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Question Number 7608 by Tawakalitu. last updated on 06/Sep/16

If x = cosΘ − sinΘ and y = cos2Θ Show that, y = x(√(2 − x^2 ))

$${If}\:\:{x}\:=\:{cos}\Theta\:−\:{sin}\Theta \\ $$$${and}\:\:{y}\:=\:{cos}\mathrm{2}\Theta \\ $$$${Show}\:{that},\:\: \\ $$$${y}\:=\:{x}\sqrt{\mathrm{2}\:−\:{x}^{\mathrm{2}} } \\ $$

Commented by sou1618 last updated on 06/Sep/16

(√(2−x^2 ))=(√(1+2cosθsinθ)) =∣cosθ+sinθ∣ so x(√(2−x))=(cosθ−sinθ)∣cosθ+sinθ∣ y=cos2θ=cos^2 θ−sin^2 θ =(cosθ−sinθ)(cosθ+sinθ) when (cosθ+sinθ)≥0 y=x(√(2−x^2 )) when (cosθ+sinθ)<0 y≠x(√(2−x^2 ))

$$\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }=\sqrt{\mathrm{1}+\mathrm{2}{cos}\theta{sin}\theta} \\ $$$$=\mid{cos}\theta+{sin}\theta\mid \\ $$$${so} \\ $$$${x}\sqrt{\mathrm{2}−{x}}=\left({cos}\theta−{sin}\theta\right)\mid{cos}\theta+{sin}\theta\mid \\ $$$$ \\ $$$${y}={cos}\mathrm{2}\theta={cos}^{\mathrm{2}} \theta−{sin}^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:=\left({cos}\theta−{sin}\theta\right)\left({cos}\theta+{sin}\theta\right) \\ $$$$ \\ $$$$ \\ $$$${when}\:\left({cos}\theta+{sin}\theta\right)\geqslant\mathrm{0} \\ $$$${y}={x}\sqrt{\mathrm{2}−{x}^{\mathrm{2}} } \\ $$$${when}\:\left({cos}\theta+{sin}\theta\right)<\mathrm{0} \\ $$$${y}\neq{x}\sqrt{\mathrm{2}−{x}^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$

Commented by Tawakalitu. last updated on 06/Sep/16

Thank you sir

$${Thank}\:{you}\:{sir} \\ $$

Commented by Tawakalitu. last updated on 06/Sep/16

i appreciate

$${i}\:{appreciate} \\ $$

Answered by sandy_suhendra last updated on 06/Sep/16

x(√(2−x^2 )) =(cosθ−sinθ)(√(2(cos^2 θ+sin^2 θ)−(cosθ−sinθ)^2 )) =(cosθ−sinθ)(√(2cos^2 θ+2sin^2 θ)−(cos^2 θ+sin^2 θ−2sinθcosθ))) =(cosθ−sinθ)(√(cos^2 θ+sin^2 θ+2sinθcosθ)) =(cosθ−sinθ)(√((cosθ+sinθ)^2 )) =(cosθ−sinθ)(cosθ+sinθ) =cos^2 θ−sin^2 θ =cos2θ =y

$${x}\sqrt{\mathrm{2}−{x}^{\mathrm{2}} } \\ $$$$=\left({cos}\theta−{sin}\theta\right)\sqrt{\mathrm{2}\left({cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta\right)−\left({cos}\theta−{sin}\theta\right)^{\mathrm{2}} } \\ $$$$\left.=\left({cos}\theta−{sin}\theta\right)\sqrt{\left.\mathrm{2}{cos}^{\mathrm{2}} \theta+\mathrm{2}{sin}^{\mathrm{2}} \theta\right)−\left({cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta−\mathrm{2}{sin}\theta{cos}\theta\right.}\right) \\ $$$$=\left({cos}\theta−{sin}\theta\right)\sqrt{{cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta+\mathrm{2}{sin}\theta{cos}\theta} \\ $$$$=\left({cos}\theta−{sin}\theta\right)\sqrt{\left({cos}\theta+{sin}\theta\right)^{\mathrm{2}} } \\ $$$$=\left({cos}\theta−{sin}\theta\right)\left({cos}\theta+{sin}\theta\right) \\ $$$$={cos}^{\mathrm{2}} \theta−{sin}^{\mathrm{2}} \theta \\ $$$$={cos}\mathrm{2}\theta \\ $$$$={y} \\ $$

Commented by Tawakalitu. last updated on 06/Sep/16

i really appreciate your effort sir. thanks so much.

$${i}\:{really}\:{appreciate}\:{your}\:{effort}\:{sir}.\:{thanks}\:{so}\:{much}. \\ $$

Commented by sou1618 last updated on 06/Sep/16

(√((cosθ+sinθ)^2 ))−(cosθ+sinθ)=^(??) 0 if θ=0 (√((1+0)^2 ))−(1+0)=0 ifθ=π (√((0−1)^2 ))−(0−1)=2 I think (√a^2 )=∣a∣ e.g. a=−2 (√((−2)^2 ))=2

$$\sqrt{\left({cos}\theta+{sin}\theta\right)^{\mathrm{2}} }−\left({cos}\theta+{sin}\theta\right)\overset{??} {=}\mathrm{0} \\ $$$${if}\:\theta=\mathrm{0} \\ $$$$\sqrt{\left(\mathrm{1}+\mathrm{0}\right)^{\mathrm{2}} }−\left(\mathrm{1}+\mathrm{0}\right)=\mathrm{0} \\ $$$${if}\theta=\pi \\ $$$$\sqrt{\left(\mathrm{0}−\mathrm{1}\right)^{\mathrm{2}} }−\left(\mathrm{0}−\mathrm{1}\right)=\mathrm{2} \\ $$$$ \\ $$$${I}\:{think}\:\sqrt{{a}^{\mathrm{2}} }=\mid{a}\mid \\ $$$${e}.{g}. \\ $$$${a}=−\mathrm{2} \\ $$$$\:\:\:\sqrt{\left(−\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{2} \\ $$

Commented by Rasheed Soomro last updated on 07/Sep/16

I agree with you.

$${I}\:{agree}\:{with}\:{you}. \\ $$

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