How do you convert #r= 5(sin(2x))# into cartesian form?

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Answer 1 · 2016-04-15T17:16:03

#x^6+2x^4y^2+x^2y^4+y^2x^4+2x^2y^4+y^6-100x^2y^2=0#

Use Formula #sin2A=2sinAcosA#

#r=5(2sinxcosx)#

#r=10sinxcosx#

#rxxr^2=10xxrsinx xx rcosx#

#sqrt(x^2+y^2) (x^2+y^2)=10 xy#

#[sqrt(x^2+y^2) (x^2+y^2)]^2=(10 xy)^2#

#(x^2+y^2)(x^2+y^2)^2=100x^2y^2#

#(x^2+y^2)(x^4+2x^2y^2+y^4)=100x^2y^2#

#x^6+2x^4y^2+x^2y^4+y^2x^4+2x^2y^4+y^6-100x^2y^2=0#