Sine minus Sine/Proof 1

Theorem

$\sin \alpha - \sin \beta = 2 \map \cos {\dfrac {\alpha + \beta} 2} \map \sin {\dfrac {\alpha - \beta} 2}$

$\text {(1)}: \quad$

$\ds \map \sin {A + B}$

$=$

$\ds \sin A \cos B + \cos A \sin B$

$\text {(2)}: \quad$

$\ds \map \sin {A - B}$

$=$

$\ds \sin A \cos B - \cos A \sin B$

$\ds \leadsto \ \ $

$\ds \map \sin {A + B} - \map \sin {A - B}$

$=$

$\ds 2 \cos A \sin B$

subtracting $(2)$ from $(1)$

$\ds \leadsto \ \ $

$\ds \sin \alpha - \sin \beta$

$=$

$\ds 2 \map \cos {\dfrac {\alpha + \beta} 2} \map \sin {\dfrac {\alpha - \beta} 2}$

setting $A + B = \alpha$ and $A - B = \beta$

$\blacksquare$