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Inverse Hyperbolic Functions

Q: Is sinh⁻¹(x) the same as 1/sinh(x)?

No, they are very different. The notation $ \sinh^{-1}(x) $ refers to the inverse function, also written as $ \text{arsinh}(x) $. The expression $ \frac{1}{\sinh(x)} $ is the reciprocal of $ \sinh(x) $, which is the hyperbolic cosecant function, $ \text{csch}(x) $.

Q: Why are these functions important? Where are they used?

Inverse hyperbolic functions appear in many areas of science and engineering. They are used in calculus to solve certain types of integrals, in physics to calculate the angle of a hanging cable (a catenary), and in special relativity to relate different observers' measurements of velocity.

Q: What is the main difference between inverse hyperbolic and inverse trigonometric functions?

The biggest difference is their algebraic form. Inverse hyperbolic functions can be expressed using natural logarithms, like $ \text{arsinh}(x) = \ln(x + \sqrt{x^2+1}) $. Inverse trigonometric functions (like $ \arcsin(x) $) do not have a similar representation with elementary functions.

Q: Do I need a special calculator to evaluate inverse hyperbolic functions?

Not necessarily. While some advanced scientific calculators have dedicated buttons for them, you can always use their logarithmic forms. As long as your calculator has a natural log (ln) button, you can find the value of any inverse hyperbolic function.

Q: Why is the domain of arcosh(x) restricted to x ≥ 1?

The range of the original function, $ y = \cosh(x) $, is $ y \ge 1 $. When we find an inverse function, the domain and range swap. Therefore, the domain of the inverse function, $ \text{arcosh}(x) $, must be $ x \ge 1 $.

Q: What does 'arsinh' stand for?

The 'ar' prefix stands for 'area'. The value of $ \text{arsinh}(u) $ is related to the area of a specific sector of the unit hyperbola $ x^2 - y^2 = 1 $. This is analogous to how 'arcsin' relates to the arc length of a sector of the unit circle.

Q: Can I find the inverse hyperbolic sine of a negative number?

Yes, you can. The domain of $ \text{arsinh}(x) $ is all real numbers, so you can input any positive, negative, or zero value. For example, $ \text{arsinh}(-1) = \ln(-1 + \sqrt{2}) $, which is a valid real number.

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Ever wondered what happens when you 'undo' a hyperbolic function? Welcome to the world of inverse hyperbolic functions! These unique functions, like $arsinh (x)$ and $arcosh (x)$ , are essential in calculus, physics, and engineering, and they have a surprising connection to natural logarithms.

Inverse Hyperbolic Functions — an original Algebra911 reference diagram defining inverse hyperbolic functions with its key formula and a worked example. — Inverse Hyperbolic Functions: The Complete Guide

What Are Inverse Hyperbolic Functions?

Inverse hyperbolic functions are the inverse functions of the hyperbolic functions, in the same way that inverse trigonometric functions (like arcsin) are the inverses of trigonometric functions (like sin). If you have a hyperbolic function that relates $x$ and $y$ , its inverse function swaps the roles of $x$ and $y$ . For example, if $y = \sinh (x)$ , then the inverse function gives us $x$ in terms of $y$ .

We denote the inverse hyperbolic sine function as $arsinh (x)$ or $\sinh^{- 1} (x)$ . The 'ar' prefix stands for 'area', because these functions can be used to calculate the area of a hyperbolic sector, similar to how inverse trigonometric functions relate to the arc length of a circle sector.

So, the fundamental relationship is:

If $y = f (x)$ is a hyperbolic function, then $x = f^{- 1} (y)$ is the corresponding inverse hyperbolic function.

