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.