How does this alternate form come about from Eulers identity?

Question

I don't understand how this identity comes about, my tutor wrote it out and I checked it on wolfram so i know its not a mistake and now I'd like to understand how to derive it.

$sin(\theta)cos(\phi) -isin(\theta)sin(\phi) = sin(\theta)e^{-i\phi}$

I know Eulers identity can be used to substitute

$e^{-i\phi} = cos(\phi) - i sin(\phi)$

but what on earth happens to the $ isin(\theta)$? Why isn't the result $(sin(\theta) -isin(\theta))e^{-i\phi}$

Answer 1

Let $\color{red}A = \color{red}{\sin \theta}$ then

$\color{red}{\sin \theta}\cos(\phi) -i\color{red}{\sin \theta}\sin(\phi)=$

$\color{red}A\cos\phi - \color{red}Ai\sin \phi = $

$\color{red}A(\cos \phi - i\sin \phi) =$

$\color{red}A(\cos (-\phi) + i\sin (-\phi))=$

$\color{red}A e^{-i\phi} =$

$\color{red}{\sin \theta} e^{-\phi}$.

but what on earth happens to the isin(θ)?

The $\sin \theta$ factored out. ANd the $i\sin \phi$ get incorporated into the conversion to polar coordinates.

Why isn't the result (sin(θ)−isin(θ))e−iϕ

Because $\cos \phi \ne \sin \phi \ne e^{-i\phi}$ and you can't factor out unlike terms.