For the main functions, this looks like:

If $y = \sinh (x)$ , then $x = arsinh (y)$
If $y = \cosh (x)$ , then $x = arcosh (y)$ (with some domain considerations)
If $y = \tanh (x)$ , then $x = artanh (y)$

A crucial warning about notation: The notation $\sinh^{- 1} (x)$ can be confusing. It means the inverse function of sinh, not the reciprocal $(\sinh (x))^{- 1}$ , which is $\frac{1}{\sinh (x)}$ or $csch (x)$ . Because of this potential confusion, many mathematicians and textbooks prefer the $arsinh (x)$ notation. In this lesson, we'll use both, but remember they mean the same thing!

The Surprising Link: Why Are Logarithms Involved?

One of the most powerful and surprising properties of inverse hyperbolic functions is that they can be expressed using natural logarithms. This is quite different from inverse trigonometric functions, which don't have such a simple algebraic form. But why does this connection exist?

The answer lies in the definitions of the hyperbolic functions themselves. Recall that hyperbolic functions are defined using the exponential function, $e^{x}$ :

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

Since the natural logarithm, $\ln (x)$ , is the inverse function of $e^{x}$ , it makes sense that when we try to 'undo' the hyperbolic functions, the natural logarithm will appear. Let's see this in action by deriving the logarithmic formula for $arsinh (x)$ .

Derivation of the formula for $arsinh (x)$ :

Let $y = arsinh (x)$ . By the definition of an inverse function, this means $x = \sinh (y)$ .
Now, substitute the exponential definition of $\sinh (y)$ :
$x = \frac{e^{y} - e^{- y}}{2}$
Our goal is to solve this equation for $y$ . First, multiply both sides by $2$ :
$2 x = e^{y} - e^{- y}$
To get rid of the negative exponent, we can write $e^{- y}$ as $\frac{1}{e^{y}}$ . Then, multiply the entire equation by $e^{y}$ to clear the fraction:
$2 x (e^{y}) = (e^{y}) (e^{y}) - (e^{- y}) (e^{y})$
$2 x e^{y} = e^{2 y} - 1$
Rearrange this equation to look like a standard quadratic equation. Let $u = e^{y}$ , which means $u^{2} = e^{2 y}$ . Substituting this in gives:
$2 x u = u^{2} - 1$
$u^{2} - 2 x u - 1 = 0$
This is a quadratic equation in the variable $u$ . We can solve for $u$ using the quadratic formula, $u = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ , where $a = 1$ , $b = - 2 x$ , and $c = - 1$ :
$u = \frac{- (- 2 x) \pm \sqrt{(- 2 x)^{2} - 4 (1) (- 1)}}{2 (1)}$
$u = \frac{2 x \pm \sqrt{4 x^{2} + 4}}{2}$
$u = \frac{2 x \pm \sqrt{4 (x^{2} + 1)}}{2}$
$u = \frac{2 x \pm 2 \sqrt{x^{2} + 1}}{2}$
$u = x \pm \sqrt{x^{2} + 1}$
Now, substitute back $u = e^{y}$ :
$e^{y} = x \pm \sqrt{x^{2} + 1}$
We need to decide whether to use the plus or minus sign. Remember that $e^{y}$ must always be positive for any real number $y$ . Let's look at the term $\sqrt{x^{2} + 1}$ . Since $x^{2}$ is always non-negative, $x^{2} + 1 > x^{2}$ , which means $\sqrt{x^{2} + 1} > \sqrt{x^{2}} = | x |$ . So, $\sqrt{x^{2} + 1}$ is always greater than $x$ . This means the term $x - \sqrt{x^{2} + 1}$ will always be negative. Since $e^{y}$ cannot be negative, we must discard the minus solution.
We are left with the positive solution:
$e^{y} = x + \sqrt{x^{2} + 1}$
Finally, to solve for $y$ , we take the natural logarithm of both sides:
$y = \ln (x + \sqrt{x^{2} + 1})$

Since we started with $y = arsinh (x)$ , we have successfully derived its logarithmic form! This process can be repeated for the other inverse hyperbolic functions.

