How do you convert #r=1+0.5cos2(theta)# into cartesian form?

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Answer 1 · 2016-09-19T09:50:39

#x^2/sqrt(x^2+y^2)-x^2/(x^2+y^2)=1/2#

The pass equations are

#((x=rcostheta),(y=rsintheta))#

and also keeping in mind that #cos2theta =1-2sin^2theta#

#r = 1+(1-2sin^2theta)/2 = 1+1/2-sin^2theta = 1/2+cos^2theta#

then

#x/costheta = 1/2 + cos^2theta#

but #costheta = x/sqrt(x^2+y^2)#

so

#x^2/sqrt(x^2+y^2)-x^2/(x^2+y^2)=1/2#