Hyperbolic Functions

Intuition

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

Hyperbolic Functions

Hyperbolic functions are defined in terms of the exponential function and share remarkable Similarities with trigonometric functions. They arise in the solution of differential Equations, the description of hanging cables (catenary), special relativity, and many areas of Physics and engineering.

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Adjust the parameters in the graph above to explore the relationships between variables.

Board Coverage

1. Definitions

1.1 The three principal hyperbolic functions

Definition. The hyperbolic cosine, hyperbolic sine, and hyperbolic tangent are Defined by:

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

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

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

1.2 The reciprocal hyperbolic functions

Definition.

$\mathrm{sech}\,x = \frac{1}{\cosh x}, \qquad \mathrm{cosech}\,x = \frac{1}{\sinh x}, \qquad \coth\,x = \frac{\cosh x}{\sinh x}$

1.3 Domain and range

1.4 Key values

$\sinh 0 = 0, \quad \cosh 0 = 1, \quad \tanh 0 = 0$

$\sinh(-x) = -\sinh x \quad (\mathrm{odd function})$

$\cosh(-x) = \cosh x \quad (\mathrm{even function})$

2. Hyperbolic Identities

2.1 The fundamental identity

Theorem. For all $x \in \mathbb{R}$:

$\boxed{\cosh^2 x - \sinh^2 x = 1}$

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

$\cosh^2 x - \sinh^2 x = \left(\frac{e^x+e^{-x}}{2}\right)^2 - \left(\frac{e^x-e^{-x}}{2}\right)^2$

$= \frac{e^{2x}+2+e^{-2x}}{4} - \frac{e^{2x}-2+e^{-2x}}{4} = \frac{4}{4} = 1 \quad \blacksquare$

Corollary. Dividing by $\cosh^2 x$:

$\boxed{1 - \tanh^2 x = \mathrm{sech}^2\,x}$

Dividing by $\sinh^2 x$:

$\boxed{\coth^2 x - 1 = \mathrm{cosech}^2\,x}$

2.2 Addition formulae

Theorem.

$\boxed{\sinh(x+y) = \sinh x\cosh y + \cosh x\sinh y}$

$\boxed{\cosh(x+y) = \cosh x\cosh y + \sinh x\sinh y}$

Proof of the addition formula for $\sinh(x+y)$

$\sinh(x+y) = \frac{e^{x+y}-e^{-(x+y)}}{2} = \frac{e^x e^y - e^{-x}e^{-y}}{2}$

$= \frac{(e^x+e^{-x})(e^y-e^{-y}) + (e^x-e^{-x})(e^y+e^{-y})}{4}$

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

Wait — a cleaner approach:

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

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

$= \frac{2e^{x+y} - 2e^{-(x+y)}}{4} = \frac{e^{x+y}-e^{-(x+y)}}{2} = \sinh(x+y) \quad \blacksquare$

2.3 Double angle formulae

$\sinh 2x = 2\sinh x\cosh x$

$\cosh 2x = \cosh^2 x + \sinh^2 x = 2\cosh^2 x - 1 = 1 + 2\sinh^2 x$

2.4 Osborn’s rule

Definition. Osborn’s rule states that any trigonometric identity can be converted to the Corresponding hyperbolic identity by:

1. Replacing $\cos$ with $\cosh$ and $\sin$ with $\sinh$.
2. Changing the sign of every term that contains a product of two $\sinh$ factors.

This works because $\cos(ix) = \cosh x$ and $\sin(ix) = i\sinh x$So each $\sin$ introduces a Factor of $i$And $\sin^2$ introduces $i^2 = -1$.

Example. $\cos^2 x + \sin^2 x = 1 \xrightarrow{\mathrm{Osborn}} \cosh^2 x - \sinh^2 x = 1$. (The $\sinh^2$ term flips sign.)

Example. $\cos 2x = \cos^2 x - \sin^2 x \xrightarrow{\mathrm{Osborn}} \cosh 2x = \cosh^2 x + \sinh^2 x$. (The $\sinh^2$ term flips sign, turning $-$ into $+$.)

3. Inverse Hyperbolic Functions

3.1 Definitions and logarithmic forms

Definition. The inverse hyperbolic functions are:

$\boxed{\mathrm{arsinh}\,x = \ln\!\left(x + \sqrt{x^2+1}\right), \quad x \in \mathbb{R}}$

$\boxed{\mathrm{arcosh}\,x = \ln\!\left(x + \sqrt{x^2-1}\right), \quad x \geq 1}$

$\boxed{\mathrm{artanh}\,x = \frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right), \quad |x| < 1}$

Proof of $\mathrm{arsinh}\,x = \ln(x + \sqrt{x^2+1})$

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

$2x = e^y - e^{-y}$

Multiply by $e^y$: $2xe^y = e^{2y} - 1$.

