In $\triangle ABC$, prove that $\sin \left(\dfrac{A - B}{2}\right) = \left(\dfrac{a - b}{c}\right) \cos \left(\dfrac{C}{2}\right)$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\implies$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$
In $\;$ $\triangle ABC$, $\;$ $A + B + C = \pi$
i.e. $\;$ $C = \pi - \left(A + B\right)$
$\therefore \;$ $\dfrac{C}{2} = \dfrac{\pi}{2} - \left(\dfrac{A + B}{2}\right)$
$\therefore \;$ $\cos \left(\dfrac{C}{2}\right) = \cos \left[\dfrac{\pi}{2} - \left(\dfrac{A + B}{2}\right)\right] = \sin \left(\dfrac{A + B}{2}\right)$ $\;\;\; \cdots \; (1)$
$\begin{aligned}
RHS & = \left(\dfrac{a - b}{c}\right) \cos \left(\dfrac{C}{2}\right) \\\\
& = \left(\dfrac{2 R \sin A - 2 R \sin B}{2 R \sin C}\right) \sin \left(\dfrac{A + B}{2}\right) \;\; \left[\text{by sine rule and equation } (1)\right] \\\\
& = \left(\dfrac{\sin A - \sin B}{\sin C}\right) \sin \left(\dfrac{A + B}{2}\right) \\\\
& = \dfrac{1}{\sin C} \times 2 \sin \left(\dfrac{A - B}{2}\right) \cos \left(\dfrac{A + B}{2}\right) \times \sin \left(\dfrac{A + B}{2}\right) \\\\
& \left[\text{Note: } 2 \sin \left(\dfrac{A + B}{2}\right) \cos \left(\dfrac{A + B}{2}\right) = \sin \left(A + B\right)\right] \\\\
& = \dfrac{1}{\sin C} \times \sin \left(\dfrac{A - B}{2}\right) \times \sin \left(A + B\right) \;\;\; \cdots \; (2)
\end{aligned}$
In $\;$ $\triangle ABC$, $\;$ $A + B = \pi - C$
$\therefore \;$ $\sin \left(A + B\right) = \sin \left(\pi - C\right) = \sin C$ $\;\;\; \cdots \; (3)$
$\therefore \;$ In view of equation $(3)$, equation $(2)$ becomes
$\begin{aligned}
RHS & = \dfrac{1}{\sin C} \times \sin \left(\dfrac{A - B}{2}\right) \times \sin C \\\\
& = \sin \left(\dfrac{A - B}{2}\right) = LHS
\end{aligned}$
Hence proved.