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Question Number 7683 by new phone last updated on 08/Sep/16

If sin θ+2cos θ=1 Prove that 2sin θ−cos θ=2

$$\mathrm{If}\:\mathrm{sin}\:\theta+\mathrm{2cos}\:\theta=\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{2sin}\:\theta−\mathrm{cos}\:\theta=\mathrm{2} \\ $$

Commented by sandy_suhendra last updated on 08/Sep/16

sinθ+^ 2cosθ=1 (sinθ+2cosθ)^2 =1^2 sin^2 θ+4sinθcosθ+4cos^2 θ=1 4sinθcosθ=1−sin^2 θ−4cos^2 θ (2sinθ−cosθ)^2 =4sin^2 θ−4sinθcosθ+cos^2 θ =4sin^2 θ−(1−sin^2 θ−4cos^2 θ)+cos^2 θ =5sin^2 θ+5cos^2 θ−1 =5(sin^2 θ+cos^2 θ)−1 =5−1=4 so (2sinθ−cosθ)=±2

$${sin}\theta+^{} \mathrm{2}{cos}\theta=\mathrm{1} \\ $$$$\left({sin}\theta+\mathrm{2}{cos}\theta\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} \\ $$$${sin}^{\mathrm{2}} \theta+\mathrm{4}{sin}\theta{cos}\theta+\mathrm{4}{cos}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$$\mathrm{4}{sin}\theta{cos}\theta=\mathrm{1}−{sin}^{\mathrm{2}} \theta−\mathrm{4}{cos}^{\mathrm{2}} \theta \\ $$$$ \\ $$$$\left(\mathrm{2}{sin}\theta−{cos}\theta\right)^{\mathrm{2}} \\ $$$$=\mathrm{4}{sin}^{\mathrm{2}} \theta−\mathrm{4}{sin}\theta{cos}\theta+{cos}^{\mathrm{2}} \theta \\ $$$$=\mathrm{4}{sin}^{\mathrm{2}} \theta−\left(\mathrm{1}−{sin}^{\mathrm{2}} \theta−\mathrm{4}{cos}^{\mathrm{2}} \theta\right)+{cos}^{\mathrm{2}} \theta \\ $$$$=\mathrm{5}{sin}^{\mathrm{2}} \theta+\mathrm{5}{cos}^{\mathrm{2}} \theta−\mathrm{1} \\ $$$$=\mathrm{5}\left({sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta\right)−\mathrm{1} \\ $$$$=\mathrm{5}−\mathrm{1}=\mathrm{4} \\ $$$${so}\:\:\left(\mathrm{2}{sin}\theta−{cos}\theta\right)=\pm\mathrm{2} \\ $$

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