How do you convert r=2 sin(θ+pi/3) from polar to rectangular form?

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Answer 1 · 2018-03-16T12:48:29

Rectangular form is $(x-sqrt3/2)^2 +(y-1/2)^2=1$

We know , $r^2=x^2+y^2 , x= rcos theta , y= r sin theta$

$r=2 sin (theta+pi/3)$ or

$r^2=2r sin (theta+pi/3)$ or

$r^2=2r(sin theta cos (pi/3)+cos theta sin (pi/3))$ or

$r^2=2r(sin theta/2+cos theta*sqrt3/2)$ or

$r^2=rsin theta+ sqrt3 rcos theta$ or

$x^2+y^2= y +sqrt3 x$ or

$x^2- sqrt3 x +y^2-y = 0$ or

$x^2- sqrt3 x +(sqrt3/2)^2+y^2-y+(1/2)^2= 3/4+1/4$ or

$(x-sqrt3/2)^2 +(y-1/2)^2=1$

Rectangular form is $(x-sqrt3/2)^2 +(y-1/2)^2=1$

graph{(x-sqrt3/2)^2+(y-1/2)^2=1 [-10, 10, -5, 5]}