If-tan-2-tan-1-then-

Question Number 62983 by hovea cw last updated on 27/Jun/19

$$\mathrm{If}\:\:\:\mathrm{tan}\:\mathrm{2}\theta\:\mathrm{tan}\:\theta\:=\:\mathrm{1},\:\mathrm{then}\:\theta\:= \\ $$

Answered by Hope last updated on 27/Jun/19

$${tan}\mathrm{2}\theta={cot}\theta={tan}\left(\frac{\pi}{\mathrm{2}}−\theta\right) \\ $$$${formula} \\ $$$$\left[{tan}\alpha={tan}\beta\:\:\:\alpha={n}\pi+\beta\right] \\ $$$${so}\:\mathrm{2}\theta={n}\pi+\left(\frac{\pi}{\mathrm{2}}−\theta\right) \\ $$$$\mathrm{3}\theta=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{2}}\rightarrow\:\theta=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{6}} \\ $$

Commented by mr W last updated on 27/Jun/19

$${please}\:{check}\:{sir}! \\ $$$${n}=\mathrm{1}:\:\theta=\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{but}\:\mathrm{tan}\:\frac{\pi}{\mathrm{2}}\:\rightarrow\:\infty\:! \\ $$

Commented by Hope last updated on 27/Jun/19

$${yes}\:{sir}…{you}\:{r}\:{right}… \\ $$

Answered by Hope last updated on 27/Jun/19

$${anothdr}\:{way} \\ $$$${a}={tan}\theta \\ $$$$\frac{\mathrm{2}{a}}{\mathrm{1}−{a}^{\mathrm{2}} }×{a}=\mathrm{1} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} =\mathrm{1}−{a}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}\rightarrow{a}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${tan}\theta={tan}\frac{\pi}{\mathrm{6}} \\ $$$$\theta={n}\pi+\frac{\pi}{\mathrm{6}} \\ $$$${tan}\theta=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}={tan}\left(\pi−\frac{\pi}{\mathrm{6}}\right) \\ $$$$\theta={n}\pi+\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$

Commented by mr W last updated on 27/Jun/19

$${good}\:{sir}! \\ $$$${that}\:{is}\:\theta={n}\pi\pm\frac{\pi}{\mathrm{6}} \\ $$

Answered by MJS last updated on 27/Jun/19

$$\mathrm{tan}\:\mathrm{2}\theta\:=\frac{\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{2cos}^{\mathrm{2}} \:\theta\:−\mathrm{1}} \\ $$$$\frac{\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{2cos}^{\mathrm{2}} \:\theta\:−\mathrm{1}}×\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}=\mathrm{1} \\ $$$$\frac{\mathrm{2sin}^{\mathrm{2}} \:\theta}{\mathrm{2cos}^{\mathrm{2}} \:\theta\:−\mathrm{1}}=\mathrm{1} \\ $$$$\mathrm{method}\:\mathrm{1} \\ $$$$\mathrm{2sin}^{\mathrm{2}} \:\theta\:=\mathrm{2cos}^{\mathrm{2}} \:\theta\:−\mathrm{1} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\theta\:−\mathrm{cos}^{\mathrm{2}} \:\theta\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}−\mathrm{2cos}^{\mathrm{2}} \:\theta\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\theta\:=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{cos}\:\theta\:=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{method}\:\mathrm{2} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\mathrm{2}\theta}=\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$… \\ $$

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