Table of Contents
Last modified on November 25th, 2024
Hyperbolic functions are similar to trigonometric functions, but instead of unit circles, they are defined using rectangular hyperbolas. In trigonometry, the coordinates on a unit circle are represented as (cos θ, sin θ), whereas in hyperbolic functions, the pair (cosh θ, sinh θ) represents points on the right half of an equilateral hyperbola.
They are used in solving linear differential equations, hyperbolic geometry, and Laplace’s equations in Cartesian coordinates.
The three basic hyperbolic functions are:
Hyperbolic functions are expressed through exponential function ex and its inverse e-x (here, e = Euler’s constant).
The hyperbolic sine function is a function f: ℝ → ℝ such that f(x) = sinh x, which is expressed as
sinh x = ${\dfrac{e^{x}-e^{-x}}{2}}$
Now, let us graph the function y = sinh x and see how it behaves.
The hyperbolic cosine function is a function f: ℝ → ℝ such that f(x) = cosh x, which is expressed as
cosh x = ${\dfrac{e^{x}+e^{-x}}{2}}$
Now, on graphing the hyperbolic cosine function y = cosh x, we get:
The hyperbolic tangent function is a function f: ℝ → ℝ such that f(x) = tanh x, which is expressed as
tanh x = ${\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}}$
Now, on graphing the hyperbolic tangent function y = tanh x, we get
In addition to the three basic hyperbolic functions, there are three other hyperbolic functions equivalent to sine, cosine, and tangent. They are secant, cosecant, and cotangent:
The hyperbolic cosecant function (denoted by cosech or csch) is the reciprocal of the hyperbolic sine function, which is given by the expression:
cosech x = ${\dfrac{1}{\sinh x}}$ = ${\dfrac{2}{e^{x}-e^{-x}}}$
Now, on graphing the hyperbolic cosecant function y = cosech x, we get
Similarly, the hyperbolic secant function (sech) is the reciprocal of the hyperbolic cosine function, which is given by the expression:
sech x = ${\dfrac{1}{\cosh x}}$ = ${\dfrac{2}{e^{x}+e^{-x}}}$
Now, on graphing the hyperbolic secant function y = sech x, we get
The hyperbolic cotangent function (coth) is the reciprocal of the hyperbolic tangent function, which is given by the expression:
coth x = ${\dfrac{1}{\tanh x}}$ = ${\dfrac{e^{x}+e^{-x}}{e^{x}-e^{-x}}}$
Now, on graphing the hyperbolic cotangent function y = coth x, we get
Here is the summary of all graphs of the hyperbolic functions:
The domain and range of the six hyperbolic functions are listed:
Hyperbolic Function | Domain | Range |
---|---|---|
sinh x | (-∞, ∞) | (-∞, ∞) |
cosh x | (-∞, ∞) | [1, ∞) |
tanh x | (-∞, ∞) | (-1, 1) |
cosech x | (-∞, 0) ∪ (0, ∞) | (-∞, 0) ∪ (0, ∞) |
sech x | (-∞, ∞) | (0, 1] |
coth x | (-∞, 0) ∪ (0, ∞) | (-∞, – 1) ∪ (1, ∞) |
Hyperbolic identities, similar to those of trigonometric functions, are related to the relationships between hyperbolic functions. These are universally true for all values of the variables involved.
The derivatives of hyperbolic functions are similar to those of trigonometric functions.
The inverse hyperbolic functions are the inverse operations of the hyperbolic functions. These functions are also known as area hyperbolic functions that provide the hyperbolic angles corresponding to a given value of the hyperbolic function.
Here is the list of all inverse hyperbolic functions in the complex plane:
Using hyperbolic functions identities, prove:
a) cosh x + sinh x = ex
b) cosh2 x – sinh2 x = 1
c) coth2x – cosech2x = 1
a) As we know,
sinh x = ${\dfrac{e^{x}-e^{-x}}{2}}$ and cosh x = ${\dfrac{e^{x}+e^{-x}}{2}}$
Here,
cosh x + sinh x
= ${\dfrac{e^{x}-e^{-x}}{2}+\dfrac{e^{x}+e^{-x}}{2}}$
= ${\dfrac{e^{x}-e^{-x}+e^{x}+e^{-x}}{2}}$
= ${\dfrac{2e^{x}}{2}}$
= ${e^{x}}$
Hence proved
b) As we know,
sinh x = ${\dfrac{e^{x}-e^{-x}}{2}}$ and cosh x = ${\dfrac{e^{x}+e^{-x}}{2}}$
Here,
cosh2 x – sinh2 x
= ${\left( \dfrac{e^{x}+e^{-x}}{2}\right) ^{2}-\left( \dfrac{e^{x}-e^{-x}}{2}\right) ^{2}}$
= ${\dfrac{\left( e^{x}+e^{-x}\right) ^{2}-\left( e^{x}-e^{-x}\right) ^{2}}{4}}$
= ${\dfrac{4\cdot e^{x}\cdot e^{-x}}{4}}$
= ${e^{x-x}}$
= ${e^{0}}$
= 1
Hence proved
c) As we know,
coth x = ${\dfrac{\cosh x}{\sinh x}}$
cosech x = ${\dfrac{1}{\sinh x}}$
Here,
coth2x – cosech2x
= ${\left( \dfrac{\cosh x}{\sinh x}\right) ^{2}-\left( \dfrac{1}{\sinh x}\right) ^{2}}$
= ${\dfrac{\cosh ^{2}x}{\sinh ^{2}x}-\dfrac{1}{\sinh ^{2}x}}$
= ${\dfrac{\cosh ^{2}x-1}{\sinh ^{2}x}}$ …..(i)
Since cosh2 x – sinh2 x = 1
⇒ cosh2 x – 1 = sinh2 x …..(ii)
From (i) and (ii),
coth2x – cosech2x = ${\dfrac{\sinh ^{2}x}{\sinh ^{2}x}}$ = 1
Hence proved
Last modified on November 25th, 2024