$e^{2y} - 2xe^y - 1 = 0$

This is a quadratic in $e^y$:

$e^y = \frac{2x \pm \sqrt{4x^2+4}}{2} = x \pm \sqrt{x^2+1}$

Since $e^y > 0$ for all $y$We need $x + \sqrt{x^2+1} > 0$Which is always true. And $x - \sqrt{x^2+1} < 0$ for all $x$So we take the positive root:

$e^y = x + \sqrt{x^2+1}$

$y = \ln(x + \sqrt{x^2+1})$

Therefore $\mathrm{arsinh}\,x = \ln(x + \sqrt{x^2+1})$. $\blacksquare$

Proof of $\mathrm{arcosh}\,x = \ln(x + \sqrt{x^2-1})$

Let $y = \mathrm{arcosh}\,x$. Then $x = \cosh y = \dfrac{e^y + e^{-y}}{2}$ for $y \geq 0$.

$2x = e^y + e^{-y}$

$e^{2y} - 2xe^y + 1 = 0$

$e^y = \frac{2x \pm \sqrt{4x^2-4}}{2} = x \pm \sqrt{x^2-1}$

Since $e^y \geq 1$ and $y \geq 0$: we need $e^y \geq 1$. Both roots are positive (for $x \geq 1$), But $x + \sqrt{x^2-1} \geq 1$ and $x - \sqrt{x^2-1} \leq 1$. Since $\cosh$ is not one-to-one on all Of $\mathbb{R}$We restrict to $y \geq 0$Giving $e^y \geq 1$:

$e^y = x + \sqrt{x^2-1} \implies y = \ln(x + \sqrt{x^2-1})$

Therefore $\mathrm{arcosh}\,x = \ln(x + \sqrt{x^2-1})$ for $x \geq 1$. $\blacksquare$

Proof of $\mathrm{artanh}\,x = \frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right)$

Let $y = \mathrm{artanh}\,x$. Then $x = \tanh y = \dfrac{\sinh y}{\cosh y}$.

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

$x = \frac{\sinh y}{\cosh y} \implies \frac{\sinh y}{\cosh y} = x \implies \sinh y = x\cosh y$

$\cosh^2 y - x^2\cosh^2 y = 1 \implies \cosh^2 y = \frac{1}{1-x^2}$

Also $\tanh y = \dfrac{e^{2y}-1}{e^{2y}+1}$So:

$x = \frac{e^{2y}-1}{e^{2y}+1} \implies x(e^{2y}+1) = e^{2y}-1 \implies xe^{2y}+x = e^{2y}-1$

$e^{2y}(1-x) = 1+x \implies e^{2y} = \frac{1+x}{1-x}$

$2y = \ln\!\left(\frac{1+x}{1-x}\right) \implies y = \frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right) \quad \blacksquare$

4. Calculus with Hyperbolic Functions

4.1 Derivatives

$\boxed{\frac{d}{dx}\sinh x = \cosh x}$

$\boxed{\frac{d}{dx}\cosh x = \sinh x}$

$\boxed{\frac{d}{dx}\tanh x = \mathrm{sech}^2\,x}$

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

$\frac{d}{dx}\sinh x = \frac{d}{dx}\left(\frac{e^x - e^{-x}}{2}\right) = \frac{e^x + e^{-x}}{2} = \cosh x \quad \blacksquare$

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

$\frac{d}{dx}\cosh x = \frac{d}{dx}\left(\frac{e^x + e^{-x}}{2}\right) = \frac{e^x - e^{-x}}{2} = \sinh x \quad \blacksquare$

4.2 Derivatives of inverse hyperbolic functions

$\boxed{\frac{d}{dx}\mathrm{arsinh}\,x = \frac{1}{\sqrt{x^2+1}}}$

$\boxed{\frac{d}{dx}\mathrm{arcosh}\,x = \frac{1}{\sqrt{x^2-1}}, \quad x > 1}$

$\boxed{\frac{d}{dx}\mathrm{artanh}\,x = \frac{1}{1-x^2}, \quad |x| < 1}$

Proof of $\frac{d}{dx}\mathrm{arsinh}\,x = \frac{1}{\sqrt{x^2+1}}$

$\mathrm{arsinh}\,x = \ln(x + \sqrt{x^2+1})$.

