How do you convert $(- 2, - 2 \sqrt{3})$ $(-2, -2 sqrt3)$ to polar form?

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Answer 1 · 2016-06-08T13:01:08

$(4, \frac{7 π}{6})$ $(4, (7pi)/6)$ .

Use $(x, y) = (- 2, - 2 \sqrt{3}) (r cos θ, r sin θ)$ $(x, y)=(-2, -2sqrt 3)(r cos theta, r sin theta)$ .

$r = \sqrt{x^{2} + y^{2}} = \sqrt{4 + 12} = 4$ $r=sqrt(x^2+y^2)=sqrt(4+12)=4$ .

$cos θ = \frac{x}{r} = - \frac{1}{2} and sin θ = \frac{y}{r} = - \frac{\sqrt{3}}{2}$ $cos theta= x/r =-1/2 and sin theta =y/r =-sqrt 3/2$ .

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

Thus, $θ = π + \frac{π}{6}$ $theta = pi+pi/6$ .so that,

$sin (π + \frac{π}{6}) = - sin (\frac{π}{6}) = - \frac{1}{2}$ $sin (pi+pi/6)=-sin (pi/6)= -1/2$ and

$cos (π + \frac{π}{6}) = - cos (\frac{π}{6}) = - \frac{\sqrt{3}}{2}$ $cos (pi+pi/6)=-cos (pi/6)=-sqrt 3/2$ .

How do you convert (-2, -2 sqrt3)(−2,−2√3) (-2, -2 sqrt3) to polar form?