Question

If A+B+C=π , prove that sin 2A + sin2B + sin2C =4sinA*sinB*sinC.

Answer 1

Answer:

Step-by-step explanation:

Given, A+B+C=π

L.H.S:

sin 2A + sin2B + sin2C  

= 2Sin(A+B)Cos(A-B) + Sin2C (∵ SinA+SinB = 2Sin(A+B/2)Cos(A-B/2))

= 2Sin(180-C)Cos(A-B) + 2SinCCosC

= 2SinCCos(A-B)+ 2SinCCos(180-(A+B)) (∵Cos(180 - (A+B)) = - Cos(A+B))

= 2SinC(Cos(A-B) - Cos(A+B)) (∵CosA - CosB = - 2Sin(A+B/2)Sin(A-B/2))

= 2SinC(-2SinA.Sin(-B))

= 2SinC(2SinASinB)

= 4SinASinBSinC.

= R.H.S

Hence Proved.