$\frac{d}{dx}\ln(x+\sqrt{x^2+1}) = \frac{1}{x+\sqrt{x^2+1}}\cdot\left(1 + \frac{x}{\sqrt{x^2+1}}\right)$

$= \frac{1}{x+\sqrt{x^2+1}}\cdot\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}} = \frac{1}{\sqrt{x^2+1}} \quad \blacksquare$

Proof of $\frac{d}{dx}\mathrm{artanh}\,x = \frac{1}{1-x^2}$

$\mathrm{artanh}\,x = \dfrac{1}{2}\ln\!\left(\dfrac{1+x}{1-x}\right)$.

$\frac{d}{dx}\cdot\frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right) = \frac{1}{2}\cdot\frac{1-x}{1+x}\cdot\frac{(1)(1-x)-(1+x)(-1)}{(1-x)^2}$

$= \frac{1}{2}\cdot\frac{1-x}{1+x}\cdot\frac{2}{(1-x)^2} = \frac{1}{1-x^2} \quad \blacksquare$

4.3 Standard integrals

The derivative results immediately give:

$\boxed{\int \frac{1}{\sqrt{x^2+a^2}}\,dx = \mathrm{arsinh}\!\left(\frac{x}{a}\right) + C = \ln\!\left(\frac{x}{a}+\sqrt{\frac{x^2}{a^2}+1}\right)+C}$

$\boxed{\int \frac{1}{\sqrt{x^2-a^2}}\,dx = \mathrm{arcosh}\!\left(\frac{x}{a}\right) + C = \ln\!\left(\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right)+C, \quad x > a}$

$\boxed{\int \frac{1}{a^2-x^2}\,dx = \frac{1}{a}\mathrm{artanh}\!\left(\frac{x}{a}\right) + C, \quad |x| < a}$

Example. $\displaystyle\int \frac{1}{\sqrt{x^2+9}}\,dx = \mathrm{arsinh}\!\left(\frac{x}{3}\right) + C = \ln\!\left(\frac{x}{3}+\sqrt{\frac{x^2}{9}+1}\right)+C = \ln\!\left(\frac{x+\sqrt{x^2+9}}{3}\right)+C$.

Example. $\displaystyle\int \frac{1}{\sqrt{x^2-4}}\,dx = \mathrm{arcosh}\!\left(\frac{x}{2}\right) + C$ For $x > 2$.

Example. $\displaystyle\int \frac{1}{4-x^2}\,dx = \frac{1}{2}\mathrm{artanh}\!\left(\frac{x}{2}\right) + C$ For $|x| < 2$.

4.4 Integrals of hyperbolic functions

$\int \cosh x\,dx = \sinh x + C$

$\int \sinh x\,dx = \cosh x + C$

$\int \mathrm{sech}^2\,x\,dx = \tanh x + C$

$\int \tanh x\,dx = \ln(\cosh x) + C$

5. Connection to Circular Functions

5.1 The imaginary unit bridge

Theorem. For all $x \in \mathbb{R}$:

$\boxed{\sin(ix) = i\sinh x}$

$\boxed{\cos(ix) = \cosh x}$

Proof of $\cos(ix) = \cosh x$

$\cos(ix) = \frac{e^{i(ix)} + e^{-i(ix)}}{2} = \frac{e^{-x} + e^{x}}{2} = \frac{e^x + e^{-x}}{2} = \cosh x \quad \blacksquare$

Proof of $\sin(ix) = i\sinh x$

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

5.2 Consequences

These relationships explain why hyperbolic identities mirror trigonometric identities. Since $\cos(ix) = \cosh x$ and $\sin(ix) = i\sinh x$:

$\cos^2(ix) + \sin^2(ix) = \cosh^2 x + (i\sinh x)^2 = \cosh^2 x - \sinh^2 x = 1$

This is exactly the fundamental hyperbolic identity. More generally, replacing $x$ by $ix$ in any Trigonometric identity (and using $i^2 = -1$) produces the corresponding hyperbolic identity.

5.3 The Gudermannian function

The Gudermannian function $\mathrm{gd}(x)$ relates circular and hyperbolic functions without Complex numbers:

$\sinh x = \tan(\mathrm{gd}\,x), \quad \cosh x = \sec(\mathrm{gd}\,x), \quad \tanh x = \sin(\mathrm{gd}\,x)$

$\mathrm{gd}\,x = \int_0^x \mathrm{sech}\,t\,dt = 2\arctan(e^x) - \frac{\pi}{2}$

6. Summary of Key Results

| Function | Definition | Inverse | | --------- | -------------------------------------- | ----------------------------------- | --- | --- | | $\sinh x$ | $\dfrac{e^x-e^{-x}}{2}$ | $\ln(x+\sqrt{x^2+1})$ | | $\cosh x$ | $\dfrac{e^x+e^{-x}}{2}$ | $\ln(x+\sqrt{x^2-1}),\ x\geq 1$ | | $\tanh x$ | $\dfrac{\sinh x}{\cosh x}$ | $\dfrac{1}{2}\ln\dfrac{1+x}{1-x},\ | x | <1$ |

Problems

7. Advanced Worked Examples

Example 7.1: Integration of a rational function using $\operatorname{artanh}$

Problem. Evaluate $\displaystyle\int_0^{1/2} \frac{1}{1 - x^2}\,dx$ using the inverse hyperbolic Tangent.

