Hyperbolic function are analogs of trigonometric function and they occur in the solution of many differential or cubic equations. In contrast to trigonometric functions who form a circle, hyperbolic functions relate to a hyperbola.
To demonstrate geometric representation of hyperbolic functions we’ll draw a hyperbola in Cartesian coordinate system.
The area closed with origin $O$, one point on hyperbola and intersection between hyperbola and $x$-axis (point $D$) is called hyperbolic sector. As we can see on the image above, a point on hyperbola has coordinates $(cosha,sinha)$. What exactly is $a$? It’s a real argument called hyperbolic angle. Hyperbolic sector and angle are directly connected. The size of hyperbolic angle is twice the area of its hyperbolic sector.
Unlike trigonometric functions, hyperbolic functions are defined in terms of another function – the exponential function $e^{x}$.
We’ll write a little bit about each hyperbolic function. Let’s start with two basic functions, hyperbolic sine and cosine.
Hyperbolic sine, $sinh$
Hyperbolic sine, $sinh:\mathbb{R} \rightarrow \mathbb{R}$, is monotonically increasing function defined with
$$\displaystyle{sinhx=\frac{e^{x}-e^{-x}}{2}}.$$
In speech, this function is pronounced like ‘shine’.
The graph of $sinhx$ looks like this:
To draw a graph of $sinhx$ e used the knowledge of graphing exponential functions. More details can be found in the lesson: Graphs of hyperbolic function.
Notice that hyperbolic sine is an odd function, $sinh(-x) = – sinhx$.
Hyperbolic cosine, $cosh$
Hyperbolic cosine is a function $cosh:\mathbb{R} \rightarrow [1, +\infty)$ defined with
$$\displaystyle{coshx=\frac{e^{x}+e^{-x}}{2}}.$$
In contrast to hyperbolic sine, $cosh$ is an even function, $cosh(-x)=cosh(x)$. Because of that, graph is symmetric with respect to the $y$-axis.
Derivatives of basic hyperbolic functions
Example. Find the derivate of $sinhx$.
Firstly, we’ll write the the definition of $sinhx$:
$$\displaystyle{sinhx=\frac{e^{x}-e^{-x}}{2}}$$
Since it’s a fraction, we’ll derivate it using the quotient rule:
$$\displaystyle{(sinhx)’=\left(\frac{e^{x}-e^{-x}}{2}\right)’}$$
$$\displaystyle{(sinhx)’=\frac{\left(e^{x}-e^{-x}\right)’\cdot 2 -\left(e^{x}-e^{-x}\right) \cdot (2)’}{4}}$$
$$\displaystyle{(sinhx)’=\frac{\left(e^{x}+e^{-x}\right)\cdot 2}{4}}$$
$$\displaystyle{(sinhx)’=\frac{\left(e^{x}+e^{-x}\right)}{2}}$$
$$\displaystyle{(sinhx)’=coshx}$$
Example. Find the derivate of $coshx$.
By following the same proces as above, we get
$$(coshx)’=sinhx$$
Be careful not to confuse hyperbolic and trigonometric functions!
For trigonometric functions we have:
$\displaystyle{\frac{d}{dx} sinx=cosx}$
$\displaystyle{\frac{d}{dx} cosx=-sinx}$
But for hyperbolic functions it’s:
$\displaystyle{\frac{d}{dx} sinhx=coshx}$
$\displaystyle{\frac{d}{dx} coshx=sinhx}$
Hyperbolic tangent, $tanh$
Hyperbolic tangent, $tanh: \mathbb{R} \rightarrow (-1,1)$, is monotonically increasing fucntion defined with
$$\displaystyle{tanhx=\frac{sinhx}{coshx}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}}$$
Notice that $tanhx$ is also an odd function, $tanh(-x)=-tanh(x)$.
Derivate of $tanhx$ is
$\displaystyle{\frac{d}{dx} tanhx = 1-tanh^{2}x}$
Hyperbolic cotangent, $coth$
Hyperbolic cotangent is a function $coth: \mathbb{R} \backslash \{0\} \rightarrow (-\infty,-1) \cup (1,+\infty)$ defined with
$$\displaystyle{cothx=\frac{coshx}{sinhx}=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}}$$
As we can see, hyperbolic cotangent is an odd function, meaning $coth(-x)=-coth(x)$.
