Question
If \cosα+\cosβ+\cosγ=0 and \sinα+\sinβ+\sinγ=0, find the value of \cos(α-β) + \cos(β-γ) + \cos(γ - α)
Answer
Given the condition,
\cosα + \cosβ + \cosγ = 0
Move \cosγ to the RHS
\cosα + \cosβ = - \cosγ
Square the both sides
(\cosα + \cosβ)^2 = \cos^2γ
Expand the brackets
\cos^2α + \cos^2β + 2 \cosα \cosβ = \cos^2γ
Move the square terms to the RHS
2 \cosα \cosβ = \cos^2γ - (\cos^2α + \cos^2β)
(1)
Similarly, we can get the following equations
2 \cosβ \cosγ = \cos^2 α - (\cos^2β +\cos^2γ)
(2)
2 \cosγ \cosα =\cos^2β- (\cos^2γ + \cos^2 α)
(3)
addition of the three equations (1) (2) and (3) gives
2( \cosα \cosβ + \cosβ \cosγ + \cosγ \cosα) = - (\cos^2 α + \cos^2β + \cos^2γ)
(4)
In the same process we can get the following equation from the give condition $$\sinα + \sinβ + \sinγ = 0 $
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 the terms gives
2[(\cosα \cosβ + \sin α \sin β) + (\cosβ \cosγ + \sin β \sin γ) + (\cosγ \cosα + \sin γ \sin α)] = - 3
2(\cos(α-β) + \cos(β-γ) + \cos(γ - α)) = - 3
or
\cos(α-β) + \cos(β-γ) + \cos(γ - α) = -\dfrac{3}{2}