I understand the fact:
$\displaystyle \frac{e^{x}+e^{-x}}{2} = \mathrm{cosh}x$
But why does:
$\displaystyle \frac{e^{jx}+e^{-jx}}{2} = \cos x$?
The reason is due to Euler's Formula:
$\displaystyle e^{jx}=\cos(x)+j\sin(x)$
$\displaystyle e^{-jx}=\cos(-x)+j\sin(-x)=\cos(x)-j\sin(x)$ [Due to the even and odd properties of sine and cosine]
Thus, $\displaystyle \frac{e^{jx}+e^{-jx}}{2}=\frac{\cos(x)+j\sin(x)+\cos(x)-j\sin(x)}{2}=\frac{2\cos(x)}{2}=\color{red}\boxed{ \cos(x)}$
Does this make sense?
--Chris