How do you convert # (-2, -2 sqrt3)# to polar form?

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Answer 1 · 2016-06-08T13:01:08

#(4, (7pi)/6)#.

Use #(x, y)=(-2, -2sqrt 3)(r cos theta, r sin theta)#.

#r=sqrt(x^2+y^2)=sqrt(4+12)=4#.

#cos theta= x/r =-1/2 and sin theta =y/r =-sqrt 3/2#.

Both cosine and sine are negative. So, #theta# is in the 4th quadrant.

Thus, #theta = pi+pi/6#.so that,

#sin (pi+pi/6)=-sin (pi/6)= -1/2# and

#cos (pi+pi/6)=-cos (pi/6)=-sqrt 3/2#.