A neat proof.

From Euler's formula $e^{i\theta} = \cos \theta + i\sin \theta$ , then $z=e^{i\alpha}$ and $w = e^{i\beta}$ and $zw = e^{i\alpha} e^{i\beta} = e^{i(\alpha+\beta)} = \cos (\alpha+\beta) + i\sin (\alpha+\beta) = A+iB$ .

$\boxed{\text{Both } A\text{ and } B}$ are parts of the product $zw$ .

Thanks to Euler we know, that: ${ e }^{ i\cdot x }=cis\left( x \right) =\cos { x } +i\cdot \sin { x }$ And therefore we get: $z\cdot w={ e }^{ i\cdot \alpha }\cdot { e }^{ i\cdot \beta }={ e }^{ i\cdot \left( \alpha +\beta \right) }=\cos { \left( \alpha +\beta \right) } +i\cdot \sin { \left( \alpha +\beta \right) }$ But there is another way to calculate the product $z\cdot w$ : $\begin{aligned}\sigma =z\cdot w=cis\left( \alpha \right) \cdot cis\left( \beta \right) &=\left( \cos { \alpha } +i\cdot \sin { \alpha } \right) \cdot \left( \cos { \beta } +i\cdot \sin { \beta } \right) \\ &=\left[ \cos { \alpha } \cdot \cos { \beta } -\sin { \alpha } \cdot \sin { \beta } \right] +i\cdot \left[ \cos { \alpha } \cdot \sin { \beta } -\sin { \alpha } \cdot \cos { \beta } \right]\end{aligned}$ We know, that both products have to be the same. $z\cdot w = \sigma$ if and only if $Re\left( z\cdot w \right) =Re\left( \sigma \right)$ and $Im\left( z\cdot w \right) =Im\left( \sigma \right)$ . Hence we get: $\begin{aligned}Re\left( \sigma \right) &=\boxed { \cos { \left( \alpha +\beta \right) } =\cos { \alpha } \cdot \cos { \beta } -\sin { \alpha } \cdot \sin { \beta } } =Re\left( z\cdot w \right) \\ Im\left( \sigma \right) &=\boxed { \sin { \left( \alpha +\beta \right) } =\cos { \alpha } \cdot \sin { \beta } -\sin { \alpha } \cdot \cos { \beta } } =Im\left( z\cdot w \right)\end{aligned}$

We have $z=e^{iα}$ and $w=e^{iβ}$ . So $z.w=(\cos {α}+i\sin {α}).(\cos {β}+i\sin {β})=e^{i(α+β)}=\cos {(α+β)}+i\sin {(α+β)}$ . Expanding the cosine and sine terms and comparing the two sides, we get both the identities.