Question

Prove:

$a\left(\cos B+\cos C\right)=2\left(b+c\right){\sin }^2\frac{A}{2}$

Answer 1

Given that,

$\begin{array}{c}a\left(\cos B+\cos C\right)=2\left(b+c\right){\sin }^2\frac{A}{2}\\ R.H.S=2\left(b+c\right){\sin }^2\frac{A}{2}\\ =2k\left(\sin B+sinc\right).{\sin }^2\frac{A}{2}\\ =2k2\sin \left(\frac{B+C}{2}\right).cos\left(\frac{B-C}{2}\right){\sin }^2\frac{A}{2}\\ =4k.\sin \left(\frac{B+C}{2}\right).cos\left(\frac{B-C}{2}\right).{\cos }^2\left(\frac{B+C}{2}\right)\\ =2k.2\sin \left(\frac{B+C}{2}\right).cos\left(\frac{B-C}{2}\right).\cos \left(\frac{B-C}{2}\right).\cos \left(\frac{B+C}{2}\right)\\ =k\sin \left(B+C\right).\left(\cos B+\cos C\right)\\ =k\sin A\left(\cos B+\cos C\right)\\ =a\left(\cos B+\cos C\right)\\ =L.H.S\\ proved.\end{array}$