Question

For a triangle ABC prove that a (cosC-cosB) = 2(b-c)cos^2A/2?

Answer 1

Then, 

a = k sinA, b = k sinB, c = k sinC

RHS

b-c/a cos A/2

= (ksinB - ksinC/k sinA )cos A/2

= {[2 cos B+C/2 Sin B-C/2]/ Sin A} cos A/2

=[2* Sin B-C/2 cos (pi/2-A/2) .cos A/2] / sinA

= [SinA* sinB-C/2] / sinA

= Sin B-C/2

= LHS