Convert the following polar coordinates to its equivalent Cartesian coordinates.
(i) `(2,pi)`
(ii) `(sqrt(3),pi//6)`

Explanation

(i) For the second quadrant point `(x,y)-=(-1,1)`, we have
`tantheta=(y)/(x)=-1rArrtheta=(3pi)/(4)`

Using positive r, we have
`r=sqrt(x^(2)+y^(2))=sqrt((-1)^(2)+(1)^(2))=sqrt(2)`
So, one set of polar coordinates is `(r, theta)-= (sqrt(2),3pi//4)`
Using negative r, we have coordinates `=(-sqrt(2),-pi//4)`
(ii) For the fourth quadrant point `(x, y)-= (2,-2)`, we have
`tan theta =(-3)/(2)`
`rArr theta = tan^(-1)((-3)/(2))=-"tan"^(-1)^(3)/(2)`
Using positive r, we have
`r=sqrt(x^(2)+y^(2))=sqrt((-3)^(2)+(2)^(2))=sqrt(13)`
So, one set of polar coordinates is
`(r, theta)-= (sqrt(13), -"tan"^(-1)(3)/(2))`

Using negative r, we have coordinates
`=(-sqrt(13),pi-"tan"^(-1)(3)/(2))`