What is the Cartesian form of #( -2 , (13pi)/3 ) #?

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Answer 1 · 2016-01-09T05:05:35

#(-1, -sqrt(3))#

For the x:

#x=r cos theta#

#x=(-2)*cos (13 pi/3)= -1#

#x=-1#

For the y:

#y=r* sin theta#

#y=(-2)*sin (13 pi/3)=-sqrt(3)#

#y=-sqrt(3)#

the required answer is #(-1, -sqrt(3))#