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Question Number 83319 by jagoll last updated on 29/Feb/20

if tan β = ((tan α + tan γ)/(1+tan αtan γ)) show that sin 2β = ((sin 2α+sin 2γ)/(1+sin 2αsin 2γ))

$$\mathrm{if}\:\mathrm{tan}\:\beta\:=\:\frac{\mathrm{tan}\:\alpha\:+\:\mathrm{tan}\:\gamma}{\mathrm{1}+\mathrm{tan}\:\alpha\mathrm{tan}\:\gamma} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{sin}\:\mathrm{2}\beta\:=\:\frac{\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{1}+\mathrm{sin}\:\mathrm{2}\alpha\mathrm{sin}\:\mathrm{2}\gamma} \\ $$

Answered by mind is power last updated on 01/Mar/20

sin(2β)=((2tg(β))/(1+tg^2 (β))) ((tan(a)+tan(γ))/(1+tan(a)tan(γ)))=((sin(a+γ))/(cos(a−γ))) sin(2β)=2(((sin(a+γ))/(cos(a−γ)))/(1+((sin^2 (a+γ))/(cos^2 (a−γ))))) =((2sin(a+γ)cos(a−γ))/(cos^2 (a−γ)+sin^2 (a+γ))) cos^2 (a−γ)+sin^2 (a+γ)=cos^2 (a)cos^2 (γ)+sin^2 (a)sin^2 (γ) +sin^2 (a)cos^2 (γ)+sin^2 (γ)cos^2 (a)+4sin(a)cos(a)sin(γ)cos(γ) =cos^2 (γ)(sin^2 (a)+cos^2 (a))+sin^2 (γ)(sin^2 (a)+cos^2 (a)) +sin(2a)sin(2γ) =1+sin(2γ)cos(2a) 2sin(a+γ)cos(a−γ)=sin(2γ)+sin(2a) i used ,2sin(a)cos(b)=sin(a+b)+sin(a−b) we get ((sin(2γ)+sin(2a))/(1+sin(2a)sin(2γ)))

$${sin}\left(\mathrm{2}\beta\right)=\frac{\mathrm{2}{tg}\left(\beta\right)}{\mathrm{1}+{tg}^{\mathrm{2}} \left(\beta\right)} \\ $$$$\frac{{tan}\left({a}\right)+{tan}\left(\gamma\right)}{\mathrm{1}+{tan}\left({a}\right){tan}\left(\gamma\right)}=\frac{{sin}\left({a}+\gamma\right)}{{cos}\left({a}−\gamma\right)} \\ $$$${sin}\left(\mathrm{2}\beta\right)=\mathrm{2}\frac{\frac{{sin}\left({a}+\gamma\right)}{{cos}\left({a}−\gamma\right)}}{\mathrm{1}+\frac{{sin}^{\mathrm{2}} \left({a}+\gamma\right)}{{cos}^{\mathrm{2}} \left({a}−\gamma\right)}} \\ $$$$=\frac{\mathrm{2}{sin}\left({a}+\gamma\right){cos}\left({a}−\gamma\right)}{{cos}^{\mathrm{2}} \left({a}−\gamma\right)+{sin}^{\mathrm{2}} \left({a}+\gamma\right)} \\ $$$${cos}^{\mathrm{2}} \left({a}−\gamma\right)+{sin}^{\mathrm{2}} \left({a}+\gamma\right)={cos}^{\mathrm{2}} \left({a}\right){cos}^{\mathrm{2}} \left(\gamma\right)+{sin}^{\mathrm{2}} \left({a}\right){sin}^{\mathrm{2}} \left(\gamma\right) \\ $$$$+{sin}^{\mathrm{2}} \left({a}\right){cos}^{\mathrm{2}} \left(\gamma\right)+{sin}^{\mathrm{2}} \left(\gamma\right){cos}^{\mathrm{2}} \left({a}\right)+\mathrm{4}{sin}\left({a}\right){cos}\left({a}\right){sin}\left(\gamma\right){cos}\left(\gamma\right) \\ $$$$={cos}^{\mathrm{2}} \left(\gamma\right)\left({sin}^{\mathrm{2}} \left({a}\right)+{cos}^{\mathrm{2}} \left({a}\right)\right)+{sin}^{\mathrm{2}} \left(\gamma\right)\left({sin}^{\mathrm{2}} \left({a}\right)+{cos}^{\mathrm{2}} \left({a}\right)\right) \\ $$$$+{sin}\left(\mathrm{2}{a}\right){sin}\left(\mathrm{2}\gamma\right) \\ $$$$=\mathrm{1}+{sin}\left(\mathrm{2}\gamma\right){cos}\left(\mathrm{2}{a}\right) \\ $$$$\mathrm{2}{sin}\left({a}+\gamma\right){cos}\left({a}−\gamma\right)={sin}\left(\mathrm{2}\gamma\right)+{sin}\left(\mathrm{2}{a}\right) \\ $$$${i}\:{used}\:\:,\mathrm{2}{sin}\left({a}\right){cos}\left({b}\right)={sin}\left({a}+{b}\right)+{sin}\left({a}−{b}\right) \\ $$$${we}\:{get}\:\frac{{sin}\left(\mathrm{2}\gamma\right)+{sin}\left(\mathrm{2}{a}\right)}{\mathrm{1}+{sin}\left(\mathrm{2}{a}\right){sin}\left(\mathrm{2}\gamma\right)} \\ $$$$ \\ $$

Commented by peter frank last updated on 01/Mar/20