Convert the rectangular coordinates to polar coordinates with r > 0 and 0 ≤ θ < 2π. (-sqrt6, -sqrt2) r=(2sqrt2), how do I get theta?
Well, as x = r cos(theta) y = r sin(theta), provided x isn't zero, then y/x = tan(theta)
i.e., for your values of x and y, \[ \tan \theta = 1/\sqrt{3} \] Now solve that and remember in which quadrant this angle should lie.
I still don't get it.
Do you understand why x = r cos(theta) y = r sin(theta) This is the definition of polar coordinates. Now, that being the case, \[ \frac{y}{x} = \frac{r \sin \theta}{r \cos \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta \] Draw a diagram with your point on it in the plane. Then draw the line to that point from the origin. The length of that line segment is r. The angle that line segment makes with the positive x-axis is the angle theta.
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