Prove that !!!!!!!!!!!!!!!!!!!!!!!!

Question

sinα + sinβ+ sinγ - sin (α+β+γ) = 4 sin{(α+β)/2} sin{(β+γ)/2} sin{(α+γ)/2}?

Answer 1

Prove that sinα +sinβ+sinY - sin (α+β+γ) = 4 sin{(α+β)/2} sin{(β+γ)/2} sin{(α+γ)/2}

Simplify the expression from left hand side

 = sinα +sinβ+sinY - sin (α+β+γ)

From the addtion formulas for sine and cosine

sinα +sinβ = 2sin(α+β)/2cos(α -β)/2     ->   (1)

sinα - sinβ = 2 sin(α-β)/2 cos (α+β)/2     ->  (2)

cosα - cosβ = 2 sin (α+β)/2 sin(β -α)/2   -> (3)

Apply the formulas from (1) and (2)

sinα +sinβ+sinY - sin (α+β+γ) =[ 2 sin (α+β)/2cos(α -β)/2] + [2 sin (γ -α -β -γ)/2 cos ( γ + (α+β)/2 ) ]

= [ 2 sin (α+β)/2cos(α -β)/2 ] - [2 sin (α+β)/2 cos (γ + (α+β)/2 ) ]

= 2 sin (α+β)/2 [ cos(α -β)/2 - cos (γ + (α+β)/2 ]

Apply the (3)rd formula

= 2 sin (α+β)/2 { 2 sin [ (α -β+2γ+ α+β)/2] / 2 sin [ ( α+β +2γ - α+β/2)] / 2

= 2 sin (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

The identity is true.