Connecting Trigonometric Functions With Hyperbolic Functions Using Complex Numbers

Using the complex exponential, we can identify the relation between the trigonometric functions ($\sin ,\cos ,\tan$) and their hyperbolic equivalent ($\sinh ,\cosh ,\tanh$)

Sin & Sinh §

We can use the definition of the complex exponential to transform $\sin$ to $\sinh$

$\{\begin{array}{ll}{e}^{i\theta }=\cos (\theta )+i\sin (\theta )& (1)\\ e^{-i\theta }=\cos \theta -i\sin (\theta )& (2)\end{array}$

$(1)-(2)\to e^{i\theta }-e^{-i\theta }$ $=\cos (\theta )-\cos (\theta )+i\sin (\theta )+i\sin (\theta )=2i\sin (\theta )$ $2i\sin (\theta )=e^{i\theta }-e^{-i\theta }$ $i\sin (\theta )=\frac{1}{2}(e^{i\theta }-e^{-i\theta })$ $\sin (\theta )=\frac{1}{2}(e^{i\theta }-e^{-i\theta })$

$i\sin (\theta )=\sinh (i\theta )$

Cos & Cosh §

A similar approach can be used for cos and cosh:

$\{\begin{array}{ll}{e}^{i\theta }=\cos (\theta )+i\sin (\theta )& (1)\\ e^{-i\theta }=\cos \theta -i\sin (\theta )& (2)\end{array}$

$(1)+(2)\to e^{i\theta }+e^{-i\theta }$ $=\cos (\theta )+\cos (\theta )+i\sin (\theta )-i\sin (\theta )=2\cos (\theta )$ $2\cos (\theta )=e^{i\theta }+e^{-i\theta }$ $\cos (\theta )=\frac{1}{2}(e^{i\theta }+e^{-i\theta })$

$cos(\theta )=\cosh (i\theta )$

Tan & Tanh §

We can combine the methods used to obtain sinh-sin and cosh-cos to get the relationship between tanh and tan.

$\{\begin{array}{ll}{e}^{i\theta }=\cos (\theta )+i\sin (\theta )& (1)\\ e^{-i\theta }=\cos \theta -i\sin (\theta )& (2)\end{array}$

$\frac{(1)-(2)}{(1)+(2)}=\frac{e^{i\theta }-e^{-i\theta }}{e^{i\theta }+e^{-i\theta }}=\tanh (i\theta )$

But we also know that:

$\frac{e^{i\theta }-e^{-i\theta }}{e^{i\theta }+e^{-i\theta }}=\frac{2i\sin (\theta )}{2\cos (\theta )}=i\tan (\theta )$

Hence:

$i\tan (\theta )=\tanh (i\theta )$