How do you convert $r = {cos}^{2} (\frac{θ}{2})$ $r= cos^2(theta/2)$ into cartesian form?

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Answer 1 · 2016-06-30T23:31:45

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

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

as we have the double angle formula

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

so

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

here that means that

$r = \frac{cos θ + 1}{2} \Rightarrow 2 r = cos θ + 1$ $r = (cos theta +1)/2 implies 2 r = cos theta +1$

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

meaning that

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

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

in cartesian, that is very very ugly

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

How do you convert r= cos^2(theta/2)r=cos2(θ2) r= cos^2(theta/2) into cartesian form?