Some combinations of \(e\)=powers are so common that they have been given special names. They are the so-called **hyperbolic functions**.

\[\begin{aligned} \sinh(x) &= \frac{\e^x-\e^{-x}}{2} \\[0.25cm] \cosh(x) &= \frac{\e^x+\e^{-x}}{2} \\[0.25cm] \tanh(x) &= \frac{\sinh(x)}{\cosh(x)} = \frac{\e^x-\e^{-x}}{\e^x+\e^{-x}} \end{aligned}\] From the definitions it immediately follows that \(\sinh\) is an odd function, \(\cosh\) is an even function, and that \(\tanh\) is an odd function. The graphs of \(\sinh\) and \(\cosh\) can be quickly sketched as the graph of the sum or difference of two exponential functions. It also follows from the definitions of \(\sinh\) and \(\cosh\) that the graphs of these functions are approaching each other for large values of \(x\).

The graph of \(\tanh\) is less easily sketched on the basis of the graphs of exponential functions and it is shown below.

The graphs above also illustrate the domain and range of the hyperpbolic functions: they are defined for all real numbers and the range of \(\sinh\), \(\cosh\) and \(\tanh\) is \((\infty,\infty)\), \([1,\infty)\) and \((-1,1)\), respectively.

Some pronounce the function names \(\sinh\) and \(cosh\) as ''shine'' and "kosh", respectively.

The naming "hyperbolic functions" suggests that these have to do with a hyperbola. Also, the names and notations of the functions resemble those of the trigonometric functions \(\sin\), \(\cos\) and \(\tan\). There is indeed a strong analogy: just as the \((x,y)\) coordinates of a point on the unit circle with equation \(x^2+y^2=1\) can be written as \(\bigl(\cos(t), \sin(t)\bigr)\), the \((x,y)\) coordinates of a point on the hyperbola with equation \(x^2-y^2=1\) can be written as \(\bigl(\cosh(t), \sinh(t)\bigr)\). But the relationship is actually more of algebraic nature, as becomes clear from the special formulas for hyperbolic functions.

The cosine hyperbolic is the function used to mathematically describe the shape of a uniform chain hanging under gravity.

Some properties of the trigonometric functions follow almost directly from the definitions.

\[\cosh^2(x) - \sinh^2(x)=1\]

\[\begin{aligned} \cosh^2(x)&= \left(\frac{\e^x+\e^{-x}}{2}\right)^2 & \blue{\text{definition of the hyperbolic function}}\\[0.25cm] &=\frac{1}{4}\left((\e^x)^2+2\cdot \e^x\cdot \e^{-x }+(\e^{-x})^2\right) &\blue{\text{expansion of special product}}\\[0.25cm] &= \frac{1}{4}\left(\e^{2x}+2\cdot \e^{x-x} +\e^{-2x}\right)& \blue{\text{rules of exponential functions}} \\[0.25cm] &= \frac{1}{4}\left(\e^{2x}+2 +\e^{-2x}\right) & \blue{\text{simplification}}\end{aligned}\] \[\begin{aligned} \sinh^2(x)&= \left(\frac{\e^x-\e^{-x}}{2}\right)^2 & \blue{\text{definition of the hyperbolic function}}\\[0.25cm] &=\frac{1}{4}\left((\e^x)^2-2\cdot \e^x\cdot \e^{-x }+(\e^{-x})^2\right) &\blue{\text{expansion of special product}}\\[0.25cm] &= \frac{1}{4}\left(\e^{2x}-2\cdot \e^{x-x} +\e^{-2x}\right)& \blue{\text{rules of exponential functions}} \\[0.25cm] &= \frac{1}{4}\left(\e^{2x}-2 +\e^{-2x}\right) & \blue{\text{simplification}}\end{aligned}\] By taking the difference of the above equations you arrive at the identity to be proved.

\[\begin{aligned}\sinh(x+y)&=\sinh(x)\cosh(y)+\cosh(x)\sinh(y)\\[0.25cm] \cosh(x+y)&=\cosh(x)\cosh(y)+\sinh(x)\sinh(y)\\[0.25cm] \tanh(x+y)&=\frac{\tanh(x)+\tanh(y)}{1+\tanh(x)\tanh(y)}\end{aligned}\]

\[\begin{aligned}\sinh(x)\cosh(y) &= \left(\frac{\e^{x}-\e^{-x}}{2}\right)\left(\frac{\e^{y}+\e^{-y}}{2}\right) & \blue{\text{definition of hyperbolic functions}}\\[0.25cm] &=\frac{1}{4}\left(\e^x\e^y+\e^x\e^{-y }-\e^{-x}\e^y-\e^{-x}\e^{-y}\right) &\blue{\text{expansion of brackets}}\\[0.25cm] &= \frac{1}{4}\left(\e^{x+y} + \e^{x-y} -\e^{y-x}- \e^{-x-y}\right)& \blue{\text{rules of exponential functions}}\end{aligned}\] \[\begin{aligned}\cosh(x)\sinh(y) &= \left(\frac{\e^{x}+\e^{-x}}{2}\right)\left(\frac{\e^{y}-\e^{-y}}{2}\right) & \blue{\text{definition of hyperbolic functions}}\\[0.25cm] &=\frac{1}{4}\left(\e^x\e^y-\e^x\e^{-y }+\e^{-x}\e^y-\e^{-x}\e^{-y}\right) &\blue{\text{expansion of brackets}}\\[0.25cm] &= \frac{1}{4}\left(\e^{x+y} - \e^{x-y} +\e^{y-x}- \e^{-x-y}\right)& \blue{\text{rules of exponential functions}}\end{aligned}\] Adding the above equations gives \[\begin{aligned}\sinh(x)\cosh(y)+\cosh(x)\sinh(y) &= \frac{\e^{x+y}-\e^{-(x+y)}}{2}\\[0.25cm] &=\sinh(x+y)\end{aligned}\]

