How do you convert #r^2=cos4theta# into cartesian form?

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Answer 1 · 2017-02-10T07:05:53

#(x^2+y^2)^3=8y^4-8y(x^2+y^2)+(x^2+y^2)^2#

As #cos2A=2cos^2A-1#, we have

#cos4A=cos2(2A)=2cos^2 2A-1#

= #2(2cos^2A-1)^2-1#

= #2(4cos^4A-4cos^2A+1)-1#

= #8cos^4A-8cos^2A+1#

Hence #r^2=cos4theta# is same as #r^2=8cos^4theta-8cos^2theta+1#

Now relation between #(r,theta)# in polar coordinates and #(x,y)# in Cartesian coordinates is given by

#x=rcostheta# and #y=rsintheta# i.e. #r^2=x^2+y^2# and hence #costheta=y/sqrt(x^2+y^2)#

and #r^2=8cos^4theta-8cos^2 2theta+1#

#hArrx^2+y^2=8(y/sqrt(x^2+y^2))^4-8(y/sqrt(x^2+y^2))^2+1#

or #x^2+y^2=(8y^4)/(x^2+y^2)^2-(8y)/(x^2+y^2)+1#

or #(x^2+y^2)^3=8y^4-8y(x^2+y^2)+(x^2+y^2)^2#