Question

Prove that sinα + sinβ + sinγ − sin(α + β+ γ) = 4sin(α+β/2)sin(β+γ/2)sin(α+γ/2)

Answer 1

Answer:

sinα + sinβ + sinγ − sin(α + β+ γ)  = 4 Sin((α + β)/2)Sin((β + γ)/2)Sin((α + γ)/2)

Step-by-step explanation:

sinα + sinβ + sinγ − sin(α + β+ γ) = 4sin(α+β/2)sin(β+γ/2)sin(α+γ/2)

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

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

LHS = sinα + sinβ + sinγ − sin(α + β+ γ)

= 2Sin((α + β)/2)Cos((α - β)/2)   + 2 Sin((-α - β)/2)Cos((α + β+ 2γ)/2)

Using Sin(-A) = - SinA

= 2Sin((α + β)/2)Cos((α - β)/2)   -2 Sin((α + β)/2)Cos((α + β+ 2γ)/2)

= 2Sin((α + β)/2) (Cos((α - β)/2) - Cos((α + β+ 2γ)/2)

Using CosA - CosB = -2Sin((a + b)/2)Sin((a-b)/2)

= 2Sin((α + β)/2)  ( -2 Sin((α + γ)/2)Sin((-β - γ)/2)

= 2Sin((α + β)/2)  ( -2 Sin((α + γ)/2) (-Sin((β + γ)/2))

= 4 Sin((α + β)/2)Sin((β + γ)/2)Sin((α + γ)/2)

= RHS

QED

Proved