What is the Cartesian form of #( 6 , ( - 16pi)/3 ) #? Calculus Polar Curves Introduction to Polar Coordinates 1 Answer Wataru Apr 10, 2017 #(x,y)=(-3,3sqrt(3))# Explanation: #(r, theta)=(6,-(16pi)/3)# #x=r cos theta=6cos(-(16pi)/3)=6cdot(-1/2)=-3# #y=r sin theta=6sin(-(16pi)/3)=6 cdot sqrt(3)/2=3sqrt(3)# Hence, #(x,y)=(-3,3sqrt(3))# I hope that this was clear. Answer link Related questions How do you find the polar coordinates of #(-4,0)# ? How do you find the polar coordinates of the point #(-2,0)# ? Find the polar coordinates of the point with rectangular coordinates #(0,1)# ? How do you find the polar coordinates of the point with rectangular coordinates #(-2,2# )? How do polar coordinates work? How do you change polar coordinates to rectangular coordinates? What are polar coordinates used for in real life? How do you change rectangular coordinates to polar coordinates? How do you find the rectangular coordinates of the point with polar coordinates #(2,pi/4)#? How do you find the rectangular coordinates of the point with polar coordinates #(-1,pi/3)#? See all questions in Introduction to Polar Coordinates Impact of this question 1357 views around the world You can reuse this answer Creative Commons License