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

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

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