Question Number 92594 by naka3546 last updated on 08/May/20
Commented by jagoll last updated on 08/May/20
$$\alpha+\beta+\gamma\:=\:\mathrm{180}^{\mathrm{o}} ? \\ $$
Commented by Prithwish Sen 1 last updated on 08/May/20
![tanβ=((sin(α+γ))/(cos(α−γ))) sin2β=((2tanβ)/(1+tan^2 β))= ((2sin(α+γ)cos(α−γ))/(cos^2 (α−γ)+sin^2 (α+γ))) =((sin2α+sin 2γ)/(1+[sin^2 (α+γ)−cos^2 (α−γ)])) = ((sin2α+sin 2γ)/(1+sin 2α.sin 2γ))](Q92598.png)
$$\mathrm{tan}\beta=\frac{\mathrm{sin}\left(\alpha+\gamma\right)}{\mathrm{cos}\left(\alpha−\gamma\right)} \\ $$$$\mathrm{sin2}\beta=\frac{\mathrm{2tan}\beta}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \beta}=\:\frac{\mathrm{2sin}\left(\alpha+\gamma\right)\mathrm{cos}\left(\alpha−\gamma\right)}{\mathrm{cos}^{\mathrm{2}} \left(\alpha−\gamma\right)+\mathrm{sin}\:^{\mathrm{2}} \left(\alpha+\gamma\right)} \\ $$$$=\frac{\mathrm{sin2}\alpha+\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{1}+\left[\mathrm{sin}\:^{\mathrm{2}} \left(\alpha+\gamma\right)−\mathrm{cos}\:^{\mathrm{2}} \left(\alpha−\gamma\right)\right]}\:=\:\frac{\mathrm{sin2}\alpha+\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{1}+\mathrm{sin}\:\mathrm{2}\alpha.\mathrm{sin}\:\mathrm{2}\gamma} \\ $$
Commented by naka3546 last updated on 08/May/20
$${I}\:\:{don}'{t}\:{know}\:\:,\:{sir} \\ $$
Commented by naka3546 last updated on 08/May/20
$${thank}\:\:{you},\:{sir} \\ $$
Commented by Tony Lin last updated on 08/May/20
![sin2α=((2tanα)/(1+tan^2 α)) sin2γ=((2tanγ)/(1+tan^2 γ)) sin2β=((2tanβ)/(1+tan^2 β)) =(((2(tanα+tanγ))/(1+tanαtanγ))/(1+(((tanα+tanγ)/(1+tanαtanγ)))^2 )) =((2(tanα+tanγ)(1+tanαtanγ))/((1+tanαtanγ)^2 +(tanα+tanγ)^2 )) =((2(tanα+tanγ+tan^2 αtanγ+tan^2 γtanα)/(1+4tanαtanγ+tan^2 αtan^2 γ+tan^2 α+tan^2 γ)) =((2[tanα(1+tan^2 γ)+tanγ(1+tan^2 α)])/((1+tan^2 α)(1+tan^2 γ)+4tanαtanγ)) =((((2tanα)/(1+tan^2 α))+((2tanγ)/(1+tan^2 γ)))/(1+ ((4tanαtanγ)/((1+tan^2 α)(1+tan^2 γ))))) =((sin2α+sin2γ)/(1+sin2αsin2γ))](Q92604.png)
$${sin}\mathrm{2}\alpha=\frac{\mathrm{2}{tan}\alpha}{\mathrm{1}+{tan}^{\mathrm{2}} \alpha} \\ $$$${sin}\mathrm{2}\gamma=\frac{\mathrm{2}{tan}\gamma}{\mathrm{1}+{tan}^{\mathrm{2}} \gamma} \\ $$$${sin}\mathrm{2}\beta=\frac{\mathrm{2}{tan}\beta}{\mathrm{1}+{tan}^{\mathrm{2}} \beta} \\ $$$$=\frac{\frac{\mathrm{2}\left({tan}\alpha+{tan}\gamma\right)}{\mathrm{1}+{tan}\alpha{tan}\gamma}}{\mathrm{1}+\left(\frac{{tan}\alpha+{tan}\gamma}{\mathrm{1}+{tan}\alpha{tan}\gamma}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\left({tan}\alpha+{tan}\gamma\right)\left(\mathrm{1}+{tan}\alpha{tan}\gamma\right)}{\left(\mathrm{1}+{tan}\alpha{tan}\gamma\right)^{\mathrm{2}} +\left({tan}\alpha+{tan}\gamma\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\left({tan}\alpha+{tan}\gamma+{tan}^{\mathrm{2}} \alpha{tan}\gamma+{tan}^{\mathrm{2}} \gamma{tan}\alpha\right.}{\mathrm{1}+\mathrm{4}{tan}\alpha{tan}\gamma+{tan}^{\mathrm{2}} \alpha{tan}^{\mathrm{2}} \gamma+{tan}^{\mathrm{2}} \alpha+{tan}^{\mathrm{2}} \gamma} \\ $$$$=\frac{\mathrm{2}\left[{tan}\alpha\left(\mathrm{1}+{tan}^{\mathrm{2}} \gamma\right)+{tan}\gamma\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right)\right]}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} \gamma\right)+\mathrm{4}{tan}\alpha{tan}\gamma} \\ $$$$=\frac{\frac{\mathrm{2}{tan}\alpha}{\mathrm{1}+{tan}^{\mathrm{2}} \alpha}+\frac{\mathrm{2}{tan}\gamma}{\mathrm{1}+{tan}^{\mathrm{2}} \gamma}}{\mathrm{1}+\:\frac{\mathrm{4}{tan}\alpha{tan}\gamma}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} \gamma\right)}} \\ $$$$=\frac{{sin}\mathrm{2}\alpha+{sin}\mathrm{2}\gamma}{\mathrm{1}+{sin}\mathrm{2}\alpha{sin}\mathrm{2}\gamma} \\ $$