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,

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

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