How does one find the rectangular coordinates of a point with polar coordinates (-2, 11pi/6)?

1 Answer
Apr 27, 2018

#(-sqrt(3),1)#

Explanation:

In general polar coordinates are #(r,theta)#
for some radius #r# and angle #theta# counter-clockwise from the #x#-axis. This forms a right triangle with hypotenuse #r# connecting the origin and #(r,theta)#.

The Pythagorean trigonometric identity is:
#sin^2(theta)+cos^2(theta)=1#
Multiplying both sides by #r# gives us :
#rsin^2(theta)+rcos^2(theta)=r#

Going back to the triangle, this tells us that if the hypotenuse is #r# then the side going vertically up (#y#-coordinate) is #rsin(theta)#
and the side going horizontal (#x#-coordinate) is #rcos(theta)#.
Therefore #(x,y) = (rcos(theta),rsin(theta))#.
Since the angle is given in a multiple of pi, I hope you mean radians.

We have #(r,theta)# = #(-2,(11pi)/6)# so #(x,y) = (-2cos((11pi)/6),-2sin((11pi)/6))# = #(-sqrt(3),1)#