Hyperbolic functions

hyperbolic functionsThe hyperbolic functions are defined through algebraic expressions that include exponential functions andx 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) = (ex + e-x) / 2

hyperbolic cosine

It is an even function.

Hyperbolic sine

senh (x) = (ex - e-x) / 2

hyperbolic sine

It is an odd function.

Hyperbolic tangent

Tanh (x) = senh (x) / cosh (x) = (ex - e-x) / (ex + e-x) = (e2x - 1) / (e2x + 1)

hyperbolic tangent

It is an odd function.

Hyperbolic cotangent

Coth (x) = cosh (x) / senh (x) = (ex + e-x) / (ex - e-x) = (e2x + 1) / (e2x - 1)

hyperbolic cotangent

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

Hyperbolic secant

Sech (x) = 1 / cosh (x) = 2 / (ex + e-x) = 2ex / (e2x + 1)

It is an even function.

Hyperbolic Cosecant

Csch (x) = 1 / senh (x) = 2 / (ex - e-x) = 2ex / (e2x - 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 x2 - Y2 = 1.

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

Cosh (t) = (et + e-t) / 2 and senh (t) = (et - e-t) / 2

We substitute in the hyperbola equation

[(andt + e-t) / 2]2 - [(et - e-t) / 2]2 = 1

[(and2t + 2etand-t + e-2t) / 4] - [(e2t - 2etand-t + e-2t) / 4] = 1

As etand-t = et - t = e0 = 1, we have:

[(and2t + 2 + e-2t) / 4] - [(e2t - 2 + e-2t) / 4] = 1

We subtract since they have the same denominator:

[(and2t + 2 + e-2t)  - (e2t - 2 + e-2t)] / 4 = 1

(and2t + 2 + e-2t  - e2t + 2 - e-2t) / 4 = 1

(and2t - 2 + e-2t  - e2t - 2 - e-2t) / 4 = 1

4 / 4 = 1

1 = 1

Which is what we wanted to demonstrate.