2012/033) If cosα + cosβ + cosγ = 0 and sinα + sinβ + sinγ = 0 …Then what is cos(β – γ) + cos(γ –α) + cos(α – β)

cosα + cosβ + cosγ = 0
=>cosα + cosβ = - cosγ
=> (cosα + cosβ)^2 = cos^2γ
=> cos^2α + cos^2β + 2 cosα cosβ = cos^2γ
=> 2 cosα cosβ = cos^2γ - (cos^2α + cos^2β) ... 1

similarly 2 cosβ cosγ = cos^2 α - (cos^2β + cos^2γ) ...2

and 2 cosγ cosα =cos^2β- (cos^2γ + cos^2 α) .. 3

adding (1) (2) and (3)

we get 2( cosα cosβ + cosβ cosγ + cosγ cosα) = - (cos^2 α + cos^2β + cos^2γ) ... 4

similarly from sinα + sinβ + sinγ = 0 replacing cos with sin we get

2( sin α sin β + sin β sin γ + sin γ sin α) = - (sin ^2 α + sin ^2β + sin ^2γ) ... 5

adding (4) and (5) and reordering we get

2((cosα cosβ + sin α sin β) + (cosβ cosγ + sin β sin γ) + (cosγ cosα + sin γ sin α)) = - 3
or
2(cos(α-β) + cos(β-γ) + cos(γ - α)) = - 3

or cos(α-β) + cos(β-γ) + cos(γ - α) = - 3/2