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Question Number 124309 by bemath last updated on 02/Dec/20
If sin θ+2cos θ=1 ; then 2sin θ−cos θ=?
$$\:{If}\:\mathrm{sin}\:\theta+\mathrm{2cos}\:\theta=\mathrm{1}\:;\:{then}\: \\ $$$$\:\mathrm{2sin}\:\theta−\mathrm{cos}\:\theta=? \\ $$
Answered by MJS_new last updated on 02/Dec/20
t=tan θ sin θ +2cos θ =((t+2)/( (√(t^2 +1))))=1 ⇒ t=−(3/4) 2sin θ −cos θ =((2t−1)/( (√(t^2 +1))))=−2
$${t}=\mathrm{tan}\:\theta \\ $$$$\mathrm{sin}\:\theta\:+\mathrm{2cos}\:\theta\:=\frac{{t}+\mathrm{2}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}=\mathrm{1}\:\Rightarrow\:{t}=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{2sin}\:\theta\:−\mathrm{cos}\:\theta\:=\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}=−\mathrm{2} \\ $$
Answered by liberty last updated on 02/Dec/20
let 2sin θ−cos θ = p ⇔ 4sin^2 θ−4sin θcos θ+cos^2 θ = p^2 ⇔ sin^2 θ+4sin θcos θ+4cos^2 θ = 1 (1)+(2)⇒ 5 = p^2 +1 ; give p = ± 2 Hence 2sin θ−cos θ = ± 2
$$\:{let}\:\mathrm{2sin}\:\theta−\mathrm{cos}\:\theta\:=\:{p}\: \\ $$$$\:\Leftrightarrow\:\mathrm{4sin}\:^{\mathrm{2}} \theta−\mathrm{4sin}\:\theta\mathrm{cos}\:\theta+\mathrm{cos}\:^{\mathrm{2}} \theta\:=\:{p}^{\mathrm{2}} \\ $$$$\Leftrightarrow\:\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{4sin}\:\theta\mathrm{cos}\:\theta+\mathrm{4cos}\:^{\mathrm{2}} \theta\:=\:\mathrm{1} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\Rightarrow\:\mathrm{5}\:=\:{p}^{\mathrm{2}} +\mathrm{1}\:;\:{give}\:{p}\:=\:\pm\:\mathrm{2} \\ $$$${Hence}\:\mathrm{2sin}\:\theta−\mathrm{cos}\:\theta\:=\:\pm\:\mathrm{2} \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 02/Dec/20
t+2(√(1−t^2 )) =1 4−4t^2 =1+t^2 −2t⇒5t^2 −2t−3=0⇒t=((2±8)/(10)).=1 ,((−3)/5) 2sinθ−cosθ=2
$${t}+\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:=\mathrm{1} \\ $$$$\mathrm{4}−\mathrm{4}{t}^{\mathrm{2}} =\mathrm{1}+{t}^{\mathrm{2}} −\mathrm{2}{t}\Rightarrow\mathrm{5}{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{3}=\mathrm{0}\Rightarrow{t}=\frac{\mathrm{2}\pm\mathrm{8}}{\mathrm{10}}.=\mathrm{1}\:,\frac{−\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{2}{sin}\theta−{cos}\theta=\mathrm{2}\: \\ $$
Commented by bemath last updated on 02/Dec/20
sir liberty . your correct but if sin θ=−(3/5) then cos θ=(4/5) ( 4^(th) quadrant) and { ((2sin θ−cos θ=−(6/5)−(4/5)=−2)),((sin θ+2cos θ=−(3/5)+(8/5)=1)) :} Thanks you sir liberty and sir dwaipayan
$${sir}\:{liberty}\:.\:{your}\:{correct}\:{but}\: \\ $$$${if}\:\mathrm{sin}\:\theta=−\frac{\mathrm{3}}{\mathrm{5}}\:{then}\:\mathrm{cos}\:\theta=\frac{\mathrm{4}}{\mathrm{5}}\:\left(\:\mathrm{4}^{{th}} \:{quadrant}\right) \\ $$$${and}\:\begin{cases}{\mathrm{2sin}\:\theta−\mathrm{cos}\:\theta=−\frac{\mathrm{6}}{\mathrm{5}}−\frac{\mathrm{4}}{\mathrm{5}}=−\mathrm{2}}\\{\mathrm{sin}\:\theta+\mathrm{2cos}\:\theta=−\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{8}}{\mathrm{5}}=\mathrm{1}}\end{cases} \\ $$$${Thanks}\:{you}\:{sir}\:{liberty}\:{and}\:{sir}\:{dwaipayan} \\ $$$$ \\ $$
Commented by liberty last updated on 02/Dec/20
if sin θ=−(3/5) ∧ cos θ=−(4/5) then 2sin θ−cos θ=−(6/5)−(−(4/5))=−(2/5) and sin θ+2cos θ=−(3/5)+2(−(4/5))=−(3/5)−(8/5) ≠ 1
$${if}\:\mathrm{sin}\:\theta=−\frac{\mathrm{3}}{\mathrm{5}}\:\wedge\:\mathrm{cos}\:\theta=−\frac{\mathrm{4}}{\mathrm{5}}\:{then}\: \\ $$$$\mathrm{2sin}\:\theta−\mathrm{cos}\:\theta=−\frac{\mathrm{6}}{\mathrm{5}}−\left(−\frac{\mathrm{4}}{\mathrm{5}}\right)=−\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${and}\:\mathrm{sin}\:\theta+\mathrm{2cos}\:\theta=−\frac{\mathrm{3}}{\mathrm{5}}+\mathrm{2}\left(−\frac{\mathrm{4}}{\mathrm{5}}\right)=−\frac{\mathrm{3}}{\mathrm{5}}−\frac{\mathrm{8}}{\mathrm{5}}\:\neq\:\mathrm{1} \\ $$
Commented by Dwaipayan Shikari last updated on 02/Dec/20
Oh ! thanking you
$${Oh}\:!\:{thanking}\:\:{you} \\ $$
Answered by Dwaipayan Shikari last updated on 02/Dec/20
sinθ+2cosθ=1 sin^2 θ+4cos^2 θ+4sinθcosθ=1⇒4sin^2 θ+cos^2 θ−4sinθcosθ=4 (2sinθ−cosθ)^2 =4⇒2sinθ−cosθ=±2
$${sin}\theta+\mathrm{2}{cos}\theta=\mathrm{1} \\ $$$${sin}^{\mathrm{2}} \theta+\mathrm{4}{cos}^{\mathrm{2}} \theta+\mathrm{4}{sin}\theta{cos}\theta=\mathrm{1}\Rightarrow\mathrm{4}{sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta−\mathrm{4}{sin}\theta{cos}\theta=\mathrm{4} \\ $$$$\left(\mathrm{2}{sin}\theta−{cos}\theta\right)^{\mathrm{2}} =\mathrm{4}\Rightarrow\mathrm{2}{sin}\theta−{cos}\theta=\pm\mathrm{2} \\ $$

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