Which quadrant is #(-4, -11pi/6)# in?

1 Answer
Nov 24, 2015

Quadrant #III#

Explanation:

To find a point #(r, theta)# in polar coordinates, imagine an arrow which rotates #theta# radians counterclockwise from the pole (what would be the positive #x# axis in rectangular coordinates) and then extends #r# in that direction.

Rotating negative radians counterclockwise is the same as rotating the same number clockwise. Thus, as #3pi/2# is a three-quarter rotation,#2pi# is a full rotation, and #3pi/2 < (11pi)/6 < 2pi#
rotating #(11pi)/6# in the clockwise direction will result in pointing to quadrant #I#

However, similarly, extending a negative value is the same as extending a positive value in the opposite direction. Thus the resulting arrow will be in quadrant #III#


Alternatively, you could note that to convert from polar coordinates to rectangular, we let #x = rcos(theta)# and #y = rsin(theta)#

Then here,
#x = (-4)cos(-(11pi)/6) = -2sqrt(3)#
#y = (-4)sin(-(11pi)/6) = -2#

and #(-2sqrt(3), -2)# is in quadrant #III#