How do you convert #(4, 4pi/3)# into rectangular forms?

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Answer 1 · 2015-12-06T10:41:42

#x=-2#
#y=-2sqrt(3)#

If you know the poalr coordinates #(\rho,\theta)#, then the transformation to have the rectangular ones is

#x=rcos(theta)#
#y=rsin(theta)#

In your case,

#x=4cos((4pi)/3)#
#y=4sin((4pi)/3)#

since #cos((4pi)/3)=-1/2#, and #sin((4pi)/3)=-sqrt(3)/2#, we have

#x=4*(-1/2)=-2#
#y=4*(-sqrt(3)/2)=-2sqrt(3)#