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

1 Answer
A. S. Adikesavan
Sep 28, 2016

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

Explanation:

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

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

gives ) r= sqrt(x^2+y^2)>=0r=√x2+y2≥0, for the principal value,

sin theta = y/sqrt(x^2+y^2) and cos theta = x/sqrt(x^2+y^2)sinθ=y√x2+y2andcosθ=x√x2+y2

Here,

sqrt(x^2+y^2)= 2(x + y)/sqrt(x^2+y^2)√x2+y2=2x+y√x2+y2,

Cross multiplying and reorganizing,

(x-1)^2+(y-1)^2=2(x−1)2+(y−1)2=2

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