# Step By Step Calculus » 7.2 - Hyperbolic Functions

Synopsis
Hyperbolic functions are analogous to the trigonometric functions. Whereas the definitions of sine and cosine are based on the unit circle x^2+y^2=1x^2+y^2=1, the definitions of the basic hyperbolic functions are based on (the right hand side of) the unit hyperbola x^2-y^2=1x^2-y^2=1. The trigonometric functions \sin(x)\sin(x) and \cos(x)\cos(x) have their hyperbolic cousins, the hyperbolic sine, written \sinh(x)\sinh(x), and the hyperbolic cosine, written \cosh(x)\cosh(x).
Hyperbolic functions are the composition of rational and exponential functions. The following table gives the definition of each, shows how each hyperbolic function is a composite function f\circ g(x)f\circ g(x) and provides the domain and range of each.
$\begin{array}{|c|c|c|c|c|c|} \hline \textrm{Function} & f\circ g(x) & f(x) & g(x) & \textrm{Domain of } f\circ g(x) & \textrm{Range of } f\circ g(x)\\ \hline \sinh x & \displaystyle\frac{e^x-e^{-x}}{2} & \displaystyle\frac{x-\frac{1}{x}}{2}& e^x &\mathbb{R} & \mathbb{R} \\ \hline \cosh x & \displaystyle\frac{e^x+e^{-x}}{2} & \displaystyle\frac{x+\frac{1}{x}}{2}& e^x & \mathbb{R} & [1,\infty ) \\ \hline \displaystyle\tanh x=\frac{\sinh x}{\cosh x} & \displaystyle\frac{e^x-e^{-x}}{e^x+e^{-x}} & \displaystyle\frac{x-\frac{1}{x}}{x+\frac{1}{x}} & e^x & \mathbb{R} & \mathbb{(-}1,1) \\ \hline \displaystyle\coth x=\frac{\cosh x}{\sinh x} & \displaystyle\frac{e^x+e^{-x}}{e^x-e^{-x}} & \displaystyle\frac{x+\frac{1}{x}}{x-\frac{1}{x}} & e^x & \mathbb{R}-\{0\} & x\notin[-1,1] \\ \hline \displaystyle \func{sech} x=\frac{1}{\cosh x} & \displaystyle\frac{2}{e^x+e^{-x}} & \displaystyle\frac{2}{x+\frac{1}{x}}& e^x &\mathbb{R} & (0,1] \\ \hline \displaystyle \func{csch} x=\frac{1}{\sinh x} & \displaystyle\frac{2}{e^x-e^{-x}} & \displaystyle\frac{2}{x-\frac{1}{x}}& e^x & \mathbb{R}-\{0\} & \mathbb{R} \\ \hline \end{array}$
The hyperbolic functions satisfy a list of identities which parallels the list of standard identities for the trigonometric functions. These hyperbolic identities are easier to prove than the corresponding trig identities because we can make direct use of the representations given in the above table:
\displaystyle \newcommand{\func}[1]{\mathop{\mathrm{#1}}}\begin{array}{l} \textrm{1. }\cosh ^{2}t-\sinh ^{2}t=1 \\ \textrm{2. }1-\tanh ^{2}t=\func{sech}^{2}t \\ \textrm{3. }\coth ^{2}t-1=\func{csch}{}^{2}t \\ \textrm{4. }\cosh \left( -t\right) =\cosh t,\quad\textrm{i.e. $\cosh$ is even}\\ \textrm{5. }\sinh \left( -t\right) =-\sinh t,\quad\textrm{i.e. $\sinh$ is odd.} \\ \textrm{6. }\cosh (t\pm u)=\cosh (t)\cosh (u)\pm \sinh (t)\sinh (u) \\ \textrm{7. }\sinh (t\pm u)=\sinh (t)\cosh (u)\pm \cosh (t)\sinh (u) \\ \textrm{8. }\cosh \left( 2t\right) =\cosh ^{2}(t)+\sinh ^{2}(t)=1+2\sinh ^{2}(t)=2\cosh ^{2}(t)-1 \\ \textrm{9. }\sinh \left( 2t\right) =2\sinh (t)\cosh (t) \\ \end{array}
\displaystyle \newcommand{\func}[1]{\mathop{\mathrm{#1}}}\begin{array}{l} \textrm{1. }\cosh ^{2}t-\sinh ^{2}t=1 \\ \textrm{2. }1-\tanh ^{2}t=\func{sech}^{2}t \\ \textrm{3. }\coth ^{2}t-1=\func{csch}{}^{2}t \\ \textrm{4. }\cosh \left( -t\right) =\cosh t,\quad\textrm{i.e. $\cosh$ is even}\\ \textrm{5. }\sinh \left( -t\right) =-\sinh t,\quad\textrm{i.e. $\sinh$ is odd.} \\ \textrm{6. }\cosh (t\pm u)=\cosh (t)\cosh (u)\pm \sinh (t)\sinh (u) \\ \textrm{7. }\sinh (t\pm u)=\sinh (t)\cosh (u)\pm \cosh (t)\sinh (u) \\ \textrm{8. }\cosh \left( 2t\right) =\cosh ^{2}(t)+\sinh ^{2}(t)=1+2\sinh ^{2}(t)=2\cosh ^{2}(t)-1 \\ \textrm{9. }\sinh \left( 2t\right) =2\sinh (t)\cosh (t) \\ \end{array}