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Question

If α+βγ=π, and sin2α+sin2βsin2γ=λsinαsinβcosγ, then write the value of λ.

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Solution

Identity
sin2Asin2B
=sin(A+B)sin(AB)
=(sin(A)+sin(B))(sin(A)sin(B))
=2sin(A+B2)cos(AB2).2cos(A+B2)sin(A+B2)
=2sin(A+B2)cos(A+B2).2sin(AB2)cos(AB2)
=sin(A+B)sin(AB)

Given equation
sin2α+[sin2βsin2γ]

=sin2α+sin(βγ)sin(β+γ)

=sin2α+sin(πα)sin(β+γ)

=sin2α+sinαsin(β+γ)

=sinα[sinα+sin(β+γ)]

=sinα[sin(π(βγ))+sin(β+γ)]

=sinα[sin(βγ)+sin(β+γ)]

=sinα[2sinβcosγ]

=2sinαsinβcosγ

λ=2

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