\[\begin{aligned}\cosh(x)\cosh(y) &= \left(\frac{\e^{x}+\e^{-x}}{2}\right)\left(\frac{\e^{y}+\e^{-y}}{2}\right) & \blue{\text{definition of hyperbolic functions}}\\[0.25cm] &=\frac{1}{4}\left(\e^x\e^y+\e^x\e^{-y }+\e^{-x}\e^y-\e^{-x}\e^{-y}\right) &\blue{\text{expansion of brackets}}\\[0.25cm] &= \frac{1}{4}\left(\e^{x+y} + \e^{x-y} +\e^{y-x}+ \e^{-x-y}\right)& \blue{\text{rules of exponential functions}}\end{aligned}\] \[\begin{aligned}\sinh(x)\sinh(y) &= \left(\frac{\e^{x}-\e^{-x}}{2}\right)\left(\frac{\e^{y}-\e^{-y}}{2}\right) & \blue{\text{definition of hyperbolic functions}}\\[0.25cm] &=\frac{1}{4}\left(\e^x\e^y-\e^x\e^{-y }-\e^{-x}\e^y+\e^{-x}\e^{-y}\right) &\blue{\text{expansion of brackets}}\\[0.25cm] &= \frac{1}{4}\left(\e^{x+y} - \e^{x-y} -\e^{y-x}+ \e^{-x-y}\right)& \blue{\text{rules of exponential functions}}\end{aligned}\] Adding the above equations gives \[\begin{aligned}\cosh(x)\cosh(y)+\sinh(x)\sinh(y) &= \frac{\e^{x+y}+\e^{-(x+y)}}{2}\\[0.25cm] &=\cosh(x+y)\end{aligned}\]

\[\begin{aligned} \tanh(x+y) &= \frac{\sinh(x+y)}{\cosh(x+y)} & \blue{\text{definition of hyperbolic functions}}\\[0.25cm] &= \frac{\sinh(x)\cosh(y)+\cosh(x)\sinh(y)}{\cosh(x)\cosh(y)+\sinh(x)\sinh(y)} & \blue{\text{addition formula}}\\[0.25cm] &= \frac{\dfrac{\sinh(x)}{\cosh(x)}+\dfrac{\sinh(y)}{\cosh(y)}}{1+\dfrac{\sinh(x)}{\cosh(x)}\dfrac{\sinh(y)}{\cosh(y)}} & \blue{\text{division by }\cosh(x)\cosh(y)}\\[0.25cm]&= \frac{\tanh(x)+\tanh(y)}{1+\tanh(x)\tanh(y)} & \blue{\text{definition of hyperbolic functions}}\end{aligned}\]

\[\begin{aligned}\sinh(2x) &=2\sinh(x)\cosh(x)\\[0.25cm] \cosh(2x) &=\cosh^2(x)+\sinh^2(x) \\[0.25cm] &=2\cosh^2(x)-1 \\[0.25cm] &=2\sinh^2(x)+1\\[0.25cm] \tanh(2x)&=\frac{2\tanh(x)}{1+\tanh^2(x)}\end{aligned} \]

These are special cases of the addition formulas.

Expressed in the natural logarithm, the inverse hyperbolic functions are defined as follows: \[\begin{aligned} \mathrm{arsinh}(x) &= \ln\left(x+\sqrt{x^2+1}\right) \\[0.25cm] \mathrm{arcosh}(x) &= \ln\left(x+\sqrt{x^2-1}\right) \\[0.25cm] \mathrm{artanh}(x) &= \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\end{aligned}\] The domain of \(\mathrm{arsinh}\), \(\mathrm{arcosh}\), and \(\mathrm{artanh}\) is \(\mathbb{R}\), \([1,\infty)\), and \((-1,1)\), respectively. The graphs of the inverse hyperbolic functions are shown below. They can be constructed by reflection of the graphs of the hyperbolic functions in the line with equation \(y=x\). This can be seen in the diagrams because the graph of the corresponding hyperbolic function is also shown as a dotted line..

graph of arsinh

graph of arcosh

graph of artanh

The convention of \(\mathrm{ar}\) at the beginning of the name\(\mathrm{arsinh}\) indicates that we are dealing with area: \(\mathrm{arsinh} (y)\) is equal to the area of the region enclosed by the hyperbola with equation \(x^2-y^2=1\) and the two halflines through the origin and the points on the hyperbola with vertical coordinates \(\pm y\). The interactive diagram below illustrates that the shaded area has an area equal to \(\mathrm{arsinh}(y)\). Likewise, \(\mathrm{arcosh}(y)\) equals the area of the region enclosed by the hyperbola with equations \(x^2-y^2=1\) and the two halflines through the origin and the points on the hyperbola with horizontal coordinate \(x\).

As an example, we prove that \[\mathrm{arsinh}(x)=\ln\left(x+\sqrt{x^2+1}\right)\] Suppose \(x=\sinh(y)\), i.e., \(y=\mathrm{arsinh}(x)\), then from \[\cosh^2(y)-\sinh^2(y)=1\] and the fact that \(\cosh(y) >0\) for all \(y\) follows that \[\cosh(y)=\sqrt{x^2+1}\] So, with the definitions of \(\sinh\) and \(\cosh\) in terms of exponential functions we get: \[\begin{aligned}\ln\left(x+\sqrt{x^2+1}\right)&=\ln\bigl(\sinh(y)+\cosh (y)\bigl) \\[0.25cm] &= \ln\left(e^y\right) \\[0.25cm] &= y \\[0.25cm] &= \mathrm{arsinh}(x) \end{aligned}\]