Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

$${sin}\theta+\mathrm{2}{cos}\theta=\mathrm{1} \\ $$$${sin}^{\mathrm{2}} \theta+\mathrm{4}{cos}^{\mathrm{2}} \theta+\mathrm{4}{sin}\theta{cos}\theta=\mathrm{1}\Rightarrow\mathrm{4}{sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta−\mathrm{4}{sin}\theta{cos}\theta=\mathrm{4} \\ $$$$\left(\mathrm{2}{sin}\theta−{cos}\theta\right)^{\mathrm{2}} =\mathrm{4}\Rightarrow\mathrm{2}{sin}\theta−{cos}\theta=\pm\mathrm{2} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com