Question

Question 6
If A, B and C are interior angles of a triangle ABC, then show that $sin\frac{(B+C)}{2}=cos\frac{A}{2}$

Answer 1

In a triangle, sum of all the interior angles is equal to 180${}^{\circ }$.
$\Rightarrow A+B+C={180}^{\circ }$
$\Rightarrow B+C={180}^{\circ }-A$
$\Rightarrow \frac{(B+C)}{2}=\frac{({180}^{\circ }-A)}{2}$
$\Rightarrow \frac{(B+C)}{2}=({90}^{\circ }-\frac{A}{2})$
$\Rightarrow sin\frac{(B+C)}{2}=sin({90}^{\circ }-\frac{A}{2})$
$\Rightarrow sin\frac{(B+C)}{2}=cos\frac{A}{2}$