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This is possible for $r=\sin(2θ)$: Polar to cartesian form of $ r = \sin(2\theta)$

Surely there is some trig identity that may substitute for $cos(2θ)$ and allow for a similar coordinates transfer. What is the cartesian form of $\cos(2\theta)$?

I found something remotely similar: $$\cos(2θ) = \cos^2θ − \sin^2θ = 2 \cos^2θ − 1 = 1 − 2 \sin^2θ$$ (source: http://www.math.ups.edu/~martinj/courses/fall2005/m122/122factsheet.pdf)

However they all use a squared form of sin or cos, which I am not certain how to convert into Cartesian coordinates.

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User3910
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2 Answers2

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By the double angle formulas, $$r=\cos(2\theta)=\cos^2\theta-\sin^2\theta=\frac{x^2}{r^2}-\frac{y^2}{r^2}=\frac{x^2-y^2}{r^2}$$ This leads, because $r^2=x^2+y^2$, to $$x^2-y^2=r^3=\left(x^2+y^2\right)^{3/2}$$ You should then be able to square, multiple terms out and find the equation in implicit form.

Wolfram Alpha gives several approximations for $y=f(x)$, but none of them are nice. I'd recommend sticking to polar coordinates when analyzing this function.

HDE 226868
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  • I squared:$(x^2-y^2)^2=(x^2+y^2)^3$ however, further expansion seems fruitless: $x^4-2x^2y^2+y^4=x^6+3x^4y^2+3x^2y^4+y^6$ – User3910 Sep 16 '16 at 15:25
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    @Dale You won't get an equation of the form $y=f(x)$; this is also the case in the question you linked to. Transforming from polar to Cartesian with functions involving trigonometric functions often leaves you with expressions like this. – HDE 226868 Sep 16 '16 at 15:26
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All polar to Cartesian / Cartesian to polar transformations derive from these simple rules

$r^2 = x^2 + y^2\\ x = r \cos \theta\\ y = r \sin \theta$

$r = cos 2\theta$ the 4 petaled rose. If it had an elegant form in Cartesian we would teach it. It will likely be a cubic or quartic equation.

$r = \cos^2 \theta - sin^2 \theta\\ r^3 = r^2 \cos^2 \theta - r^2 \sin^2 \theta\\ (x^2+y^2)^{\frac32} = x^2 - y^2\\ (x^2 + y^2)^3 = (x^2 - y^2)^2$

I suppose we could keep multiplying that out. But I think that looks pretty elegant.

user317176
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  • I just wanted to add that it is possible to scale the cartesian equation by dividing the equation so that it is equal to a constant: $\frac{(x^2-y^2)^2}{(x^2+y^2)^3}=constant$ – User3910 Sep 16 '16 at 15:45