I am trying to prove the following
$$\frac{a - b}{b} = \frac{2 \sin {\frac{1}{2}C} \sin {\frac{1}{2}(A - B)}}{\sin{B}}$$
with this approach:
From law of sines
$$\frac{c}{\sin {C}} = \frac{a}{\sin{A}},\qquad \frac{c}{\sin{C}} = \frac{b}{\sin{B}}$$ $$ \implies \qquad \frac{a}{c} = \frac{\sin{A}}{\sin{C}}, \qquad \frac{b}{c} = \frac{\sin{B}}{\sin{C}} \qquad \qquad $$ $$\implies \qquad \frac{\frac{a - b}{c}}{\frac{b}{c}} \quad = \quad \frac{\frac{\sin{A} - \sin{B}}{\sin{C}}}{\frac{\sin{B}}{\sin{C}}} \qquad \qquad$$ $$\implies \qquad \frac{a-b}{b} \quad = \quad \frac{\sin{A}-\sin{B}}{\sin{B}} \qquad $$
then,
$$\frac{a-b}{b} \quad = \quad \frac{2 \cos{\frac{1}{2}(A + B)} \sin {\frac{1}{2}(A - B)} }{\sin{B}}. \qquad (1)$$
After arriving at this equation, I no longer knew how to proceed to the next steps. Help is greatly appreciated.
You're almost there $$ \cos \left [ \frac 12 (A+B) \right ] = \cos \left [ \frac 12 (\pi - C) \right ] = \sin \left [ \frac 12 C \right ] $$