Write the equation $r = 2 cos 2 θ$ $r=2cos2theta$ in polar form in Cartesian or rectangular form?

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Answer 1 · 2017-12-06T16:10:47

${(x^{2} + y^{2})}^{\frac{3}{2}} = 2 (x^{2} - y^{2})$ $(x^2+y^2)^(3/2)=2(x^2-y^2)$

The relation between polar coordinates $(r, θ)$ $(r,theta)$ and rectangular or Cartesian coordinates $(x, y)$ $(x,y)$ is given by

$x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ i.e. $x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

Hence $r = 2 cos 2 θ = 2 ({cos}^{2} θ - {sin}^{2} θ)$ $r=2cos2theta=2(cos^2theta-sin^2theta)$

or $r^{3} = 2 (r^{2} {cos}^{2} θ - r^{2} {sin}^{2} θ)$ $r^3=2(r^2cos^2theta-r^2sin^2theta)$

i.e. ${(x^{2} + y^{2})}^{\frac{3}{2}} = 2 (x^{2} - y^{2})$ $(x^2+y^2)^(3/2)=2(x^2-y^2)$

The graph appears as shown below.

graph{(x^2+y^2)^(3/2)=2(x^2-y^2) [-2.5, 2.5, -1.25, 1.25]}

Write the equation r=2cos2thetar=2cos2θr=2cos2theta in polar form in Cartesian or rectangular form?