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Question Number 158391 by mr W last updated on 03/Nov/21

if α,β,γ are the angles of a triangle, find ((sin 2𝛂+sin 2𝛃+sin 2𝛄)/(sin 𝛂 sin 𝛃 sin 𝛄))=?

$${if}\:\alpha,\beta,\gamma\:{are}\:{the}\:{angles}\:{of}\:{a}\:{triangle}, \\ $$$${find}\:\frac{\boldsymbol{\mathrm{sin}}\:\mathrm{2}\boldsymbol{\alpha}+\boldsymbol{\mathrm{sin}}\:\mathrm{2}\boldsymbol{\beta}+\boldsymbol{\mathrm{sin}}\:\mathrm{2}\boldsymbol{\gamma}}{\boldsymbol{\mathrm{sin}}\:\boldsymbol{\alpha}\:\boldsymbol{\mathrm{sin}}\:\boldsymbol{\beta}\:\boldsymbol{\mathrm{sin}}\:\boldsymbol{\gamma}}=? \\ $$

Commented by MJS_new last updated on 03/Nov/21

4

$$\mathrm{4} \\ $$

Answered by puissant last updated on 03/Nov/21

((sin2α+sin2β+sin2γ)/(sinα sinβ sinγ)) = ((4sinαsinβsinγ)/(sinαsinβsinγ)) = 4.

$$\frac{{sin}\mathrm{2}\alpha+{sin}\mathrm{2}\beta+{sin}\mathrm{2}\gamma}{{sin}\alpha\:{sin}\beta\:{sin}\gamma}\:=\:\frac{\mathrm{4}{sin}\alpha{sin}\beta{sin}\gamma}{{sin}\alpha{sin}\beta{sin}\gamma}\:=\:\mathrm{4}. \\ $$

Answered by mr W last updated on 03/Nov/21

2π−2γ=2α+2β −sin 2γ=sin 2α cos 2β+sin 2β cos 2α −sin 2γ=sin 2α (1−2 sin^2 β)+sin 2β (1−2 sin^2 α) sin 2α+sin 2β+sin 2γ=2 sin 2α sin^2 β+2 sin 2β sin^2 α sin 2α+sin 2β+sin 2γ=4 sin α sin β(cos α sin β+cos α sin α) sin 2α+sin 2β+sin 2γ=4 sin α sin β sin (α+β) sin 2α+sin 2β+sin 2γ=4 sin α sin β sin γ ⇒((sin 2α+sin 2β+sin 2γ)/(sin α sin β sin γ))=4

$$\mathrm{2}\pi−\mathrm{2}\gamma=\mathrm{2}\alpha+\mathrm{2}\beta \\ $$$$−\mathrm{sin}\:\mathrm{2}\gamma=\mathrm{sin}\:\mathrm{2}\alpha\:\mathrm{cos}\:\mathrm{2}\beta+\mathrm{sin}\:\mathrm{2}\beta\:\mathrm{cos}\:\mathrm{2}\alpha \\ $$$$−\mathrm{sin}\:\mathrm{2}\gamma=\mathrm{sin}\:\mathrm{2}\alpha\:\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\beta\right)+\mathrm{sin}\:\mathrm{2}\beta\:\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\alpha\right) \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\beta+\mathrm{sin}\:\mathrm{2}\gamma=\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\alpha\:\:\mathrm{sin}^{\mathrm{2}} \:\beta+\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\beta\:\:\mathrm{sin}^{\mathrm{2}} \:\alpha \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\beta+\mathrm{sin}\:\mathrm{2}\gamma=\mathrm{4}\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\left(\mathrm{cos}\:\alpha\:\:\mathrm{sin}\:\beta+\mathrm{cos}\:\alpha\:\:\mathrm{sin}\:\alpha\right) \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\beta+\mathrm{sin}\:\mathrm{2}\gamma=\mathrm{4}\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\beta+\mathrm{sin}\:\mathrm{2}\gamma=\mathrm{4}\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma \\ $$$$\Rightarrow\frac{\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\beta+\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma}=\mathrm{4} \\ $$

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