Derivate of $cothx$ is
$\displaystyle{\frac{d}{dx} cothx = 1-coth^{2}x}$
Hyperbolic secant, $sech$
Hyperbolic secant is a function $sech: \mathbb{R} \rightarrow (0,1) $ defined with
$$\displaystyle{sechx=\frac{1}{coshx}=\frac{2}{e^{x}+e^{-x}}}$$
Graph looks like this:
The second even function in this family of functions is hyperbolic secant, $sech(-x)=sech(x)$. That’s because it’s directly connected to $coshx$.
Derivate of $sechx$ is
$\displaystyle{\frac{d}{dx}sechx =-tanhx \cdot sechx}$
Hyperbolic cosecant, $csch$ or $cosech$
Hyperbolic cosecant is a function $csch: (-\infty,-1) \cup (1,+\infty) \rightarrow \mathbb{R} \backslash \{0\}$
$$\displaystyle{cschx=\frac{1}{sinhx}=\frac{2}{e^{x}-e^{-x}}}$$
As we can see from the graph, it’s an odd function, meaning $csch(-x)=-csch(x)$
Derivate of $cschx$ is
$\displaystyle{\frac{d}{dx}cschx =-cothx \cdot cschx}$
Identities
The relations between hyperbolic functions are similar to those between trigonometric functions.
Since hyperbolic sine and cosine are basic functions they satisfy
$coshx+sinhx=e^{x}$
$coshx-sinhx=e^{-x}$
$cosh^{2}x-sinh^{2}x=1$
with the last one being the most important one and similar to Pythagorean trigonometric identity.
Example. Prove $cosh^{2}x-sinh^{2}x=1$.
We know that $\displaystyle{sinhx=\frac{e^{x}-e^{-x}}{2}}$ and $\displaystyle{coshx=\frac{e^{x}+e^{-x}}{2}}$ but what do $cosh^{2}x$ and $sinh^{2}x$ look like?
$\displaystyle{cosh^{2}x=\left(\frac{e^{x}+e^{-x}}{2}\right)^{2}=\frac{e^{2x}+2e^{x-x}+e^{-2x}}{4}=\frac{e^{2x}+2+e^{-2x}}{4}}$
$\displaystyle{sinh^{2}x=\left(\frac{e^{x}-e^{-x}}{2}\right)^{2}=\frac{e^{2x}-2e^{x-x}+e^{-2x}}{4}=\frac{e^{2x}-2+e^{-2x}}{4}}$
Now let’s combine them together.
$\displaystyle{cosh^{2}x – sinh^{2}x=\frac{e^{2x}+2+e^{-2x}}{4} -\frac{e^{2x}-2+e^{-2x}}{4} = \frac{e^{2x}+2+e^{-2x}-e^{2x}+2-e^{-2x}}{4}}$
When we simplify the fraction, we get $\displaystyle{cosh^{2}x – sinh^{2}x=\frac{4}{4}=1}$, and identity is proven.
For the rest of the hyperbolic functions, we have:
$tanh{x}\cdot coth{x}=1$
$sech^{2}x=1-tanh^{2}x$
$csch^{2}x=coth^{2}x-1$
Example. Given $sinhx=\displaystyle{\frac{12}{35}}$ find $coshx$.
The first thing that comes to our mind is the definition of $sinhx$
$\displaystyle{\frac{e^{x}-e^{-x}}{2} = \frac{12}{35}}$.
By solving this equation we get
$\displaystyle{e^{x} = \frac{24}{35}} + e^{-x}$
By doing that, we can see we don’t get much. We could continue with substitution, but it is a more difficult path. We know there is an identity that connects $sinhx$ and $cosh$, let’s try that.
$cosh^{2}x=1+sinh^{2}x$
$= 1+ \displaystyle{\left(\frac{e^{x}-e^{-x}}{2}\right)^{2}}$
$=1+\displaystyle{\left(\frac{12}{35}\right)^{2}}$
$=\displaystyle{\frac{35^{2}+12^{2}}{35^{2}}}$
$coshx=\displaystyle{\sqrt{\frac{35^{2}+12^{2}}{35^{2}}}}$
Finally, we by taking the square root of the expression, we get:
$coshx=\displaystyle{\frac{37}{35}}$
Remember, $coshx$ is always positive, that’s why we only took the positive square root.
