proving that \(\displaystyle{\sin{\beta}}{\cos{{\left(\beta+\theta\right)}}}=-{\sin{\theta}}\) implies \(\displaystyle{\tan{\theta}}=-{\tan{\beta}}\)

ab0utfallingm1z2

ab0utfallingm1z2

Answered question

2022-03-29

proving that sinβcos(β+θ)=sinθ implies tanθ=tanβ

Answer & Explanation

tabido8uvt

tabido8uvt

Beginner2022-03-30Added 16 answers

Separating out sinθ,cosθ after using

formula
sinθ(sin2β1)=sinβcosβcosθ
sinθcosθ=sinβcosβsin2β1=?
Drahthaare89c

Drahthaare89c

Beginner2022-03-31Added 19 answers

Indeed, cos(β+θ)=cosβcosθsinβsinθ your equation becomes
sinβ(cosβcosθsinβsinθ)=sinθ
Dividing by cosθ
sinβ(cosβ-sinβtanθ)=-tanθsinβcosβ-sin2βtanθ=-tanθ
Dividing by cos2β=11+tan2β
tanβ-tan2βtanθ=-tanθ(1+tan2β)tanβ=-tanθ

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