# How do you convert r^2=cos4theta into cartesian form?

Feb 10, 2017

${\left({x}^{2} + {y}^{2}\right)}^{3} = 8 {y}^{4} - 8 y \left({x}^{2} + {y}^{2}\right) + {\left({x}^{2} + {y}^{2}\right)}^{2}$

#### Explanation:

As $\cos 2 A = 2 {\cos}^{2} A - 1$, we have

$\cos 4 A = \cos 2 \left(2 A\right) = 2 {\cos}^{2} 2 A - 1$

= $2 {\left(2 {\cos}^{2} A - 1\right)}^{2} - 1$

= $2 \left(4 {\cos}^{4} A - 4 {\cos}^{2} A + 1\right) - 1$

= $8 {\cos}^{4} A - 8 {\cos}^{2} A + 1$

Hence ${r}^{2} = \cos 4 \theta$ is same as ${r}^{2} = 8 {\cos}^{4} \theta - 8 {\cos}^{2} \theta + 1$

Now relation between $\left(r , \theta\right)$ in polar coordinates and $\left(x , y\right)$ in Cartesian coordinates is given by

$x = r \cos \theta$ and $y = r \sin \theta$ i.e. ${r}^{2} = {x}^{2} + {y}^{2}$ and hence $\cos \theta = \frac{y}{\sqrt{{x}^{2} + {y}^{2}}}$

and ${r}^{2} = 8 {\cos}^{4} \theta - 8 {\cos}^{2} 2 \theta + 1$

$\Leftrightarrow {x}^{2} + {y}^{2} = 8 {\left(\frac{y}{\sqrt{{x}^{2} + {y}^{2}}}\right)}^{4} - 8 {\left(\frac{y}{\sqrt{{x}^{2} + {y}^{2}}}\right)}^{2} + 1$

or ${x}^{2} + {y}^{2} = \frac{8 {y}^{4}}{{x}^{2} + {y}^{2}} ^ 2 - \frac{8 y}{{x}^{2} + {y}^{2}} + 1$

or ${\left({x}^{2} + {y}^{2}\right)}^{3} = 8 {y}^{4} - 8 y \left({x}^{2} + {y}^{2}\right) + {\left({x}^{2} + {y}^{2}\right)}^{2}$