What is the Cartesian form of #( 6,-pi/3 )#?

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Answer 1 · 2018-07-18T14:45:54

#(3, -3\sqrt3)#

The Cartesian coordinates #(x, y)# of the point #(6, -{\pi}/3)\equiv(r, \theta)# are given as follows

#x=r\cos\theta#

#=6\cos(-{\pi}/3)#

#=6\cos(\pi/3)#

#=6\cdot 1/2#

#=3#

#y=r\sin\theta#

#=6\sin(-{\pi}/3)#

#=-6\sin(\pi/3)#

#=-6\cdot \sqrt3/2#

#=-3\sqrt3#

hence, the Cartesian coordinates are #(3, -3\sqrt3)#