How do you convert #r=sin2(theta) # in rectangular form?

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Answer 1 · 2018-05-13T14:45:13

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

# r = sin2theta #

#=> r = 2sintheta costheta #

We know # rcos theta = x #

#=> cos theta = x / r = x / ( sqrt(x^2 + y^2 ) #

#=> sqrt(x^2 + y^2) = 2 * x/sqrt(x^2 +y^2) * y/sqrt(x^2 + y^2 ) #

#=> sqrt(x^2 + y^2 ) = (2xy)/(x^2 + y^2 ) #

#=> (x^2 + y^2) ^ (3/2) = 2xy #

#=> (x^2 + y^2 ) ^3 = 4x^2y^2 #