What is the Cartesian equivalent of polar coordinates #(2, pi/6)#?

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Answer 1 · 2014-10-02T00:15:14

#(r,theta)->(2,pi/6)#

#(x,y)->(rcos(theta),rsin(theta))#

Substitute in #r# and #theta#

#(x,y)->(2cos(pi/6),2sin(pi/6))#

Remember back to the unit circle and special triangles.

#pi/6=30^circ#

#cos(pi/6)=sqrt(3)/2#

#sin(pi/6)=1/2#

Substitute in those values.

#(x,y)->(2*sqrt(3)/2,2*1/2)#

#(x,y)->(sqrt(3),1)#