Given:
\[ \mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \frac{A}{2}} \]
Using the relation:
\[ \cos \frac{A}{2} = \sin \left( \frac{A + \delta_m}{2} \right) \]
We get:
\[ \delta_m = \pi - 2A \]
Therefore:
\[ \delta_m = 180^\circ - 2A \]