Question

I am having a bit of trouble finding the polar of (6,-6sqrt3). I have r=12 and the tan (-6sqrt3)/6 however I loose it there.

Answer 1

you have \( x =6, y = -6\sqrt3\) hence \(r^2 = x^2 +y^2 = 144\) then r =12.
the easiest way is taking 12 out. That is
\((6,-6\sqrt3)=12(\dfrac{1}{2}, \dfrac{-\sqrt3}{2})\)
Now which value of \(\theta\) make \(cos \theta =1/2? \) and \(sin\theta =-\sqrt3/2\)