## 17Calculus Precalculus - Hyperbolic Functions

Hyperbolic Functions

Hyperbolic functions are not usually discussed in detail until calculus 1 and may even be delayed until calculus 2. However, we give you a comprehensive tutorial here to give you a head start on these important functions. You can also use this page to learn about hyperbolic functions when you reach this topic in calculus.

Hyperbolic functions are functions formed from exponentials. They appear so often that they are given the special name hyperbolic and they seem to work similar to trig functions, so they are also called hyperbolic trig functions.

In trigonometry we have sine, cosine, tangent, etc. Hyperbolic functions are called hyperbolic sine, hyperbolic cosine, hyperbolic tangent and the abbreviations are written $$\sinh(x), \cosh(x), \tanh(x) . . .$$, with an 'h' after the same name as the trig functions. Here are the definitions.

 $$\displaystyle{ \sinh(x) = \frac{e^x-e^{-x}}{2} }$$ $$\displaystyle{ \cosh(x) = \frac{e^x+e^{-x}}{2} }$$ $$\displaystyle{ \csch(x) = \frac{1}{\sinh(x)} }$$ $$\displaystyle{ \sech(x) = \frac{1}{\cosh(x)} }$$ $$\displaystyle{ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} }$$ $$\displaystyle{ \coth(x) = \frac{\cosh(x)}{\sinh(x)} }$$

So, you can see why we call them hyperbolic trig functions. Other than $$\sinh(x)$$ and $$\cosh(x)$$, we define the other hyperbolic functions similar to the corresponding definitions for trig functions. Now let's look at some identities and I think the relationship will become clearer.

 $$\displaystyle{ \sinh(-x) = -\sinh(x) }$$ $$\displaystyle{ \cosh(-x) = \cosh(x) }$$ $$\displaystyle{ \cosh^2(x) - \sinh^2(x) = 1 }$$ $$\displaystyle{ 1-\tanh^2(x) = \sech^2(x) }$$

Be careful here. Notice that although they are similar, they are NOT always identical, usually with a change in sign or some other subtle difference. So don't gloss over this. Take a few minutes to notice the similarities AND the differences.

Let's start with a very good video discussing some of these equations and working with them to prove some identities. These are a couple of great videos to get you started with hyperbolic functions.

### PatrickJMT - Hyperbolic Functions - The Basics [10min-0secs]

video by PatrickJMT

### blackpenredpen - Introduction to Hyperbolic Trig Functions

video by blackpenredpen

Okay, let's take a look at another quick video working with hyperbolic functions. Using the definitions for $$\cosh(x)$$ and $$\sinh(x)$$, this presenter proves the identity $$\displaystyle{ \cosh^2(x) - \sinh^2(x) = 1 }$$. Although the end result is not particularly significant, this video shows how you can work with hyperbolic trig functions using the definitions.

### Krista King Math - Hyperbolic identities [5min-14secs]

video by Krista King Math

Remember, there is nothing special or magical going on here. We just define $$\sinh(x)$$ and $$\cosh(x)$$ using exponentials and everything else builds from there.

Plots of Hyperbolic Functions

Here are the plots of the two basic hyperbolic functions. Plot 1 - $$\sinh(x)$$ Plot 2 - $$\cosh(x)$$

Notice that the plot for hyperbolic cosine looks a lot like a parabola, but it is NOT a parabola.
Now let's combine the plots for hyperbolic sine and hyperbolic cosine and see if anything interesting shows up. Plot 3 shows this. Plot 3 - $$\sinh(x)$$ and $$\cosh(x)$$

Notice the right side of both functions are getting closer and closer. This makes sense from the equations since $$e^{-x}$$ gets closer to zero as $$x$$ gets larger. Consequently, both graphs are approaching $$e^x / 2$$ . If this doesn't make sense, just think about it for a bit to get a general idea. You will learn to think about functions in this way in more detail in calculus.

The other hyperbolic functions are plotted using GeoGebra and can be found at this link.

To get used to working with hyperbolic functions, let's work these practice problems.

Practice

Unless otherwise instructed, use the definitions of the hyperbolic functions in terms of $$e^x$$ to solve these problems.

Use the definitions of the hyperbolic functions in terms of $$e^x$$ to show that $$\cosh^2(x) - \sinh^2(x) = 1$$ holds.

Problem Statement

Use the definitions of the hyperbolic functions in terms of $$e^x$$ to show that $$\cosh^2(x) - \sinh^2(x) = 1$$ holds.

Solution

Here are two video solutions by two different instructors.

### 2845 video

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Use the definitions of the hyperbolic functions in terms of $$e^x$$ to show that $$\tanh^2(x) + \text{sech}^2(x) = 1$$ holds.

Problem Statement

Use the definitions of the hyperbolic functions in terms of $$e^x$$ to show that $$\tanh^2(x) + \text{sech}^2(x) = 1$$ holds.

Solution

### 2846 video

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Use the definitions of the hyperbolic functions in terms of $$e^x$$ to find expressions for $$\sinh(2x)$$ and $$\cosh(2x)$$.

Problem Statement

Use the definitions of the hyperbolic functions in terms of $$e^x$$ to find expressions for $$\sinh(2x)$$ and $$\cosh(2x)$$.

Solution

### 2847 video

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Use the definitions of the hyperbolic functions in terms of $$e^x$$ to find expressions for when the argument of each of the 6 hyperbolic functions is negative, i.e. find an expression for $$\sinh(-x)$$, etc.

Problem Statement

Use the definitions of the hyperbolic functions in terms of $$e^x$$ to find expressions for when the argument of each of the 6 hyperbolic functions is negative, i.e. find an expression for $$\sinh(-x)$$, etc.

Solution

### 2848 video

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Use the definitions of the hyperbolic functions in terms of $$e^x$$ to show that $$\sinh^2(x) = (-1+\cosh(2x))/2$$ holds.

Problem Statement

Use the definitions of the hyperbolic functions in terms of $$e^x$$ to show that $$\sinh^2(x) = (-1+\cosh(2x))/2$$ holds.

Solution

Here are two video solutions by two different instructors.

### 2849 video

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Use the definitions of the hyperbolic functions in terms of $$e^x$$ to show that $$\cosh^2(x) = (1+\cosh(2x))/2$$ holds.

Problem Statement

Use the definitions of the hyperbolic functions in terms of $$e^x$$ to show that $$\cosh^2(x) = (1+\cosh(2x))/2$$ holds.

Solution

### 2850 video

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Use the definitions of the hyperbolic functions in terms of $$e^x$$ to show that $$\sinh(x+y) = \sinh(x) \cosh(y) + \cosh(x) \sinh(y)$$ holds.

Problem Statement

Use the definitions of the hyperbolic functions in terms of $$e^x$$ to show that $$\sinh(x+y) = \sinh(x) \cosh(y) + \cosh(x) \sinh(y)$$ holds.

Solution

### 2851 video

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Really UNDERSTAND Precalculus

 exponentials trigonometry

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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