Question

What is the Cartesian form of `(-16,(-23pi)/4))`?

Answer 1

(-`16/sqrt2, -16/sqrt2`)

Explanation:

The given point is r= -16 and `theta=(-23pi)/4`

The cartesian coordinates would be x= `rcostheta` and y = `r sin theta`

Accordingly, x= `-16 cos ((-23pi)/4)` and y= `-16sin((-23pi)/4)`

Hence x= -16 cos `-(6pi-pi/4)` and y=-16sin `-(6pi-pi/4)`

x= -16cos `(6pi-pi/4)` and y= 16sin `(6pi-pi/4)`

x=-16 cos`(-pi/4)` and y= 16 sin `-(pi/4)`

x= -16 cos `pi/4` and y= -16 `sin pi/4`

x=-`16/sqrt2` and y= -`16/sqrt2`