The Main Inverse Hyperbolic Functions: Formulas and Domains

Being able to express inverse hyperbolic functions as logarithms is incredibly useful for solving equations and for calculus operations like differentiation and integration. Here is a reference table of the most common inverse hyperbolic functions, their logarithmic forms, and the domains for which they are defined.

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

arcosh (x) = \ln (x + \sqrt{x^{2} - 1})

artanh (x) = \frac{1}{2} \ln (\frac{1 + x}{1 - x})

The table below summarizes these and others.

Function	Logarithmic Formula	Domain (Input $x$ )	Range (Output $y$ )
$arsinh (x)$	$\ln (x + \sqrt{x^{2} + 1})$	All real numbers $(- \infty, \infty)$	All real numbers $(- \infty, \infty)$
$arcosh (x)$	$\ln (x + \sqrt{x^{2} - 1})$	$x \geq 1$	$y \geq 0$
$artanh (x)$	$\frac{1}{2} \ln (\frac{1 + x}{1 - x})$	$- 1 < x < 1$	All real numbers $(- \infty, \infty)$
$arcoth (x)$	$\frac{1}{2} \ln (\frac{x + 1}{x - 1})$	$\| x \| > 1$ (i.e., $x > 1$ or $x < - 1$ )	All real numbers except $0$
$arsech (x)$	$\ln (\frac{1 + \sqrt{1 - x^{2}}}{x})$	$0 < x \leq 1$	$y \geq 0$
$arcsch (x)$	$\ln (\frac{1}{x} + \frac{\sqrt{1 + x^{2}}}{\| x \|})$	All real numbers except $0$	All real numbers except $0$

Pay close attention to the domains! You cannot, for example, take the $arcosh$ of $0.5$ or the $artanh$ of $2$ , because those values are outside the functions' domains. This is a common source of errors.

How Do You Calculate Values of Inverse Hyperbolic Functions?

Calculating the value of an inverse hyperbolic function might seem intimidating, but once you have the logarithmic formulas, it's just a matter of substitution and careful arithmetic. You don't need a special button on your calculator (though some scientific calculators have them); you just need the natural log (ln) button.

Example 1

Problem: Find the exact value of $arsinh (1)$ .

Solution:

We use the formula for $arsinh (x)$ :
$arsinh (x) = \ln (x + \sqrt{x^{2} + 1})$

Substitute $x = 1$ into the formula:

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

Simplify the expression inside the logarithm:

$arsinh (1) = \ln (1 + \sqrt{1 + 1})$

$arsinh (1) = \ln (1 + \sqrt{2})$

This is the exact answer. If you need a decimal approximation, you can use a calculator: $\ln (1 + \sqrt{2}) \approx \ln (1 + 1.414) = \ln (2.414) \approx 0.881$ .

Example 2

Problem: Evaluate $arcosh (2)$ .

Solution:

First, check the domain. The domain for $arcosh (x)$ is $x \geq 1$ . Since $2 \geq 1$ , we can proceed.

Use the formula for $arcosh (x)$ :
$arcosh (x) = \ln (x + \sqrt{x^{2} - 1})$

Substitute $x = 2$ :

$arcosh (2) = \ln (2 + \sqrt{2^{2} - 1})$

Simplify the expression:

$arcosh (2) = \ln (2 + \sqrt{4 - 1})$

$arcosh (2) = \ln (2 + \sqrt{3})$

This is the exact value. The decimal approximation is $\ln (2 + \sqrt{3}) \approx \ln (2 + 1.732) = \ln (3.732) \approx 1.317$ .

Example 3

Problem: Find the value of $artanh (- 0.5)$ .

Solution:

First, check the domain. The domain for $artanh (x)$ is $- 1 < x < 1$ . Since $- 1 < - 0.5 < 1$ , our input is valid.

