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Question Number 139609 by bemath last updated on 29/Apr/21

If tan 14° = x then tan 18° =?

$$\mathrm{If}\:\mathrm{tan}\:\mathrm{14}°\:=\:\mathrm{x}\:\mathrm{then}\:\mathrm{tan}\:\mathrm{18}°\:=? \\ $$

Answered by qaz last updated on 29/Apr/21

18°=((18×14°)/(14))=(9/7)tan^(−1) x ⇒tan 18°=tan ((9/7)tan^(−1) x)

$$\mathrm{18}°=\frac{\mathrm{18}×\mathrm{14}°}{\mathrm{14}}=\frac{\mathrm{9}}{\mathrm{7}}\mathrm{tan}^{−\mathrm{1}} {x} \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{18}°=\mathrm{tan}\:\left(\frac{\mathrm{9}}{\mathrm{7}}\mathrm{tan}^{−\mathrm{1}} {x}\right) \\ $$

Answered by mr W last updated on 29/Apr/21

recall tan 3α=((tan α (3−tan^2 α))/(1−3 tan^2 α)) tan 42°=tan (3×14°)=((x(3−x^2 ))/(1−3x^2 )) tan 18°=tan (60°−42°)=((tan 60°−tan 42°)/(1+tan 60°tan 42°)) =(((√3)−((x(3−x^2 ))/(1−3x^2 )))/(1+(√3)×((x(3−x^2 ))/(1−3x^2 ))))=(((√3)−3x−3(√3)x^2 +x^3 )/(1+3(√3)x−3x^2 −(√3)x^3 ))

$${recall}\:\mathrm{tan}\:\mathrm{3}\alpha=\frac{\mathrm{tan}\:\alpha\:\left(\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:\alpha\right)}{\mathrm{1}−\mathrm{3}\:\mathrm{tan}^{\mathrm{2}} \:\alpha} \\ $$$$ \\ $$$$\mathrm{tan}\:\mathrm{42}°=\mathrm{tan}\:\left(\mathrm{3}×\mathrm{14}°\right)=\frac{{x}\left(\mathrm{3}−{x}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\mathrm{18}°=\mathrm{tan}\:\left(\mathrm{60}°−\mathrm{42}°\right)=\frac{\mathrm{tan}\:\mathrm{60}°−\mathrm{tan}\:\mathrm{42}°}{\mathrm{1}+\mathrm{tan}\:\mathrm{60}°\mathrm{tan}\:\mathrm{42}°} \\ $$$$\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}}−\frac{{x}\left(\mathrm{3}−{x}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{1}+\sqrt{\mathrm{3}}×\frac{{x}\left(\mathrm{3}−{x}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} }}=\frac{\sqrt{\mathrm{3}}−\mathrm{3}{x}−\mathrm{3}\sqrt{\mathrm{3}}{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{3}}{x}−\mathrm{3}{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}^{\mathrm{3}} } \\ $$

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