i was looking at an example and it stated this:
$e^{i(α+β)} = (\cosα\cosβ−\sinα\sinβ)+i(\sinα\cosβ+\cos α\sin β)$
using de moivre's method: $\cos(α+β)+i\sin(α+β)=(\cosα\cosβ−\sinα\sinβ)+i(\sinα\cosβ+\cosα\sinβ)$
i am a bit unsure of how $e^{i(α+β)} = \cos(α+β)+i\sin(α+β)$ and am looking for a point in the right direction. thanks.
It is because $$e^{ix}=\cos x+i\sin x.$$
Indeed observe what happens when you expand out both sides of the identity $$e^{i(x+y)}=e^{ix} e^{iy}.$$