The **hyperbolic functions** are defined through algebraic expressions that include exponential functions and^{x} and its inverse function e^{-x }, where e is Euler's constant (or as it is commonly known as “number e”), whose approximate value is 2.718281. The basic hyperbolic functions are **hyperbolic sine** (sinh) and the **hyperbolic cosine** (cosh), from these the function of **hyperbolic tangent** (tanh). The other functions: cotangent (coth), secant (sech) and cosecant (csch), are the inverse of the previous three respectively.

**Definition**

They are called **hyperbolic functions **to the hyperbolic cosine (cosh), hyperbolic sine (senh) and the functions obtained from them, such as the hyperbolic tangent (tanh) and their respective inverses:

**Hyperbolic cosine**

Cosh (x) = (e^{x }+ e^{-x}) / 2

It is an even function.

**Hyperbolic sine**

senh (x) = (e^{x }- e^{-x}) / 2

It is an odd function.

**Hyperbolic tangent**

Tanh (x) = senh (x) / cosh (x) = (e^{x }- e^{-x}) / (e^{x }+ e^{-x}) = (e^{2x }- 1) / (e^{2x }+ 1)

It is an odd function.

**Hyperbolic cotangent**

Coth (x) = cosh (x) / senh (x) = (e^{x }+ e^{-x}) / (e^{x }- e^{-x}) = (e^{2x }+ 1) / (e^{2x }- 1)

Defined on R * and more generally on C *; it is an odd function.

**Hyperbolic secant**

Sech (x) = 1 / cosh (x) = 2 / (e^{x }+ e^{-x}) = 2e^{x }/ (e^{2x }+ 1)

It is an even function.

**Hyperbolic Cosecant**

Csch (x) = 1 / senh (x) = 2 / (e^{x }- e^{-x}) = 2e^{x }/ (e^{2x }- 1)

Defined on R * and more generally on C *; it is an odd function.

**Geometric proof**

The functions sinh and cosh satisfy the equation of the hyperbola x^{2 }- Y^{2 }= 1.

Assuming that x = cosh (t) and y = sinh (t) and considering that:

Cosh (t) = (e^{t }+ e^{-t}) / 2 and senh (t) = (e^{t }- e^{-t}) / 2

We substitute in the hyperbola equation

[(and^{t }+ e^{-t}) / 2]^{2 }- [(e^{t }- e^{-t}) / 2]^{2 }= 1

[(and^{2t }+ 2e^{t}and^{-t} + e^{-2t}) / 4] - [(e^{2t }- 2e^{t}and^{-t }+ e^{-2t}) / 4] = 1

As e^{t}and^{-t} = e^{t - t} = e^{0} = 1, we have:

[(and^{2t }+ 2 + e^{-2t}) / 4] - [(e^{2t }- 2 + e^{-2t}) / 4] = 1

We subtract since they have the same denominator:

[(and^{2t }+ 2 + e^{-2t}) ^{ }- (e^{2t }- 2 + e^{-2t})] / 4 = 1

(and^{2t }+ 2 + e^{-2t} ^{ }- e^{2t }+ 2 - e^{-2t}) / 4 = 1

(and^{2t }- 2 + e^{-2t} ^{ }- e^{2t }- 2 - e^{-2t}) / 4 = 1

4 / 4 = 1

1 = 1

Which is what we wanted to demonstrate.