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Question

Let α,β,γ,δ ϵ(0,π) be such that 2cosα+6cosβ+7cosγ+9cosδ=0 and 2sinα6sinβ+7sinγ9sinδ=0, then the value of λ for which λcos(α+δ)=7cos(β+γ) is

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Solution

Given α,β,γ,δ ϵ(0,π)
2cosα+6cosβ+7cosγ+9cosδ=0
2sinα6sinβ+7sinγ9sinδ=0,

Now,
2cosα+6cosβ+7cosγ+9cosδ=0 [given]
2cosα+9cosδ=(6cosβ+7cosγ)
(2cosα+9cosδ)2=((6cosβ+7cosγ))2 [Squaring on both sides]
4cos2α+81cos2δ+36cosαcosδ=36cos2β+49cos2γ+84cosβcosγ (i)

and
2sinα6sinβ+7sinγ9sinδ=0 [given]

2sinα9sinδ=6sinβ7sinγ
(2sinα9sinδ)2=(6sinβ7sinγ)2 [Squaring on both sides]
4sin2α+81sin2δ36sinαsinδ=36sin2β+49sin2γ84sinβsinγ (ii)

Add (i),(ii)
4cos2α+81cos2δ+36cosαcosδ+4sin2α+81sin2δ36sinαsinδ
=36cos2β+49cos2γ+84cosβcosγ+36sin2β+49sin2γ84sinβsinγ
4cos2α+4sin2α+81sin2δ+81cos2δ+36cosαcosδ36sinαsinδ
=36cos2β+36sin2β+49cos2γ+49cos2γ+84cosβcosγ84sinβsinγ
4[cos2α+sin2α]+81[sin2δ+cos2δ]+36[cosαcosδsinαsinδ]
=36[cos2β+sin2β]+49[cos2γ+cos2γ]+84[cosβcosγsinβsinγ]
4(1)+81(1)+36(cos(α+δ))=36(1)+49(1)+84(cos(β+γ))
85+36(cos(α+δ))=85+84(cos(β+γ))
36(cos(α+δ))=84(cos(β+γ))
3(cos(α+δ))=7(cos(β+γ))

by comparing with λcos(α+δ)=7cos(β+γ)

we get λ=3

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