Use the formula for $artanh (x)$ :
$artanh (x) = \frac{1}{2} \ln (\frac{1 + x}{1 - x})$

Substitute $x = - 0.5$ :

$artanh (- 0.5) = \frac{1}{2} \ln (\frac{1 + (- 0.5)}{1 - (- 0.5)})$

Simplify the fraction inside the logarithm:

$artanh (- 0.5) = \frac{1}{2} \ln (\frac{1 - 0.5}{1 + 0.5})$

$artanh (- 0.5) = \frac{1}{2} \ln (\frac{0.5}{1.5})$

$artanh (- 0.5) = \frac{1}{2} \ln (\frac{1}{3})$

Using the logarithm property $\ln (a / b) = \ln (a) - \ln (b)$ or $\ln (1 / c) = - \ln (c)$ , we can also write this as $- \frac{1}{2} \ln (3)$ . This is the exact answer. The decimal approximation is $- \frac{1}{2} \times 1.0986 \approx - 0.549$ .

What Do the Graphs of Inverse Hyperbolic Functions Look Like?

Understanding the graph of a function is key to understanding its behavior. The graphs of the inverse hyperbolic functions are reflections of the hyperbolic function graphs across the line $y = x$ . Let's look at the three main ones.

Graph of $y = arsinh (x)$

The graph of $y = arsinh (x)$ is a reflection of $y = \sinh (x)$ across the line $y = x$ . Since $\sinh (x)$ has a domain and range of all real numbers, so does its inverse, $arsinh (x)$ . The graph passes through the origin $(0, 0)$ and increases continuously. It looks a bit like a gentle 'S' curve that has been stretched vertically.

Graph of $y = arcosh (x)$

This one is a bit trickier. The function $y = \cosh (x)$ is an even function (like $y = x^{2}$ ) and fails the horizontal line test, meaning it doesn't have a simple inverse. To create an inverse, we first restrict the domain of $\cosh (x)$ to $x \geq 0$ . By doing this, we get a one-to-one function whose range is $y \geq 1$ .

The graph of $y = arcosh (x)$ is the reflection of this restricted $y = \cosh (x)$ graph. Its domain is $x \geq 1$ and its range is $y \geq 0$ . The graph starts at the point $(1, 0)$ and increases from there, curving upwards.

Graph of $y = artanh (x)$

The graph of $y = \tanh (x)$ has two horizontal asymptotes at $y = 1$ and $y = - 1$ . When we reflect this graph across the line $y = x$ to get $y = artanh (x)$ , these horizontal asymptotes become vertical asymptotes at $x = 1$ and $x = - 1$ . Therefore, the domain of $artanh (x)$ is $- 1 < x < 1$ . The graph passes through the origin and rises steeply as it approaches the vertical asymptotes.

Key formulas for inverse hyperbolic functions by Algebra911.

What Are Some Common Mistakes to Avoid?

Inverse hyperbolic functions have a few tricky spots where students often make mistakes. Being aware of them is the first step to avoiding them!

Notation Confusion: The biggest mistake is confusing $\sinh^{- 1} (x)$ with $(\sinh (x))^{- 1}$ . Remember, $\sinh^{- 1} (x)$ is the inverse function (arsinh), while $(\sinh (x))^{- 1} = \frac{1}{\sinh (x)} = csch (x)$ . They are completely different. To be safe, use the $arsinh (x)$ notation.
Domain Errors for $arcosh (x)$ : Forgetting that the domain of $arcosh (x)$ is $x \geq 1$ . You cannot plug a number like $0.7$ or $- 2$ into $arcosh (x)$ . Your calculator will give you an error, and the logarithmic formula would involve the square root of a negative number.
Domain Errors for $artanh (x)$ : Similarly, forgetting that the domain of $artanh (x)$ is strictly between $- 1$ and $1$ (i.e., $- 1 < x < 1$ ). Trying to calculate $artanh (1)$ or $artanh (2)$ is not possible. The logarithmic formula would involve taking the log of zero or a negative number.
Formula Mix-ups: The logarithmic formulas for $arsinh (x)$ and $arcosh (x)$ are very similar: one has $\sqrt{x^{2} + 1}$ and the other has $\sqrt{x^{2} - 1}$ . It's easy to mix them up. A good way to remember is that $arcosh (x)$ has the minus sign, and its domain $x \geq 1$ ensures that the term inside the square root, $x^{2} - 1$ , is never negative.

