(a) Prove that $\dfrac{d}{dx}(\operatorname{arcosh}\,x) = \dfrac{1}{\sqrt{x^2 - 1}}$ for $x > 1$.

(b) Evaluate $\displaystyle\int_2^3 \frac{dx}{\sqrt{x^2 - 1}}$ in exact form.

A curve $C$ has equation $y = \sinh^{-1}(2x - 1)$.

(a) Find $\dfrac{dy}{dx}$.

(b) Find the equation of the tangent to $C$ at the point where $x = 1$.

Problem. Solve $3\sinh x + 4\cosh x = 5$ Giving your answer in exact logarithmic form.

Solution. Using the exponential definitions:

Multiplying by $e^x$: $7e^{2x} + 1 = 10e^x$

This is a quadratic in $e^x$:

Since $\dfrac{5 - 3\sqrt{2}}{7} \approx 0.109 > 0$ Both solutions are valid.

Problem. Evaluate $\displaystyle\int \sqrt{x^2 - 9}\,dx$ for $x \geq 3$.

Solution. Use the substitution $x = 3\cosh u$, $dx = 3\sinh u\,du$:

Using $\sinh^2 u = \dfrac{\cosh 2u - 1}{2}$:

Since $x = 3\cosh u$: $\cosh u = \dfrac{x}{3}$ $\sinh u = \sqrt{\dfrac{x^2}{9} - 1} = \dfrac{\sqrt{x^2 - 9}}{3}$.

$\sinh 2u = 2\sinh u\cosh u = \dfrac{2x\sqrt{x^2 - 9}}{9}$.

$u = \operatorname{arcosh}\!\left(\dfrac{x}{3}\right) = \ln\!\left(\dfrac{x}{3} + \sqrt{\dfrac{x^2}{9} - 1}\right)$.

Problem. Prove that $\cosh 3x = 4\cosh^3 x - 3\cosh x$.

Solution. From the trigonometric identity $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ Applying Osborne’s rule: since $\cos^3\theta$ contains no products of $\sin$ It remains unchanged. Therefore:

Direct proof. Starting from $\cosh 3x = \cosh(2x + x)$:

Problem. Find $\dfrac{dy}{dx}$ where $y = \dfrac{\sinh x}{1 + \cosh x}$ and simplify Your answer.

Solution. Using the quotient rule:

Using $\cosh^2 x - \sinh^2 x = 1$:

The final simplification uses the identity $1 + \cosh x = 2\cosh^2(x/2)$.

Problem. Evaluate $\displaystyle\int_0^1 \frac{dx}{\sqrt{1 + 4x^2}}$.

Solution. Write $\sqrt{1 + 4x^2} = \sqrt{4(x^2 + 1/4)} = 2\sqrt{x^2 + (1/2)^2}$.

Problem. A curve is given parametrically by $x = 2\cosh t$, $y = 3\sinh t$. Find the value of $\dfrac{dy}{dx}$ at $t = \ln 2$.

Solution. $\dfrac{dx}{dt} = 2\sinh t$, $\dfrac{dy}{dt} = 3\cosh t$.

At $t = \ln 2$: $\cosh(\ln 2) = \dfrac{2 + 1/2}{2} = \dfrac{5}{4}$ $\sinh(\ln 2) = \dfrac{2 - 1/2}{2} = \dfrac{3}{4}$.

Problem. Let $I_n = \displaystyle\int_0^{\pi/2} \cosh^n(\sin x)\,dx$ is not well-formed. Instead, verify that $\displaystyle\int \tanh^3 x\,dx = \ln(\cosh x) - \dfrac{1}{2}\mathrm{sech}^2\,x\tanh x + C$ is Incorrect, and find the correct integral.

Solution. Let us compute $\displaystyle\int \tanh^3 x\,dx$ correctly.

Write $\tanh^3 x = \tanh x \cdot \tanh^2 x = \tanh x(1 - \mathrm{sech}^2\,x) = \tanh x - \tanh x\,\mathrm{sech}^2\,x$.

The first integral: $\displaystyle\int\tanh x\,dx = \ln(\cosh x) + C$.

For the second integral, let $u = \tanh x$, $du = \mathrm{sech}^2\,x\,dx$:

Therefore:

Problem. Find the arc length of $y = a\cosh(x/a)$ from $x = 0$ to $x = b$.

Solution. $\dfrac{dy}{dx} = \sinh(x/a)$.

The differential equation $\dfrac{d^2y}{dx^2} - y = 0$ has general solution $y = A\cosh x + B\sinh x$ Which can also be written $y = Ce^x + De^{-x}$. See Differential Equations.

The identities $\cosh x = \cos(ix)$ and $\sinh x = -i\sin(ix)$ connect the two topics. See Complex Numbers.