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.