Quick Summary and Reference

This lesson covered a lot of ground. Here are the most important points to remember about inverse hyperbolic functions:

They are Inverses: They 'undo' the hyperbolic functions. If $y = \sinh (x)$ , then $x = arsinh (y)$ .
Logarithmic Forms: Their most powerful feature is that they can be written using natural logarithms. This makes them easy to work with in algebra and calculus.
The Three Main Formulas: You should be most familiar with these three functions and their domains.

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

Domain: All real numbers.

arcosh (x) = \ln (x + \sqrt{x^{2} - 1})

Domain:

x \geq 1

artanh (x) = \frac{1}{2} \ln (\frac{1 + x}{1 - x})

Domain:

- 1 < x < 1

Mastering these formulas and being mindful of their domains will give you a solid foundation for using inverse hyperbolic functions in more advanced math courses.

Frequently Asked Questions

Is sinh⁻¹(x) the same as 1/sinh(x)?

No, they are very different. The notation $\sinh^{- 1} (x)$ refers to the inverse function, also written as $arsinh (x)$ . The expression $\frac{1}{\sinh (x)}$ is the reciprocal of $\sinh (x)$ , which is the hyperbolic cosecant function, $csch (x)$ .

Why are these functions important? Where are they used?

Inverse hyperbolic functions appear in many areas of science and engineering. They are used in calculus to solve certain types of integrals, in physics to calculate the angle of a hanging cable (a catenary), and in special relativity to relate different observers' measurements of velocity.

What is the main difference between inverse hyperbolic and inverse trigonometric functions?

The biggest difference is their algebraic form. Inverse hyperbolic functions can be expressed using natural logarithms, like $arsinh (x) = \ln (x + \sqrt{x^{2} + 1})$ . Inverse trigonometric functions (like $\arcsin (x)$ ) do not have a similar representation with elementary functions.

Do I need a special calculator to evaluate inverse hyperbolic functions?

Not necessarily. While some advanced scientific calculators have dedicated buttons for them, you can always use their logarithmic forms. As long as your calculator has a natural log (ln) button, you can find the value of any inverse hyperbolic function.

Why is the domain of arcosh(x) restricted to x ≥ 1?

The range of the original function, $y = \cosh (x)$ , is $y \geq 1$ . When we find an inverse function, the domain and range swap. Therefore, the domain of the inverse function, $arcosh (x)$ , must be $x \geq 1$ .

What does 'arsinh' stand for?

The 'ar' prefix stands for 'area'. The value of $arsinh (u)$ is related to the area of a specific sector of the unit hyperbola $x^{2} - y^{2} = 1$ . This is analogous to how 'arcsin' relates to the arc length of a sector of the unit circle.

Can I find the inverse hyperbolic sine of a negative number?

Yes, you can. The domain of $arsinh (x)$ is all real numbers, so you can input any positive, negative, or zero value. For example, $arsinh (- 1) = \ln (- 1 + \sqrt{2})$ , which is a valid real number.

Inverse Hyperbolic Functions

What Are Inverse Hyperbolic Functions?

The Surprising Link: Why Are Logarithms Involved?

The Main Inverse Hyperbolic Functions: Formulas and Domains

How Do You Calculate Values of Inverse Hyperbolic Functions?

What Do the Graphs of Inverse Hyperbolic Functions Look Like?

Graph of y=arsinh(x)

Graph of y=arcosh(x)

Graph of y=artanh(x)

What Are Some Common Mistakes to Avoid?

Quick Summary and Reference

Frequently Asked Questions

Graph of $y = arsinh (x)$

Graph of $y = arcosh (x)$

Graph of $y = artanh (x)$