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Question Number 109337 by ZiYangLee last updated on 22/Aug/20
x=cosθ, where ((3π)/2)<θ<2π, and that 2cosθ−sinθ=2, show that (√(1−x^2 ))=2(1−x). Hence or otherwise, find x and deduce that tan2θ=((24)/7)
$${x}=\mathrm{cos}\theta,\:\mathrm{where}\:\frac{\mathrm{3}\pi}{\mathrm{2}}<\theta<\mathrm{2}\pi,\:\mathrm{and}\:\mathrm{that}\:\mathrm{2cos}\theta−\mathrm{sin}\theta=\mathrm{2}, \\ $$$$\mathrm{show}\:\mathrm{that}\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }=\mathrm{2}\left(\mathrm{1}−{x}\right). \\ $$$$\mathrm{Hence}\:\mathrm{or}\:\mathrm{otherwise},\:\mathrm{find}\:{x}\:\mathrm{and}\:\mathrm{deduce}\:\mathrm{that}\:\mathrm{tan2}\theta=\frac{\mathrm{24}}{\mathrm{7}} \\ $$
Answered by Aziztisffola last updated on 22/Aug/20
(√(1−x^2 ))=(√(1−cos^2 θ))=−sinθ =2−2cosθ=2(1−cosθ)=2(1−x) (√(1−x^2 ))=2(1−x) ⇔1−x^2 =4(x^2 −2x+1) 5x^2 −8x+3=0 △=4 ⇒x_1 =1 ∧ x_2 =(3/5) θ≠2π ⇒x≠1 ⇒x=(3/5) tan2θ=((2tan(θ))/(1−tan^2 (θ))) tanθ=((−(4/5))/(3/5))=−(4/3) tan2θ=((2×((−4)/3))/(1−(((−4)/3))^2 ))=(((−8)/3)/(1−((16)/9))) =((8×3)/7)=((24)/7)
$$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }=\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta}=−\mathrm{sin}\theta \\ $$$$\:=\mathrm{2}−\mathrm{2cos}\theta=\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\theta\right)=\mathrm{2}\left(\mathrm{1}−{x}\right) \\ $$$$\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }=\mathrm{2}\left(\mathrm{1}−{x}\right)\:\Leftrightarrow\mathrm{1}−{x}^{\mathrm{2}} =\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$\:\mathrm{5}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{3}=\mathrm{0} \\ $$$$\:\bigtriangleup=\mathrm{4}\:\Rightarrow{x}_{\mathrm{1}} =\mathrm{1}\:\wedge\:{x}_{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\theta\neq\mathrm{2}\pi\:\Rightarrow{x}\neq\mathrm{1}\:\Rightarrow{x}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\:\mathrm{tan2}\theta=\frac{\mathrm{2tan}\left(\theta\right)}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \left(\theta\right)} \\ $$$$\:\mathrm{tan}\theta=\frac{−\frac{\mathrm{4}}{\mathrm{5}}}{\frac{\mathrm{3}}{\mathrm{5}}}=−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\:\mathrm{tan2}\theta=\frac{\mathrm{2}×\frac{−\mathrm{4}}{\mathrm{3}}}{\mathrm{1}−\left(\frac{−\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} }=\frac{\frac{−\mathrm{8}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{16}}{\mathrm{9}}} \\ $$$$\:=\frac{\mathrm{8}×\mathrm{3}}{\mathrm{7}}=\frac{\mathrm{24}}{\mathrm{7}} \\ $$
Commented by ZiYangLee last updated on 23/Aug/20
NICE!
