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Question Number 157388 by naka3546 last updated on 22/Oct/21
If m tan (θ − 30°) = n tan (θ + 12°) cos 2θ = ?
$${If}\:\:{m}\:\mathrm{tan}\:\left(\theta\:−\:\mathrm{30}°\right)\:=\:{n}\:\mathrm{tan}\:\left(\theta\:+\:\mathrm{12}°\right) \\ $$$$\mathrm{cos}\:\mathrm{2}\theta\:\:=\:? \\ $$
Commented by cortano last updated on 22/Oct/21
mtan (θ−30°)=ntan (30°+θ−18°) ⇒m(((3tan θ−(√3))/(3+(√3) tan θ)))=n(((tan (30°+θ)−tan 18°)/(1+tan (30°+θ) tan 18°))) tan 18°=((sin 18°)/(cos 18°))=((((√5)−1)/4)/( (√(1−((((√5)−1)/4))^2 )))) = (((√5)−1)/( (√(10+2(√5)))))
$$\:{m}\mathrm{tan}\:\left(\theta−\mathrm{30}°\right)={n}\mathrm{tan}\:\left(\mathrm{30}°+\theta−\mathrm{18}°\right) \\ $$$$\Rightarrow{m}\left(\frac{\mathrm{3tan}\:\theta−\sqrt{\mathrm{3}}}{\mathrm{3}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta}\right)={n}\left(\frac{\mathrm{tan}\:\left(\mathrm{30}°+\theta\right)−\mathrm{tan}\:\mathrm{18}°}{\mathrm{1}+\mathrm{tan}\:\left(\mathrm{30}°+\theta\right)\:\mathrm{tan}\:\mathrm{18}°}\right) \\ $$$$\mathrm{tan}\:\mathrm{18}°=\frac{\mathrm{sin}\:\mathrm{18}°}{\mathrm{cos}\:\mathrm{18}°}=\frac{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}}{\:\sqrt{\mathrm{1}−\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:=\:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\:\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}\: \\ $$
Commented by naka3546 last updated on 23/Oct/21
thank you, sir.
$${thank}\:\:{you},\:\:{sir}. \\ $$
Answered by mr W last updated on 23/Oct/21
(m/n)tan (θ−30)=tan (θ−30+18)=((tan (θ−30)+tan 18)/(1−tan (θ−30)tan 18)) μ=(m/n), t=tan (θ−30) μt=((t+tan 18)/(1−t tan 18)) μtan 18 t^2 −(μ−1)t+tan 18=0 t=((μ−1±(√((μ−1)^2 −4μtan^2 18)))/(2μtan 18)) t=((tan θ−(1/( (√3))))/(1+((tan θ)/( (√3)))))=(((√3)tan θ−1)/(tan θ+(√3))) tan θ=((1+(√3)t)/( (√3)−t)) cos 2θ=(2/(1+tan^2 θ))−1
$$\frac{{m}}{{n}}\mathrm{tan}\:\left(\theta−\mathrm{30}\right)=\mathrm{tan}\:\left(\theta−\mathrm{30}+\mathrm{18}\right)=\frac{\mathrm{tan}\:\left(\theta−\mathrm{30}\right)+\mathrm{tan}\:\mathrm{18}}{\mathrm{1}−\mathrm{tan}\:\left(\theta−\mathrm{30}\right)\mathrm{tan}\:\mathrm{18}} \\ $$$$\mu=\frac{{m}}{{n}},\:{t}=\mathrm{tan}\:\left(\theta−\mathrm{30}\right) \\ $$$$\mu{t}=\frac{{t}+\mathrm{tan}\:\mathrm{18}}{\mathrm{1}−{t}\:\mathrm{tan}\:\mathrm{18}} \\ $$$$\mu\mathrm{tan}\:\mathrm{18}\:{t}^{\mathrm{2}} −\left(\mu−\mathrm{1}\right){t}+\mathrm{tan}\:\mathrm{18}=\mathrm{0} \\ $$$${t}=\frac{\mu−\mathrm{1}\pm\sqrt{\left(\mu−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\mu\mathrm{tan}^{\mathrm{2}} \:\mathrm{18}}}{\mathrm{2}\mu\mathrm{tan}\:\mathrm{18}} \\ $$$${t}=\frac{\mathrm{tan}\:\theta−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}+\frac{\mathrm{tan}\:\theta}{\:\sqrt{\mathrm{3}}}}=\frac{\sqrt{\mathrm{3}}\mathrm{tan}\:\theta−\mathrm{1}}{\mathrm{tan}\:\theta+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{1}+\sqrt{\mathrm{3}}{t}}{\:\sqrt{\mathrm{3}}−{t}} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}−\mathrm{1} \\ $$
Commented by naka3546 last updated on 23/Oct/21
Thank you, sir.
$$\mathrm{Thank}\:\:\mathrm{you},\:\:\mathrm{sir}. \\ $$

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