What is the Cartesian form of r^2+r = 2theta-2sectheta-csctheta ?

1 Answer
Oct 27, 2016

The conversion maps to 3 equations. Please see the explanation.

Explanation:

Before we begin the conversion, please observe that the secant function has a division by zero issue at theta = pi/2 + npi. The same is true for the cosecant function and theta = npi. This translates into the Cartesian as a restriction of x !=0 and y != 0

For csc(theta), begin with:

y = rsin(theta)

1/sin(theta) = r/y

1/sin(theta) = sqrt(x^2 + y^2)/y

csc(theta) = sqrt(x^2 + y^2)/y

A similar substitution exists for the secant function:

sec(theta) = sqrt(x^2 + y^2)/x

Substitute x^2 + y^2 for r^2 and sqrt(x^2 + y^2) for r:

x^2 + y^2 + sqrt(x^2 + y^2) = 2theta - 2sqrt(x^2 + y^2)/x - sqrt(x^2 + y^2)/y; x != 0 and y!=0

The substitution for theta breaks the equation into 3 equations:

x^2 + y^2 + sqrt(x^2 + y^2) = 2tan^-1(y/x) - 2sqrt(x^2 + y^2)/x - sqrt(x^2 + y^2)/y; x > 0 and y>0

x^2 + y^2 + sqrt(x^2 + y^2) = 2(tan^-1(y/x) + pi) - 2sqrt(x^2 + y^2)/x - sqrt(x^2 + y^2)/y; x < 0 and y!=0

x^2 + y^2 + sqrt(x^2 + y^2) = 2(tan^-1(y/x) + 2pi) - 2sqrt(x^2 + y^2)/x - sqrt(x^2 + y^2)/y; x > 0 and y<0

Undefined elsewhere.