How do you convert #r= 2 sin r + 2 cos r# into cartesian form?

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Answer 1 · 2016-09-28T05:09:50

Perhaps, the equation is #r=2(sin theta + cos theta)#. If so, the cartesian form is #(x-1)^2+(y-1)^2=2#.

I solve this for the corrected version #r = 2 (sin theta + cos theta )#

The conversion equation is #r)cos theta, sin theta ) = (x, y)# that

gives ) #r= sqrt(x^2+y^2)>=0#, for the principal value,

#sin theta = y/sqrt(x^2+y^2) and cos theta = x/sqrt(x^2+y^2)#

Here,

#sqrt(x^2+y^2)= 2(x + y)/sqrt(x^2+y^2)#,

Cross multiplying and reorganizing,

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