Hyperbolic Functions Solver

Evaluate Hyperbolic Functions

Select a hyperbolic function and enter a value for $x$ to calculate its value step-by-step.

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Solve Equations Involving Hyperbolic Functions

Equations involving hyperbolic functions can often be solved by converting them to their exponential forms or by using inverse hyperbolic functions.

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Understanding Hyperbolic Functions

Hyperbolic functions are analogs of the ordinary trigonometric (circular) functions, but defined using the hyperbola rather than the circle. Just as the points $(\cos t, \sin t)$ form a circle with a unit radius, the points $(\cosh t, \sinh t)$ form the right half of the unit hyperbola $x^2 - y^2 = 1$.

Practical Applications:

Catenary Curves: The shape of a hanging chain or cable under its own weight is described by $\cosh x$. This is seen in suspension bridges and power lines.
Engineering: Used in electrical engineering (transmission line theory), civil engineering (architecture of arches and domes), and aerospace engineering.
Physics: Appear in the theory of relativity (Lorentz transformations), and in describing the motion of objects subject to certain forces.
Calculus: They simplify certain integrals, much like trigonometric substitutions.

Definitions of Hyperbolic Functions

Hyperbolic functions are defined using the exponential function $e^x$:

Hyperbolic Sine:
$$ \sinh x = \frac{e^x - e^{-x}}{2} $$
Hyperbolic Cosine:
$$ \cosh x = \frac{e^x + e^{-x}}{2} $$
Hyperbolic Tangent:
$$ \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}} $$
Hyperbolic Cosecant:
$$ \text{cosech } x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}} \quad (x \neq 0) $$
Hyperbolic Secant:
$$ \text{sech } x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}} $$
Hyperbolic Cotangent:
$$ \coth x = \frac{1}{\tanh x} = \frac{\cosh x}{\sinh x} = \frac{e^x + e^{-x}}{e^x - e^{-x}} \quad (x \neq 0) $$

Osborne's Rule

Osborne's rule provides a simple way to convert trigonometric identities into corresponding hyperbolic identities.

The rule states: If you have a valid trigonometric identity involving powers of $\sin x$ and $\cos x$ (or other trigonometric functions expressed in terms of $\sin x$ and $\cos x$), you can obtain a corresponding hyperbolic identity by:

Replacing every trigonometric function with its corresponding hyperbolic function (e.g., $\sin x \to \sinh x$, $\cos x \to \cosh x$, $\tan x \to \tanh x$).
Changing the sign of any term that contains a product of two $\sinh$ functions (or an even power of $\sinh$ if it's $\sinh^2 x$, $\sinh^4 x$, etc., effectively). More precisely, change the sign of any term involving $\sinh^2 x$, or $\sinh A \sinh B$, or $\tanh^2 x$, or $\tanh A \tanh B$, etc.

Example: Trigonometric identity: $\cos^2 x + \sin^2 x = 1$. Applying Osborne's rule:

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

(The $\sinh^2 x$ term changes sign).

Example: Trigonometric identity: $\sin(A+B) = \sin A \cos B + \cos A \sin B$. Applying Osborne's rule:

$$ \sinh(A+B) = \sinh A \cosh B + \cosh A \sinh B $$

(No product of two sines).

Example: Trigonometric identity: $\cos(A+B) = \cos A \cos B - \sin A \sin B$. Applying Osborne's rule:

$$ \cosh(A+B) = \cosh A \cosh B + \sinh A \sinh B $$

(The term $\sin A \sin B$ becomes $\sinh A \sinh B$ and its sign changes from - to +).

Common Hyperbolic Identities

$$ \cosh^2 x - \sinh^2 x = 1 $$
Proof:
$$ \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} = \frac{4}{4} = 1 $$
$$ \text{sech}^2 x = 1 - \tanh^2 x $$
Proof: Divide $\cosh^2 x - \sinh^2 x = 1$ by $\cosh^2 x$:
$$ 1 - \frac{\sinh^2 x}{\cosh^2 x} = \frac{1}{\cosh^2 x} \implies 1 - \tanh^2 x = \text{sech}^2 x $$
$$ \text{cosech}^2 x = \coth^2 x - 1 $$
Proof: Divide $\cosh^2 x - \sinh^2 x = 1$ by $\sinh^2 x$:
$$ \frac{\cosh^2 x}{\sinh^2 x} - 1 = \frac{1}{\sinh^2 x} \implies \coth^2 x - 1 = \text{cosech}^2 x $$
$$ \sinh(2x) = 2 \sinh x \cosh x $$
$$ \cosh(2x) = \cosh^2 x + \sinh^2 x = 2 \cosh^2 x - 1 = 1 + 2 \sinh^2 x $$

Series Expansions for cosh x and sinh x

The Maclaurin series for $e^x$ is

$$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$

And for $e^{-x}$ is

$$ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \dots = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!} $$

Derivation for cosh x:

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

$$ = \frac{1}{2} \left[ \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \right) + \left(1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots \right) \right] $$

$$ = \frac{1}{2} \left[ 2 + 0x + 2\frac{x^2}{2!} + 0x^3 + 2\frac{x^4}{4!} + \dots \right] $$

$$ = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \dots = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} $$

Derivation for sinh x:

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

$$ = \frac{1}{2} \left[ \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \right) - \left(1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots \right) \right] $$

$$ = \frac{1}{2} \left[ 0 + 2x + 0\frac{x^2}{2!} + 2\frac{x^3}{3!} + 0\frac{x^4}{4!} + \dots \right] $$

$$ = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \dots = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} $$