What is the Cartesian form of rtheta+r = 2costheta-csctheta rθ+r=2cosθcscθ?

1 Answer
Oct 25, 2016

There are 5 conditions that yield 4 equations and a division by 0. Please see the explanation.

Explanation:

Before we begin. Please notice that thetaθ must not be a 2npi2nπ multiple of 0 or pi0orπ, because the cosecant function is undefined at the these points. This converts to the Cartesian restriction y != 0y0

Eventually, we will substitute tan^-1(y/x)tan1(yx) for thetaθ, therefore, we must look at the pristine function, r = (2cos(theta) - csc(theta))/(theta + 1)r=2cos(θ)csc(θ)θ+1, and see what happens at theta = pi/2 and (3pi)/2θ=π2and3π2

r = -2/(pi + 2) and r = -2/(3pi + 2)r=2π+2andr=23π+2

These are the y coordinates at x = 0.

y = -2/(pi + 2) and y = -2/(3pi + 2)y=2π+2andy=23π+2

Now, we may proceed with the conversion.

For csc(theta)csc(θ):

y = rsin(theta)y=rsin(θ)

1/sin(theta) = r/y1sin(θ)=ry

csc(theta) = r/ycsc(θ)=ry

csc(theta) = sqrt(x^2 + y^2)/ycsc(θ)=x2+y2y

r = (2cos(theta) - sqrt(x^2 + y^2)/y)/(theta + 1)r=2cos(θ)x2+y2yθ+1

Multiply both sides by r:

r^2 = (2rcos(theta) - (x^2 + y^2)/y)/(theta + 1)r2=2rcos(θ)x2+y2yθ+1

Substitute x for rcos(theta)rcos(θ)

r^2 = (2x - (x^2 + y^2)/y)/(theta + 1)r2=2xx2+y2yθ+1

Substitute x^2 + y^2x2+y2 for r^2r2

(x^2 + y^2) = (2x - (x^2 + y^2)/y)/((theta + 1)(x2+y2)=2xx2+y2y(θ+1)

Multiply the right side by y/yyy:

(x^2 + y^2) = (2xy - x^2 - y^2)/(y(theta + 1))(x2+y2)=2xyx2y2y(θ+1)

The substitution for thetaθ breaks the equation into 3 equations plus the 2 for x = 0 and y = 0:

Undefined for y = 0

y = -2/(pi + 2) and y = -2/(3pi + 2); x = 0y=2π+2andy=23π+2;x=0

(x^2 + y^2) = (2xy - x^2 - y^2)/(y(tan^-1(y/x) + 1)); x > 0 and y > 0(x2+y2)=2xyx2y2y(tan1(yx)+1);x>0andy>0

(x^2 + y^2) = (2xy - x^2 - y^2)/(y(tan^-1(y/x) + pi + 1)); x < 0 and y != 0(x2+y2)=2xyx2y2y(tan1(yx)+π+1);x<0andy0

(x^2 + y^2) = (2xy - x^2 - y^2)/(y(tan^-1(y/x) + 2pi + 1)); x > 0 and y < 0(x2+y2)=2xyx2y2y(tan1(yx)+2π+1);x>0andy<0