What is the Cartesian form of rsin(theta) = 2theta+cot(theta)-tan(theta) ?

1 Answer
Feb 18, 2018

y = 2tan^-1(y/x)+x/y-y/x

Explanation:

Given: rsin(theta) = 2theta+cot(theta)-tan(theta)

Substitute theta = tan^-1(y/x)

rsin(theta) = 2tan^-1(y/x)+cot(theta)-tan(theta)

Substitute cot(theta) = cos(theta)/sin(theta)

rsin(theta) = 2tan^-1(y/x)+cos(theta)/sin(theta)-tan(theta)

Substitute tan(theta) = sin(theta)/cos(theta)

rsin(theta) = 2tan^-1(y/x)+cos(theta)/sin(theta)-sin(theta)/cos(theta)

Multiply the last 2 terms by in the form of r/r:

rsin(theta) = 2tan^-1(y/x)+(rcos(theta))/(rsin(theta))-(rsin(theta))/(rcos(theta))

Substitute rsin(theta) = y:

y = 2tan^-1(y/x)+(rcos(theta))/y-y/(rcos(theta))

Substitute rcos(theta) = x:

y = 2tan^-1(y/x)+x/y-y/x

Done.