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I've been studying the unit circle and inverse trig functions on Khan Academy. One of the questions asked, is, what is the arctan of $-\frac{\sqrt{3}}{3}$.

The solution is $-\frac{1}{6}\pi$, I don't understand why?

If I pull up $-\frac{1}{6}\pi$ on a unit circle tool (in this case on Khan Academy). The $y$ (the sine) value on the same angle $-\frac{1}{6}\pi$ is $-\frac{1}{2}$. I see that the $x$ value (the cosine) on unit circle of $-\frac{1}{6}\pi$ is $\frac{\sqrt{3}}{2}$.

Tan is opposite side / adjacent side, or in the unit circle's case sine/cosine. Which suggests to me that the tan of $-\frac{1}{6}\pi$ is $-(1/2)/(\sqrt{3}/2)$ which equals $-1/\sqrt{3}$ not $-\frac{1}{6}\pi$.

Any insight would be great!

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    @drc: Indeed, which is why "3/4 pi" is a poor example. Elsewhere it asks for the arctangent of $-1$, again giving "3/4 pi" as an example, confusingly in my view.2011-12-08

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Maybe you're just missing the fact that $1/\sqrt{3}$ is the same as $\sqrt{3}/3$. You get that by rationalizing the denominator:

$ \frac{1}{\sqrt{3}} = \frac{1\cdot\sqrt{3}}{\sqrt{3}\sqrt{3}} = \frac{\sqrt{3}}{3}. $

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    Wow. Yes I completely missed that. That seems to have been my error. Thank you Michael.2011-12-08
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Maybe you're just messing with the concept of inverse function... Remember:

$\arctan x=y$ if and only if $\tan y =x$ and $y\in ]-\pi/2,\pi/2[$.

Now, you have found $\tan (-\pi/6) =-1/\sqrt{3}=-\sqrt{3}/3$ and you also have $-\pi/6 \in ]-\pi/2, \pi/2[$, hence previous statement applies and it yields $\arctan (-\sqrt{3}/3)=-\pi/6$ as claimed.

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    @drc: If $y\notin ]-\pi/2 ,\pi/2[$ then nothing "happens", in the sense that you cannot give any meaning to the formula $\arctan x=y$. In fact, as you should know from your book, the range of $\mathbb{R}\ni x\mapsto \arctan x\in \mathbb{R}$ is the open interval $]-\pi/2, \pi/2[$.2011-12-08