Question

In any Δ prove that 1. asin(A/2+B)=(b+c)sinA/2 2. a 2 sin(B-C)/sinB+sinC +b 2 sin(C-A)/sinC+sinA +c 2 sin(A-B)/sinA+sinB =0

Answer 1

Actual question is:- If A + B + C = π, p rove that: ( sin2A + sin2B + sin2C ) / ( sinA + sinB + sinC ) = 8 sin(A/2) sin(B/2) sin(C/2). ANSWER— [Sin 2A + Sin 2B + Sin 2C]/[sin A+ sin B + sin C] = [2sin (A+B) cos (A-B) + 2sin C cos C]/[2Sin {(A+B)/2} cos {(A-B)/2}+ 2sin C/2 cos C/2] = [2sin C {cos (A-B) - cos (A+B)}] / [2cos C/2 {cos (A-B)/2 - cos (A+B)/2}] = [2sin C (2sinA sinB)] /[2cos C/2 2cos A/2 cos B/2]= [sinA sinB sinC]/[cosA/2 cosB/2 cosC/2] = 8sin A/2 sin B/2 sin C/2.