Question

formulas to solve sin2a+sin2b+sin2c=4sinasinbsinc

Answer 1

sin2A + sin2B + sin2C = 2 sin (2A + 2B)/2 . cos (2A - 2B)/2 + sin2C = 2sin(A + B).cos(A - B) + 2 sinC.cosC = 2sin(A + B).cos(A - B) + 2 sin (Pie - (A + B)) cos (Pie - (A + B)) = 2 sin(A + B) (cos (A - B) - cos (A + B)) = 2 sin(A + B).2sinA.sinB = 4 sinA.sinB.sinC