Question

Prove that:
$a(\cos B-\cos C)=2(c-b){\cos }^2\frac{A}{2}$

Answer 1

We have, for any triangle $\Delta$ ABC,

$a\cos B+b\cos A=c$......(1)

and

$a\cos C+c\cos A=b$.......(2).

Now subtracting (2) from (1) we get,

$a(\cos B-\cos C)+\cos A(b-c)=(c-b)$

or, $a(\cos B-\cos C)=(c-b)(1+\cos A)$

or, $a(\cos B-\cos C)=2(c-b){\cos }^2\frac{A}{2}$