question_answer

In A, B and C are interior angles of \[\Delta \text{ }ABC\], then  prove that: \[\sin \frac{(A+C)}{2}=\cos \frac{B}{2}\].

Answer:

In \[\Delta \text{ }ABC\]

\[\angle A+\angle B+\angle C=180{}^\circ \]

\[\angle A+\angle C=180{}^\circ -\angle B\]

Divide by 2 on both sides

\[\frac{\angle A+\angle C}{2}=\frac{180{}^\circ -\angle B}{2}\]

\[\frac{\angle A+\angle C}{2}=90{}^\circ -\frac{\angle B}{2}\]

\[\sin \left( \frac{\angle A+\angle C}{2} \right)=\sin \left( 90{}^\circ -\frac{\angle B}{2} \right)\]

\[\sin \left( \frac{\angle A+\angle C}{2} \right)=\cos \frac{\angle B}{2}\]

\[\sin \frac{(A+c)}{2}=\cos \frac{B}{2}\]            Hence Proved.