Polar equation to rectangular equation

Aug 2018
1
0
NORWAY
Hey, Im having trouble with solving

r=1-cos2(Θ)


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HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
First, is that cosine squared rather than 2 times the angle?
The first would be better written "cos^2(Θ)" and the second "cos(2Θ)". I am going to assume you mean \(\displaystyle cos^2(\theta)\).

So what do you know about converting between Cartesian and polar coordinates?

Do you, for example, know that \(\displaystyle x= r cos(\theta)\) and \(\displaystyle y= r sin(\theta)\)?
Do you know that \(\displaystyle r= \sqrt{x^2+ y^2}\) and \(\displaystyle \theta= arctan\left(\frac{y}{x}\right)\)?

If so you might think about multiplying both sides by \(\displaystyle r^2\) so that you have \(\displaystyle r^3= r^2- r^2cos^2(\theta)\). That is, then, \(\displaystyle (x^2+ y^2)^{3/2}= x^2+ y^2- x^2\) which simplifies to \(\displaystyle (x^2+ y^2)^{3/2}= y^2\).
 
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