I must be messing up with my algebra somewhere but I don't understand where and I can't figure out what is going on:
I defined a function $f(a,b): \mathbb{R}^{2} \rightarrow \mathbb{C}$
$\begin{array}{l}{f(a, b):=\frac{1}{2}\left(e^{(a+b i)}+e^{-(a+b i)}\right)=\frac{1}{2}\left(e^{a} e^{b i}+e^{-a} e^{-b i}\right)} \\ {=\frac{1}{2}\left(e^{a}(\cos (b)+i \sin (b))+e^{-\alpha}(\cos (-b)+i \sin (-b))\right)} \\ {=\frac{1}{2}\left(e^{a} \cos (b)+ie^{a} \sin (b)+e^{-a} \cos (b)-i e^{-a} \sin (b) \right)} \\ {=\frac{1}{2}\left(\cos (b)\left(e^{a}+e^{-a}\right)+i \sin \left(e^{a}-e^{-a}\right)\right)} \\ {=\cosh (a) \cos (b)+i \sinh (a) \sin (b)}\end{array}$
Now, the problem is when I try to input $0$ for $a$ and $x-\frac{\pi}{2}$ for $b$. The initial definition of $f$ seems to give a different answer from the derived last line:
$\begin{aligned} f\left(0,\left(x-\frac{\pi}{2}\right)\right)=\frac{1}{2}\left(e^{\left(x-\frac{\pi}{2}\right) i}+e^{-\left(x-\frac{\pi}{2}\right) i}\right) &=\frac{1}{2}\left(e^{i x}(-i)+e^{-ix}(i)\right) \\ &=-i \sin (x) \end{aligned}$
$\begin{aligned} f\left(0,\left(x-\frac{\pi}{2}\right)\right) &=\cosh (0) \cos \left(x-\frac{\pi}{2}\right)+i \sinh (0) \sin \left(x-\frac{\pi}{2}\right) \\ &=\sin (x)+0 \end{aligned}$
Leading to $\sin(x)=-i\sin(x)$. Like I said, I don't know where I went wrong with the algebra. Is nothing really wrong with the algebra - is this just caused by some technical issue with defining complex functions?
You should have $\dfrac12 (e^{ix}(-i)+e^{-ix}{i})=\dfrac{e^{ix}-e^{-ix}}{2i}=\sin x$.