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Question Number 34647 by JOHNMASANJA last updated on 09/May/18

if   xsin^3 θ + ycos^3 θ=sinθcosθ  and xsinθ −ycosθ=0  prove that x^2  + y^2 =1

$${if}\:\:\:{xsin}^{\mathrm{3}} \theta\:+\:{ycos}^{\mathrm{3}} \theta={sin}\theta{cos}\theta \\ $$$${and}\:{xsin}\theta\:−{ycos}\theta=\mathrm{0} \\ $$$${prove}\:{that}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} =\mathrm{1} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/May/18

xsinθ=ycosθ  (x/(cosθ))=(y/(sinθ))=k(assumed)  x=kcosθ ,y=ksinθ  putting in first equation    kcosθsin^3 θ+ksinθcos^3 θ=sinθcosθ  ksinθcosθ(sin^2 θ+cos^2 θ)=sinθcosθ  so k=1  hence x^2 +y^2   =cos^2 θ+sin^2 θ  =1

$${xsin}\theta={ycos}\theta \\ $$$$\frac{{x}}{{cos}\theta}=\frac{{y}}{{sin}\theta}={k}\left({assumed}\right) \\ $$$${x}={kcos}\theta\:,{y}={ksin}\theta \\ $$$${putting}\:{in}\:{first}\:{equation}\:\: \\ $$$${kcos}\theta{sin}^{\mathrm{3}} \theta+{ksin}\theta{cos}^{\mathrm{3}} \theta={sin}\theta{cos}\theta \\ $$$${ksin}\theta{cos}\theta\left({sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta\right)={sin}\theta{cos}\theta \\ $$$${so}\:{k}=\mathrm{1} \\ $$$${hence}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$$={cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta \\ $$$$=\mathrm{1} \\ $$

Answered by ajfour last updated on 09/May/18

xsin θ=ycos θ  ⇒   xsin θ(sin^2 θ)+ycos θ(cos^2 θ)                          = ((xsin θcos θ)/x)  or    xsin^2 θ+xcos^2 θ=cos θ  ⇒      x=cos θ     similarly           ysin^2 θ+ycos^2 θ=sin θ  ⇒      y=sin θ       hence   x^2 +y^2  = 1  .

$${x}\mathrm{sin}\:\theta={y}\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\:\:{x}\mathrm{sin}\:\theta\left(\mathrm{sin}\:^{\mathrm{2}} \theta\right)+{y}\mathrm{cos}\:\theta\left(\mathrm{cos}\:^{\mathrm{2}} \theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{x}\mathrm{sin}\:\theta\mathrm{cos}\:\theta}{{x}} \\ $$$${or}\:\:\:\:{x}\mathrm{sin}\:^{\mathrm{2}} \theta+{x}\mathrm{cos}\:^{\mathrm{2}} \theta=\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\:\:\:\:\:{x}=\mathrm{cos}\:\theta \\ $$$$\:\:\:{similarly} \\ $$$$\:\:\:\:\:\:\:\:\:{y}\mathrm{sin}\:^{\mathrm{2}} \theta+{y}\mathrm{cos}\:^{\mathrm{2}} \theta=\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:\:\:\:\:{y}=\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:{hence}\:\:\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} \:=\:\mathrm{1}\:\:. \\ $$

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