`a sin ((A)/(2)+B) = (b+c) sin ""(A)/(2)`

In any triangle ABC, prove that : a sin (A/2+B)= (b +c) sin frac (A)(2) .

If A+B+C= pi and (sin 2 A + sin 2B + sin 2 C)/(sin A + sin B + sin C ) = lamda sin ((A)/(2)) sin ((B)/(2)) sin ((C )/(2)) , then the value of lamda must be

(ix) (a^(2) sin (B-C))/(sinA) + (b^(2) sin (C-A))/(sin B) + (c^(2) sin (A-B))/(sin C)=0

(x) (a sin(B-C))/(b^(2)-c^(2)) = (b sin (C-A))/(c^(2)-a^(2)) = (c sin(A-B))/(a^(2)-b^(2))

(sin2A + sin2B + sin2C) / (sin A + sin B + sin C) = 8sin ((A) / (2)) sin ((B) / (2)) sin ((C) / (2))

Show that: sin A + sin B +sin C - sin(A+B+C)=4 sin ((A+B)/2)sin ((B+C)/2)sin ((C+A)/2)

Prove that, sinA+sinB+sinC-sin(A+B+C)= 4sin((A+B)/(2))sin((B+C)/(2))sin((C+A)/(2))