How do you convert $r^{2} = sin 2 (θ)$ $r^2=sin2(theta)$ to rectangular form?

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Answer 1 · 2016-02-08T01:18:27

Use the following identities:

$sin 2 θ = 2 cos θ sin θ$ $sin2theta=2costhetasintheta$
$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$
$x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

Using the above identities ...

$r^{2} = sin 2 (θ)$ $r^2=sin2(theta)$

$x^{2} + y^{2} = 2 (\frac{x}{r}) (\frac{y}{r}) = \frac{2 x y}{r^{2}} = \frac{2 x y}{x^{2} + y^{2}}$ $x^2+y^2=2(x/r)(y/r)=(2xy)/r^2=(2xy)/(x^2+y^2)$

Now simplify ...

${(x^{2} + y^{2})}^{2} = 2 x y$ $(x^2+y^2)^2=2xy$

$x^{4} + 2 x^{2} y^{2} + y^{4} - 2 x y = 0$ $x^4+2x^2y^2+y^4-2xy=0$

hope that helped

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How do you convert r^2=sin2(theta)r2=sin2(θ)r^2=sin2(theta) to rectangular form?