Example. Suppose $sinhx=\displaystyle{\frac{3}{4}}$. Find $x$.
At the first glance, our first step isn’t obvious. Since we have $3$ identities which connect $sinh$ and $cosh$, let’s find $coshx$.
From $cosh^{2}x=1+sinh^{2}x$ and because $coshx$ is always positive, we get that $coshx=\displaystyle{\frac{5}{4}}$ like in the example above.
By looking at the rest of the identities, we can notice that identity $coshx+sinhx=e^{x}$ is the one we need to use to get $x$. We just substitute $sinhx$ and $coshx$ with their values and use logarithm to get $x$.
$coshx+sinhx=e^{x}$
$\displaystyle{\frac{3}{4} + \frac{5}{4} = e^x}$
By adding the fractions we get
$2=e^x$
Now just use natural logarithm to get $x$:
$x=ln2$
Addition and subtraction of arguments
$sinh(x \pm y)=sinhxcoshy \pm coshxsinhy$
$cosh(x\pm y)=coshxcoshy \pm sinhxsinhy$
$tanh(x+y)=\displaystyle{\frac{tanhx+tanhy}{1+tanhxtanhy}}$
Example. Let’s prove $cosh(x- y)=coshxcoshy – sinhxsinhy$
Let’s use definitions of $sinh$ and $cosh$ to see what do products on the right side look like.
$coshxcoshy=\displaystyle{\frac{e^{x}+e^{-x}}{2} \cdot \frac{e^{y}+e^{-y}}{2} = \frac{e^{x+y}+e^{x-y}+e^{-x+y}+e^{-x-y}}{4}}$
And we do the same for $sinhxsinhy$
$sinhxsiny=\displaystyle{\frac{e^{x}-e^{-x}}{2} \cdot \frac{e^{y}-e^{-y}}{2} = \frac{e^{x+y}-e^{x-y}-e^{-x+y}+e^{-x-y}}{4}}$
Now by substracting these expressions, we get the identity $cosh(x-y)$
$coshxcoshy-sinhxsiny=\displaystyle{\frac{e^{x+y}+e^{x-y}+e^{-x+y}+e^{-x-y}}{4} -\frac{e^{x+y}-e^{x-y}-e^{-x+y}+e^{-x-y}}{4}}$
$=\displaystyle{\frac{e^{x+y}+e^{x-y}+e^{-x+y}+e^{-x-y}-e^{x+y}+e^{x-y}+e^{-x+y}-e^{-x-y}}{4}}$
$=\displaystyle{\frac{2 \cdot e^{x-y}+2 \cdot e^{-(x-y)}}{4}}$
$=cosh(x-y)$
$cosh(x+y)$ and $sinh(x \pm y)$ are proven analogously.
Addition and subtraction formulas
$sinhx+sinhy=\displaystyle{2sinh\left(\frac{x+y}{2}\right) cosh\left (\frac{x-y}{2}\right)}$
$coshx+coshy=\displaystyle{2cosh\left(\frac{x+y}{2}\right) cosh\left (\frac{x-y}{2}\right)}$
$sinhx-sinhy=\displaystyle{2cosh\left(\frac{x+y}{2}\right) sinh\left (\frac{x-y}{2}\right)}$
$coshx-coshy=\displaystyle{2sinh\left(\frac{x+y}{2}\right) sinh\left (\frac{x-y}{2}\right)}$
Double and half angle formulas
$cosh(2x)=sinh^{2}x+cosh^{2}x=2sinh^{2}x+1=2cosh^{2}x-1$
$sinh(2x)=2sinhsxcoshx$
$sinh\displaystyle{\left(\frac{x}{2}\right)=\frac{sinhx}{\sqrt{2(coshx+1)}}}$
$cosh\displaystyle{\left(\frac{x}{2}\right)=\sqrt{\frac{coshx+1}{2}}}$
$tanh\displaystyle{\left(\frac{x}{2}\right)=\frac{sinhx}{coshx+1}}$
But also if $x\ne 0$
$tanh\displaystyle{\left(\frac{x}{2}\right)=\frac{coshx-1}{sinhx}} = cothx – cschx$
