﻿ If \cosα+\cosβ+\cosγ=0 and \sinα+\sinβ+\sinγ=0, find the value of \cos(α-β) + \cos(β-γ) + \cos(γ - α)

#### Question

If \cosα+\cosβ+\cosγ=0 and \sinα+\sinβ+\sinγ=0, find the value of \cos(α-β) + \cos(β-γ) + \cos(γ - α)

Collected in the board: Trigonometry

Steven Zheng posted 1 week ago

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}

Steven Zheng posted 1 week ago

Scroll to Top