Polar coordinates clarification why is the radians portion wrong despite right angle triangle? Converting $(x,y) = (-6, 2\sqrt{3})$

Question

Trying to plot polar coordinates and got all the right steps until the end where I need to evaluate $\arctan \frac{\sqrt{3}}{-3} = \frac{\pi}{-6}$, but that puts the point on the wrong quadrant. Where did I go wrong?

Answer 1

The inverse tangent function gives an angle $\theta$ satisfying $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ (the right half of the circle when we measure angles in standard position). Sure enough, $$ \theta = \arctan \biggl( -\frac{\sqrt{3}}{3} \biggr) = -\arctan \biggl( \frac{\sqrt{3}}{3} \biggr) = -\frac{\pi}{6} $$ lies on that interval and has the correct tangent value. However, there are many—infinitely many—angles that have that tangent value. In fact, there are two different angles every time you go around the circle! The upshot: you can add or subtract $\pi$ to $\theta$ as many times as you like without changing the value of its tangent.

Look at the $x$ and $y$ coordinates of the point: it’s in the second quadrant, i.e. $\frac{\pi}{2} \leq \theta \leq \pi$.

With a little bit of effort you can see that $$ \theta + \pi = \frac{5\pi}{6} $$ gives an angle that lands in the second quadrant, so this is an appropriate angle coordinate for a positive radial coordinate.

Answer 2

Computed result is $-30^\circ$.

Note that $\tan (180^\circ + x) = \tan x$.

Note also that $\theta =\arctan \left(- \dfrac ab\right)$ is equal to $\arctan \left( \dfrac {-a}b\right)$ or $\arctan \left( \dfrac a{-b}\right)$ and we have to choose the right one, the one that matches the original given.

Answer 3

I would do it solving a system of linear equations: \begin{cases} \cos\theta=-\dfrac{6}{4\sqrt 3}=-\dfrac{\sqrt 3}2 \iff \theta \equiv \pm \dfrac{5\pi} 6\pmod{2\pi},\\ \sin\theta=\dfrac{2\sqrt 3}{4\sqrt 3}=\dfrac12\iff \theta \equiv \dfrac\pi 6,\dfrac{5\pi}6\pmod{2\pi}. \end{cases} So the solution is $\;\theta\equiv\dfrac{5\pi} 6\pmod{2\pi}$, and on the interval $[0,2\pi)$, it is $\;\dfrac{5\pi}6$.