How do you convert  r= cos^2(theta/2) into cartesian form?

Jun 30, 2016

$2 \left({x}^{2} + {y}^{2}\right) = x + \sqrt{{x}^{2} + {y}^{2}}$

Explanation:

With $r = {\cos}^{2} \left(\frac{\theta}{2}\right)$, we can get that $\frac{\theta}{2}$ back into a $\theta$

as we have the double angle formula

$\cos 2 Q = 2 {\cos}^{2} Q - 1$

so

${\cos}^{2} Q = \frac{\cos 2 Q + 1}{2}$

here that means that

$r = \frac{\cos \theta + 1}{2} \implies 2 r = \cos \theta + 1$

now we have $x = r \cos \theta$ so $\cos \theta = \frac{x}{r}$

meaning that

$2 r = \frac{x}{r} + 1$

$2 {r}^{2} = x + r$

in cartesian, that is very very ugly

$2 \left({x}^{2} + {y}^{2}\right) = x + \sqrt{{x}^{2} + {y}^{2}}$