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

1 Answer
Feb 18, 2018

y = 2tan^-1(y/x)+x/y-y/xy=2tan1(yx)+xyyx

Explanation:

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

Substitute theta = tan^-1(y/x)θ=tan1(yx)

rsin(theta) = 2tan^-1(y/x)+cot(theta)-tan(theta)rsin(θ)=2tan1(yx)+cot(θ)tan(θ)

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

rsin(theta) = 2tan^-1(y/x)+cos(theta)/sin(theta)-tan(theta)rsin(θ)=2tan1(yx)+cos(θ)sin(θ)tan(θ)

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

rsin(theta) = 2tan^-1(y/x)+cos(theta)/sin(theta)-sin(theta)/cos(theta)rsin(θ)=2tan1(yx)+cos(θ)sin(θ)sin(θ)cos(θ)

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

rsin(theta) = 2tan^-1(y/x)+(rcos(theta))/(rsin(theta))-(rsin(theta))/(rcos(theta))rsin(θ)=2tan1(yx)+rcos(θ)rsin(θ)rsin(θ)rcos(θ)

Substitute rsin(theta) = yrsin(θ)=y:

y = 2tan^-1(y/x)+(rcos(theta))/y-y/(rcos(theta))y=2tan1(yx)+rcos(θ)yyrcos(θ)

Substitute rcos(theta) = xrcos(θ)=x:

y = 2tan^-1(y/x)+x/y-y/xy=2tan1(yx)+xyyx

Done.