Question

If A, B and C are interior angles of a triangle ABC, then what is $\sin \left(\frac{B+C}{2}\right)$ in terms of cosine trigonometric function?

Answer 1

ABC is a triangle.
So,
$\begin{array}{cc}A+B+C& ={180}^{\circ }\\ B+C& ={180}^{\circ }-A\\ \frac{B+C}{2}& =\frac{{180}^{\circ }-A}{2}\\ & =\frac{{180}^{\circ }}{2}-\frac{A}{2}\\ & ={90}^{\circ }-\frac{A}{2}\end{array}$

Taking sin on both sides,
$\begin{array}{cc}\sin \frac{B+C}{2}& =\sin ({90}^{\circ }-\frac{A}{2})\\ & =\cos \frac{A}{2}\end{array}$