$$\mathrm{NICE}! \\ $$
Answered by Don08q last updated on 22/Aug/20
x = cosθ x^2 = cos^2 θ 1 − x^2 = sin^2 θ ± (√(1 − x^2 )) = sinθ But for the interval, ((3π)/2)<θ<2π sine has a negative value. ⇒ − (√(1 − x^2 )) = sinθ so (√(1 − x^2 )) = −sinθ ............ (1) But, 2cosθ − sinθ = 2 ⇒ 2 − 2x = −sinθ 2(1 − x) = −sinθ ........... (2) from (1) and (2), (√(1 − x^2 )) = 2(1 − x) qed It follows that 1 − x^2 = 4(1 − 2x + x^2 ) 5x^2 − 8x + 3 = 0 x = 1 or x = (3/5) x cannot be 1, since θ < 2π ∴ x = (3/5) tanθ = ((sin θ)/(cos θ)) tanθ = ((−2(1 − x))/x) tanθ = ((2(x − 1))/x) tanθ = ((2((3/5) − 1))/(3/5)) tanθ = − (4/3) But tan 2θ = ((2tan θ)/(1 − tan^2 θ)) tan 2θ = ((2(−(4/3)))/(1 − (−(4/3))^2 )) ∴ tan 2θ = ((24)/7) qed
$$\:{x}\:=\:\mathrm{cos}\theta \\ $$$$\:{x}^{\mathrm{2}} \:=\:\mathrm{cos}^{\mathrm{2}} \theta \\ $$$$\:\mathrm{1}\:−\:{x}^{\mathrm{2}} \:=\:\mathrm{sin}^{\mathrm{2}} \theta \\ $$$$\:\pm\:\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }\:=\:\mathrm{sin}\theta\: \\ $$$${But}\:{for}\:{the}\:{interval},\:\frac{\mathrm{3}\pi}{\mathrm{2}}<\theta<\mathrm{2}\pi\:\mathrm{sine}\: \\ $$$${has}\:{a}\:{negative}\:{value}. \\ $$$$\Rightarrow\:\:−\:\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }\:=\:\mathrm{sin}\theta \\ $$$$\:{so}\:\:\:\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }\:=\:−\mathrm{sin}\theta\:…………\:\left(\mathrm{1}\right) \\ $$$$\:\:\: \\ $$$${But},\:\mathrm{2cos}\theta\:−\:\mathrm{sin}\theta\:=\:\mathrm{2} \\ $$$$\Rightarrow\:\:\:\:\:\:\mathrm{2}\:−\:\mathrm{2}{x}\:=\:−\mathrm{sin}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\left(\mathrm{1}\:−\:{x}\right)\:=\:−\mathrm{sin}\theta\:………..\:\left(\mathrm{2}\right) \\ $$$$\:{from}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right),\: \\ $$$$\:\:\:\:\:\:\:\:\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }\:=\:\mathrm{2}\left(\mathrm{1}\:−\:{x}\right)\:\:\:\:\:\:{qed} \\ $$$$ \\ $$$$\:{It}\:{follows}\:{that} \\ $$$$\:\:\:\:\:\:\mathrm{1}\:−\:{x}^{\mathrm{2}} \:=\:\mathrm{4}\left(\mathrm{1}\:−\:\mathrm{2}{x}\:+\:{x}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\mathrm{5}{x}^{\mathrm{2}} \:−\:\mathrm{8}{x}\:+\:\mathrm{3}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:{x}\:=\:\mathrm{1}\:{or}\:{x}\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\:\:{x}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{1},\:\mathrm{since}\:\theta\:<\:\mathrm{2}\pi \\ $$$$\:\therefore\:\:{x}\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\:\:\mathrm{tan}\theta\:=\:\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta} \\ $$$$\:\:\mathrm{tan}\theta\:=\:\frac{−\mathrm{2}\left(\mathrm{1}\:−\:{x}\right)}{{x}} \\ $$$$\:\:\mathrm{tan}\theta\:=\:\frac{\mathrm{2}\left({x}\:−\:\mathrm{1}\right)}{{x}} \\ $$$$\:\:\mathrm{tan}\theta\:=\:\frac{\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{5}}\:−\:\mathrm{1}\right)}{\frac{\mathrm{3}}{\mathrm{5}}} \\ $$$$\:\:\mathrm{tan}\theta\:=\:−\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${But}\:\mathrm{tan}\:\mathrm{2}\theta\:=\:\frac{\mathrm{2tan}\:\theta}{\mathrm{1}\:−\:\mathrm{tan}^{\mathrm{2}} \theta} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{tan}\:\mathrm{2}\theta\:=\:\frac{\mathrm{2}\left(−\frac{\mathrm{4}}{\mathrm{3}}\right)}{\mathrm{1}\:−\:\left(−\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$\:\therefore\:\:\:\:\mathrm{tan}\:\mathrm{2}\theta\:=\:\:\frac{\mathrm{24}}{\mathrm{7}}\:\:\:\:\:\:\:\:\:\:\:\:{qed} \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$$$\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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