# A calculus of the absurd

#### 16.3 Relationship to trig functions

Using Euler’s formula, it is possible to write both $$\cos (\theta )$$ and $$\sin (\theta )$$ in terms of $$e^{x}$$. As $$e^{i\theta } = \cos (\theta ) + i\sin (\theta )$$, and $$e^{-ix} = \cos (-\theta ) + i\sin (-\theta ) = \cos (\theta ) - i\sin (\theta )$$ we can either add or subtract these two quantities in order to write both trigonometric functions in terms of $$e$$.

For cos, we can add $$e^{ix}$$ and $$e^{-ix}$$.

\begin{align} e^{ix} + e^{-ix} &= \cos (\theta ) - i\sin (\theta ) + \cos (\theta ) + i\sin (\theta ) \\ &= 2\cos (\theta ) \end{align}

Thus we can say that

\begin{equation} \cos (x) = \frac {e^{ix} + e^{-ix}}{2} \end{equation}

for all values of x. 105105 Which looks remarkably like a hyperbolic function!.

We can do a similar thing for $$\sin (x)$$.

\begin{align} e^{ix} - e^{-ix} &= \cos (\theta ) - i\sin (\theta ) - (\cos (\theta ) + i\sin (\theta )) \\ &= -2i\sin (\theta ) \end{align}

Which means that

\begin{equation} \sin (x) = \frac {e^{ix} - e^{-ix}}{-2i} \end{equation}

If, in either of these equations we set $$x=i\theta$$ we can write $$\sin (i\theta )$$ and $$\cos (i\theta )$$ in terms of $$\sinh (x)$$ and $$\cosh (x)$$, respectively.

\begin{align} \sin (i\theta ) &= \frac {e^{i(i\theta )} - e^{-i(i\theta )}}{-2i} \\ &= \frac {e^{-\theta } - e^{\theta }}{-2i} \\ &= i\frac {e^\theta - e^{-\theta }}{2} \text { multiplying by $\frac {i}{i}=1$} \\ &= i\sinh (\theta ) \end{align}

\begin{align} \cos (i\theta ) &= \frac {e^{i(i\theta )} + e^{-i(i\theta )}}{2} \\ &= \frac {e^{-\theta } + e^{\theta }}{2} \\ &= \cosh (\theta ) \end{align}