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Question Number 114108 by bemath last updated on 17/Sep/20
prove that 2 tan^(−1) ((2/3))=sin^(−1) (((12)/(13)))
$${prove}\:{that}\:\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{13}}\right) \\ $$
Answered by bobhans last updated on 17/Sep/20
let tan^(−1) ((2/3)) = x → tan x = (2/3) and { ((sin x= (2/( (√(13)))))),((cos x = (3/( (√(13)))))) :} ⇔ 2x = sin^(−1) (((12)/(13))) ⇔sin (2x) = sin (sin^(−1) (((12)/(13)))) ⇒ 2sin x cos x = ((12)/(13)) ⇒2((2/( (√(13)))))((3/( (√(13))))) = ((12)/(13))
$${let}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:=\:{x}\:\rightarrow\:\mathrm{tan}\:{x}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${and}\:\begin{cases}{\mathrm{sin}\:{x}=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}}\\{\mathrm{cos}\:{x}\:=\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}}\end{cases} \\ $$$$\Leftrightarrow\:\mathrm{2}{x}\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{13}}\right) \\ $$$$\Leftrightarrow\mathrm{sin}\:\left(\mathrm{2}{x}\right)\:=\:\mathrm{sin}\:\left(\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{13}}\right)\right) \\ $$$$\Rightarrow\:\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}\:=\:\frac{\mathrm{12}}{\mathrm{13}}\: \\ $$$$\Rightarrow\mathrm{2}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}\right)\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}\right)\:=\:\frac{\mathrm{12}}{\mathrm{13}} \\ $$
Answered by Dwaipayan Shikari last updated on 17/Sep/20
tan^(−1) ((((2/3)+(2/3))/(1−(4/9))))=tan^(−1) ((12)/5) tanθ=((12)/5) secθ=((13)/5) cosθ=(5/(13)) and sinθ=(√(1−(5^2 /(13^2 )))) =((12)/(13)) θ=sin^(−1) ((12)/(13)) θ=2tan^(−1) (2/3) (Which is true)
$${tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{9}}}\right)={tan}^{−\mathrm{1}} \frac{\mathrm{12}}{\mathrm{5}} \\ $$$${tan}\theta=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$${sec}\theta=\frac{\mathrm{13}}{\mathrm{5}} \\ $$$${cos}\theta=\frac{\mathrm{5}}{\mathrm{13}}\:\:\:{and}\:{sin}\theta=\sqrt{\mathrm{1}−\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{13}^{\mathrm{2}} }}\:=\frac{\mathrm{12}}{\mathrm{13}} \\ $$$$\theta={sin}^{−\mathrm{1}} \frac{\mathrm{12}}{\mathrm{13}} \\ $$$$\theta=\mathrm{2}{tan}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}\:\:\:\left({Which}\:{is}\:{true}\right) \\ $$
Answered by physicstutes last updated on 17/Sep/20
if sin θ = ((12)/(13)), then tan θ = ((12)/5) tan((θ/2)) = ((sin θ)/(1 + cos θ)) cos θ = (5/(13)) ⇒ tan ((θ/2)) = (((12)/(13))/(1+ (5/(13)))) = ((12)/(18)) =(2/3) hence tan ((θ/2) ) = (2/3) ⇒ θ = 2 tan^(−1) ((2/3)) thus, θ = sin^(−1) (((12)/(13))) =2 tan^(−1) ((2/3))
$$\mathrm{if}\:\mathrm{sin}\:\theta\:=\:\frac{\mathrm{12}}{\mathrm{13}},\:\mathrm{then}\:\mathrm{tan}\:\theta\:=\:\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\mathrm{tan}\left(\frac{\theta}{\mathrm{2}}\right)\:=\:\frac{\mathrm{sin}\:\theta}{\mathrm{1}\:+\:\mathrm{cos}\:\theta} \\ $$$$\mathrm{cos}\:\theta\:=\:\frac{\mathrm{5}}{\mathrm{13}}\:\:\:\Rightarrow\:\mathrm{tan}\:\left(\frac{\theta}{\mathrm{2}}\right)\:=\:\frac{\frac{\mathrm{12}}{\mathrm{13}}}{\mathrm{1}+\:\frac{\mathrm{5}}{\mathrm{13}}}\:=\:\frac{\mathrm{12}}{\mathrm{18}}\:=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{hence}\:\mathrm{tan}\:\left(\frac{\theta}{\mathrm{2}}\:\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow\:\theta\:=\:\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$$\mathrm{thus},\:\:\theta\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{13}}\right)\:=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$

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