Wait ! How can $i$ be inside?

$cos(x)=1-\frac{x^2}2+\frac{x^4}{24}-\cdots$ $cos(ix)=1+\frac{x^2}2+\frac{x^4}{24}+\cdots$

We can use a fact that $cos (xi) = cosh (x)$ .

$cosh (x)=\frac {e^{x}+e^{-x}}{2}$ is always real for real $x$ .

Proof :

Using Euler's Identity we obtain that $e^x=e^{i(-ix)}= \cos (-ix) + i \sin (-ix) = \cos (ix) - i \sin (ix)$

Aplying the above to the definitionn of $\cosh x$ :

$\cosh x = \frac {e^{x}+e^{-x}}{2} = \frac{ \cos (ix) - i \sin (ix) + \cos (-ix) - i \sin (-ix)}{2} = \frac{ 2\cos (ix)}{2} = \cos (ix)$

e^(iy) = cos(y) + I sin(y). letting y = ix, e^i(ix) cos(ix) + I sin(ix). But e^i(ix) = e^(-x), which is real. Ed Gray