Hyperbolic and Inverse 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.
\(\displaystyle{ \sinh(x) = \frac{e^xe^{x}}{2} }\) 
\(\displaystyle{ \cosh(x) = \frac{e^x+e^{x}}{2} }\)  
\(\displaystyle{ \csch(x) = \frac{1}{\sinh(x)} }\) 
\(\displaystyle{ \sech(x) = \frac{1}{\cosh(x)} }\)  
\(\displaystyle{ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} }\) 
\(\displaystyle{ \coth(x) = \frac{\cosh(x)}{\sinh(x)} }\) 
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.
video by PatrickJMT 

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.
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, work these practice 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.
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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 

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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 

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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 

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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.
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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 

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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 

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Inverse Hyperbolic Functions
As you probably expect, we can define inverse hyperbolic functions. Some of the hyperbolic functions require a restricted domain since only onetoone functions can have inverses. (For more detail, see the inverse functions page.)
Let's think about what the equations might look like. Since the hyperbolic functions are defined in terms of \(e^x\), we would except the inverse hyperbolic functions to be some form involving \(\ln(x)\). This is indeed the case. Here is a table listing the inverse hyperbolic functions.
\(\arcsinh(x) = \ln\left[ x + \sqrt{x^2+1} \right]\) 
Domain: all real numbers  
\(\arccosh(x) = \ln\left[ x + \sqrt{x^21} \right]\) 
Domain: \( [1, +\infty) \)  
\(\displaystyle{ \arctanh(x) = \frac{1}{2} \ln\left[ \frac{1+x}{1x} \right] }\) 
Domain: \( (1,1) \)  
\(\displaystyle{ \arccoth(x) = \frac{1}{2}\ln\left[ \frac{x+1}{x1} \right] }\) 
Domain: \( (\infty, 1) \cup ( 1, \infty ) \)  
\(\displaystyle{ \arcsech(x) = \ln \left[ \frac{1}{x} + \sqrt{\frac{1}{x^2}1} \right] }\) 
Domain: \((0,1]\)  
\(\displaystyle{ \arccsch(x) = \ln \left[ \frac{1}{x} + \sqrt{\frac{1}{x^2}+1} \right] }\) 
Domain: all real numbers except for zero  
This GeoGebra page shows the graphs of the inverse hyperbolic functions. 

This Wikipedia page contains more information about inverse hyperbolic functions. 
To get used to working with inverse hyperbolic functions, work these practice problems.
Use the definitions of the hyperbolic function \(\sinh(x)\) in terms of \(e^x\) to get the equation for \(\text{arcsinh}(t)\) involving the natural logarithm.
Problem Statement 

Use the definitions of the hyperbolic function \(\sinh(x)\) in terms of \(e^x\) to get the equation for \(\text{arcsinh}(t)\) involving the natural logarithm.
Solution 

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Use the definitions of the hyperbolic function \(\cosh(x)\) in terms of \(e^x\) to get the equation for \(\text{arccosh}(t)\) involving the natural logarithm.
Problem Statement 

Use the definitions of the hyperbolic function \(\cosh(x)\) in terms of \(e^x\) to get the equation for \(\text{arccosh}(t)\) involving the natural logarithm.
Solution 

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Use the definitions of the hyperbolic function \(\tanh(x)\) in terms of \(e^x\) to get the equation for \(\text{arctanh}(t)\) involving the natural logarithm.
Problem Statement 

Use the definitions of the hyperbolic function \(\tanh(x)\) in terms of \(e^x\) to get the equation for \(\text{arctanh}(t)\) involving the natural logarithm.
Solution 

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Here is a playlist of the videos on this page.
Really UNDERSTAND Precalculus
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  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle 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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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