Question

In a triangle $ABC,$ write $\cos \left(\frac{B+C}{2}\right)$ in terms of angle $A$

Answer 1

$\begin{array}{c}Weknowthat\\ A+B+C={180}^o\left[allangleofanytriangleis{180}^{\circ }\right]\\ \Rightarrow B+C={180}^o-A\\ \Rightarrow \cos \left(\frac{B+C}{2}\right)2=\frac{{180}^o-A}{2}\\ \cos \left(\frac{{180}^o}{2}-\frac{A}{2}\right)\\ =\cos \left({90}^o-\frac{A}{2}\right)\\ =\sin \frac{A}{2}\\ Hence,\\ \frac{\cos \left(B+C\right)}{2}=\sin \frac{A}{2}.\end{array}$