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Answer 1 · 2018-02-07T19:11:49

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

I am assuming -3.14 is supposed to be #-pi#.

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

Where #r= "radius"#, in this case #1#

#x=1*cos(-pi/6)=sqrt(3)/2#

#y=1*sin(-pi/6)=-1/2#

Coordinates of point on unit circle are:

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

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