As you know, hyperbolic functions are defined in terms of $e$. For example, the hyperbolic cosine:
$$\cosh x = \frac{e^x + e^{-x}}{2}$$
There is a result in complex analysis that looks uncannily similar.
If $z=1$, we can write it as a complex number in modulus-argument form: $\cos\theta + i\sin\theta$
If we then raise $z$ to an integer power $n$ and add to it $z^{-n}$, we get:
\begin{align}
z^n + z^{-n} &= (\cos\theta + i\sin\theta)^n + (\cos\theta + i\sin\theta)^{-n} \\
&= (\cos n\theta + i\sin n\theta) + (\cos (-n\theta) + i\sin (-n\theta)) && \text{Using de Moivre's theorem}\\
&= \cos n\theta + i\sin n\theta + \cos n\theta – i\sin n\theta \\
&= 2\cos n\theta
\end{align}
Using $n\theta=x$, the cosine can be exressed in terms of the sum of two complex numbers in the exponential form:
$$\cos x = \frac{e^{ix} + e^{-ix}}{2}$$
How does the hyperbolic cosine relate to the normal cosine? Is it possible to elegantly derive one from another, by somehow cancelling out the $i$s from the powers of $e$?
Best Answer
They are related by
$$\cos (ix)=\frac{e^{i^2x} + e^{-i^2x}}{2}=\frac{e^{x} + e^{-x}}{2}=\cosh x$$
$$\sin (ix)=\frac{e^{i^2x} - e^{-i^2x}}{2i}=i\frac{e^{x} - e^{-x}}{2}=i\sinh x$$
and also
$$\cosh (ix)=\frac{e^{ix} + e^{-ix}}{2}=\cos x$$
$$\sinh (ix)=\frac{e^{ix} - e^{-ix}}{2}=i\frac{e^{x} - e^{-x}}{2i}=i\sin x$$