Question

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

• A. True
• B. False

Answer 1

The correct option is B False

We know that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

$\therefore a=k\sin A,b=k\sin B,c=k\sin C$

Consider the RHS$=2(b+c){\cos }^2\frac{A}{2}$

$=2k(\sin B+\sin C){\cos }^2\frac{A}{2}$

$=2k\times 2\sin \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right){\cos }^2\frac{A}{2}$

$=2k\times 2\sin (\frac{\pi }{2}-\frac{A}{2})\cos \left(\frac{B-C}{2}\right){\cos }^2\frac{A}{2}$ since $A+B+C=\pi$

$=4k\cos \frac{A}{2}\cos \left(\frac{B-C}{2}\right){\cos }^2\frac{A}{2}$

$=4k{\cos }^3\frac{A}{2}\cos \left(\frac{B-C}{2}\right)\ne LHS$