Question Number 191901 by mehdee42 last updated on 03/May/23
$${if}\:\:\mathrm{0}<\theta<\mathrm{45}^{\mathrm{0}} \:\:{which}\:{is}\:{bigger}\:? \\ $$$$\mathrm{2}{tan}\theta\:\:{or}\:\:{tan}\mathrm{2}\theta \\ $$
Answered by mr W last updated on 03/May/23
Commented by mr W last updated on 03/May/23
$${BC}=\sqrt{{BD}^{\mathrm{2}} +{DC}^{\mathrm{2}} }\:>\:{BD}={AB} \\ $$$${AB}=\mathrm{tan}\:\theta \\ $$$${AC}=\mathrm{tan}\:\mathrm{2}\theta={AB}+{BC}\:>{AB}+{AB}=\mathrm{2}\:{AB}=\mathrm{2}\:\mathrm{tan}\:\theta \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{2}\theta\:>\:\mathrm{2}\:\mathrm{tan}\:\theta \\ $$
Commented by mehdee42 last updated on 03/May/23
$${very}\:{good} \\ $$
Answered by Subhi last updated on 04/May/23
$${tan}\left(\mathrm{2}\theta\right)=\:\frac{\mathrm{2}{tan}\left(\theta\right)}{\mathrm{1}−\left({tan}\left(\theta\right)\right)^{\mathrm{2}} } \\ $$$$\mathrm{0}<\theta<\mathrm{45} \\ $$$${tan}^{\mathrm{2}} \left(\mathrm{0}\right)<{tan}^{\mathrm{2}} \left(\theta\right)<\mathrm{1} \\ $$$$\mathrm{0}<{tan}^{\mathrm{2}} \left(\theta\right)<\mathrm{1} \\ $$$$\mathrm{1}>\mathrm{1}−{tan}^{\mathrm{2}} \left(\theta\right)>\mathrm{0} \\ $$$$\frac{\mathrm{2}{tan}\left(\theta\right)}{\mathrm{1}−{tan}^{\mathrm{2}} \left(\theta\right)}>\mathrm{2}{tan}\left(\theta\right) \\ $$$$ \\ $$
Commented by mehdee42 last updated on 03/May/23
$${very}\:{nice} \\ $$
Answered by ajfour last updated on 04/May/23
$${f}\left(\theta\right)=\mathrm{tan}\:\mathrm{2}\theta−\mathrm{2tan}\:\theta \\ $$$$\:\:\:\:\:\:\:=\mathrm{2tan}\:\theta\left\{\frac{\mathrm{1}−\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta\right)}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta}\right\} \\ $$$$\:\:\:\:\:=\frac{\mathrm{2tan}\:^{\mathrm{3}} \theta}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta} \\ $$$${f}\left(\theta\right)>\mathrm{0}\:\:\:{if}\:\:\:\mathrm{0}<\mathrm{tan}\:\theta<\mathrm{1} \\ $$$$\Rightarrow\:\:\:\mathrm{tan}\:\mathrm{2}\theta>\mathrm{2tan}\:\theta \\ $$