Question

Prove that sin α* sin (α + 2β) - sin β* sin (β+2α) = sin (α + β) * sin(α - β)

Answer 1

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Answer 2

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Prove that sin α* sin (α + 2β) - sin β* sin (β+2α) = sin (α + β) * sin(α - β)

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Proof:

$\begin{gathered}\text{We need to prove that,}\\\\:\longrightarrow2\sin^2\beta + 4\cos(\alpha+\beta)\times\sin\alpha\sin\beta+\cos2(\alpha+\beta)=\cos2\alpha\\\\\text{Taking LHS,}\\\\:\implies2\sin^2\beta+4\cos(\alpha+\beta)\times\sin\alpha\sin\beta+\cos2(\alpha+\beta)\\\\:\implies2\sin^2\beta+4\cos(\alpha+\beta)\times\sin\alpha\sin\beta+\cos(2\alpha+2\beta)\\\\\text{We know that,}\\\\:\hookrightarrow\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi\end{gathered}$

We need to prove that,

:⟶2sin

2

β+4cos(α+β)×sinαsinβ+cos2(α+β)=cos2α

Taking LHS,

:⟹2sin

2

β+4cos(α+β)×sinαsinβ+cos2(α+β)

:⟹2sin

2

β+4cos(α+β)×sinαsinβ+cos(2α+2β)

We know that,

:↪cos(θ+ϕ)=cosθcosϕ−sinθsinϕ

$\begin{gathered}\text{So,}\\\\:\implies2\sin^2\beta+4\cos(\alpha+\beta)\times\sin\alpha\sin\beta+\cos(2\alpha+2\beta)\\\\:\implies2\sin^2\beta+4(\cos\alpha\cos\beta-\sin\alpha\sin\beta)\times\sin\alpha\sin\beta+(\cos2\alpha\cos2\beta-\sin2\alpha\sin2\beta)\end{gathered} So,:⟹2sin 2 β+4cos(α+β)×sinαsinβ+cos(2α+2β):⟹2sin 2 β+4(cosαcosβ−sinαsinβ)×sinαsinβ+(cos2αcos2β−sin2αsin2β)$

$\begin{gathered}:\implies2\sin^2\beta+4\cos\alpha\cos\beta\sin\alpha\sin\beta-4\sin^2\alpha\sin^2\beta+\cos2\alpha\cos2\beta-\sin2\alpha\sin2\beta\\\\:\implies2\sin^2\beta+(2\sin\alpha\cos\alpha)(2\sin\beta\cos\beta)-4\sin^2\alpha\sin^2\beta+\cos2\alpha\cos2\beta-\sin2\alpha\sin2\beta\end{gathered} :⟹2sin$

HENCE PROVED!!!