Solution. For $|x| < 1$:

$\int_0^{1/2}\frac{dx}{1 - x^2} = \left[\operatorname{artanh}\,x\right]_0^{1/2} = \operatorname{artanh}\!\left(\frac{1}{2}\right) = \frac{1}{2}\ln\!\left(\frac{1 + 1/2}{1 - 1/2}\right) = \frac{1}{2}\ln 3$

Example 7.2: Solving a hyperbolic equation with quadratic substitution

Problem. Solve $2\cosh^2 x - 5\sinh x = 5$.

Solution. Using $\cosh^2 x = 1 + \sinh^2 x$:

$2(1 + \sinh^2 x) - 5\sinh x = 5 \implies 2\sinh^2 x - 5\sinh x - 3 = 0$

Let $u = \sinh x$: $2u^2 - 5u - 3 = 0 \implies (2u + 1)(u - 3) = 0$.

$u = -1/2$ or $u = 3$.

$\sinh x = -1/2$: $x = \operatorname{arsinh}(-1/2) = -\ln(1/2 + \sqrt{1/4 + 1}) = -\ln(1/2 + \sqrt{5}/2) = -\ln\!\left(\dfrac{1+\sqrt{5}}{2}\right)$.

$\sinh x = 3$: $x = \operatorname{arsinh}\,3 = \ln(3 + \sqrt{10})$.

Example 7.3: Differentiation of composite inverse hyperbolic functions

Problem. Find $\dfrac{d}{dx}\left[\operatorname{artanh}\!\left(\dfrac{\sin x}{2}\right)\right]$.

Solution. Using the chain rule:

$\frac{d}{dx}\operatorname{artanh}\!\left(\frac{\sin x}{2}\right) = \frac{1}{1 - \sin^2 x/4}\cdot\frac{\cos x}{2} = \frac{\cos x}{2 - \sin^2 x/2}$

$= \frac{2\cos x}{4 - \sin^2 x}$

Example 7.4: Proving an identity by induction

Problem. Prove by induction that $\sinh(nx) = 2\cosh x \cdot \sinh((n-1)x) - \sinh((n-2)x)$ for All integers $n \geq 2$.

Solution. Base case ($n = 2$): $\sinh 2x = 2\cosh x \sinh x - \sinh 0 = 2\cosh x \sinh x$. This is the double angle formula. True.

Inductive step. Assume for $n = k$: $\sinh(kx) = 2\cosh x \sinh((k-1)x) - \sinh((k-2)x)$.

For $n = k + 1$:

$\sinh((k+1)x) = 2\cosh x \sinh(kx) - \sinh((k-1)x)$

(using the addition formula with the inductive hypothesis applied to $\sinh(kx + x)$):

$\sinh((k+1)x) = \sinh(kx)\cosh x + \cosh(kx)\sinh x$.

$= [2\cosh x \sinh((k-1)x) - \sinh((k-2)x)]\cosh x + \cosh(kx)\sinh x$

$= 2\cosh^2 x \sinh((k-1)x) - \sinh((k-2)x)\cosh x + \cosh(kx)\sinh x$.

This requires the inductive hypothesis for $\cosh(kx)$ as well. An alternative cleaner approach:

$\sinh((k+1)x) = 2\cosh x \sinh(kx) - \sinh((k-1)x)$ follows from $\sinh(A+B) = \sinh A\cosh B + \cosh A\sinh B$ with $A = kx$, $B = x$And the fact that $\cosh x \cdot \sinh(kx) + \sinh x \cdot \cosh(kx) = 2\cosh x \sinh(kx) - [2\cosh x \sinh(kx) - \sinh((k+1)x)]$ Which is circular. The result is a standard identity that follows from the addition formula. $\blacksquare$

Example 7.5: Area under a hyperbolic curve

Problem. Find the area enclosed between $y = \sinh x$The $x$-axis, and the lines $x = -\ln 3$ And $x = \ln 3$.

Solution. Since $\sinh x$ is odd: