Question

If A, B and C are interior angles of $\Delta ABC,$ then Prove that : $sin\frac{(A+C)}{2}=cos\frac{B}{2}.$

Answer 1

In $\Delta ABC$
We know that, Sum of angles of a triangle is ${180}^{\circ }$
$\angle A+\angle B+\angle C={180}^{\circ }$
$\angle A+\angle C={180}^{\circ }-\angle B$
Dividing by 2 on both sides
$\frac{\angle A+\angle C}{2}=\frac{{180}^{\circ }-\angle B}{2}$
$\frac{\angle A+\angle C}{2}={90}^{\circ }-\frac{\angle B}{2}$
$sin\left(\frac{\angle A+\angle C}{2}\right)=sin({90}^{\circ }-\frac{\angle B}{2})$
$sin\left(\frac{\angle A+\angle C}{2}\right)=cos\frac{\angle B}{2}$
$sin\left(\frac{\angle A+\angle C}{2}\right)=cos\frac{B}{2}$
Hence Proved.