What is the Cartesian form of #(12,(5pi )/3)#?

1 Answer
256
Jan 20, 2018

#(6,-6sqrt(3))#

Explanation:

Remember that polar coordinates are of the form #(r, theta)#.

Also, remember that our #x#-values correspond with cosine and #y#-values with sine.

Then, remember that our sine and cosine values come from the unit circle, where #r=1#. So, when changing our coordinates from polar to cartesian coordinates we are taking #(r, theta) -> (rcos(theta), rsin(theta))#.

Notice that #(5pi)/3=2pi-pi/3#. So, we can say that #theta=-pi/3#, which is in the fourth quadrant. This means that cosine is positive, and sine is negative.

Then we can essentially say that #(12, (5pi)/3) -> (12cos(pi/3),-12sin(pi/3))#
#=(12(1/2),-12((sqrt(3))/2))=(6,-6sqrt(3))#

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