Menu Close

Show-that-tan-1-1-3-sin-1-1-3-pi-4-

Tinku Tara 0 Comments
Pin




Question Number 155898 by Tawa11 last updated on 05/Oct/21
Show that tan^(− 1) ((1/3)) + sin^(− 1) ((1/3)) = (π/4)
$$\mathrm{Show}\:\mathrm{that}\:\:\:\:\:\mathrm{tan}^{−\:\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:\:\:+\:\:\:\mathrm{sin}^{−\:\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:\:\:=\:\:\:\frac{\pi}{\mathrm{4}} \\ $$
Answered by immortel last updated on 05/Oct/21
L=tan(tan^(−1) ((1/3))+sin^(−1) ((1/3)))=((tan(arctan((1/3)))+tan(arcsin(1/3)))/(1−tan(arctan((1/3)))tan(arcsin(1/3)))) or tan(arcsinx)=(x/( (√(1−x^2 )))) donc tan(arcsin(1/3))=((1/3)/( (√(1−(1/9))))) =(1/( (√8)))=((√2)/4) ainsi L=(((1/3)+((√2)/4))/(1−((√2)/(12)))) =(((4+3(√2)))/(12))×((12)/(12−(√2))) =(((4+3(√2))(12+(√2)))/((12−(√2))(12+(√2)))) =((48+4(√2)+36(√2)+6)/(144−2)) =((54+40(√2))/(142))=((27+20(√2))/(71))<1 donc L≠1 ainsi tan^(− 1) ((1/3)) + sin^(− 1) ((1/3)) ≠ (π/4) ..........l′immortel........
$${L}={tan}\left({tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right)=\frac{{tan}\left({arctan}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right)+{tan}\left({arcsin}\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{1}−{tan}\left({arctan}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right){tan}\left({arcsin}\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$${or}\:{tan}\left({arcsinx}\right)=\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${donc}\:\:{tan}\left({arcsin}\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{8}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${ainsi}\:{L}=\frac{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}}{\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{12}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}\right)}{\mathrm{12}}×\frac{\mathrm{12}}{\mathrm{12}−\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}\right)\left(\mathrm{12}+\sqrt{\mathrm{2}}\right)}{\left(\mathrm{12}−\sqrt{\mathrm{2}}\right)\left(\mathrm{12}+\sqrt{\mathrm{2}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{48}+\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{36}\sqrt{\mathrm{2}}+\mathrm{6}}{\mathrm{144}−\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{54}+\mathrm{40}\sqrt{\mathrm{2}}}{\mathrm{142}}=\frac{\mathrm{27}+\mathrm{20}\sqrt{\mathrm{2}}}{\mathrm{71}}<\mathrm{1} \\ $$$${donc}\:\:{L}\neq\mathrm{1}\:{ainsi}\: \\ $$$$\:\mathrm{tan}^{−\:\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:\:\:+\:\:\:\mathrm{sin}^{−\:\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:\:\:\neq\:\:\frac{\pi}{\mathrm{4}} \\ $$$$\: \\ $$$$\:\:……….{l}'{immortel}…….. \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa11 last updated on 05/Oct/21
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 05/Oct/21
α=tan^(−1) (1/3) ⇒tan α=(1/3) β=sin^(−1) (1/3) ⇒sin β=(1/3) ⇒tan β=(1/( 2(√2))) tan (α+β)=((tan α+tan β)/(1−tan α tan β)) =(((1/3)+(1/( 2(√2))))/(1−(1/( 6(√2)))))=((3+2(√2))/( 6(√2)−1))=((27+20(√2))/(71)) ≈0.778≠1 ⇒α+β≠45° in fact tan^(− 1) ((1/3))+sin^(− 1) ((1/3)) ≈38°
$$\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\beta=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{sin}\:\beta=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{tan}\:\beta=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta} \\ $$$$=\frac{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{1}−\frac{\mathrm{1}}{\:\mathrm{6}\sqrt{\mathrm{2}}}}=\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\:\mathrm{6}\sqrt{\mathrm{2}}−\mathrm{1}}=\frac{\mathrm{27}+\mathrm{20}\sqrt{\mathrm{2}}}{\mathrm{71}} \\ $$$$\approx\mathrm{0}.\mathrm{778}\neq\mathrm{1} \\ $$$$\Rightarrow\alpha+\beta\neq\mathrm{45}° \\ $$$${in}\:{fact}\:\mathrm{tan}^{−\:\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{sin}^{−\:\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:\approx\mathrm{38}° \\ $$
Commented by Tawa11 last updated on 05/Oct/21
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *