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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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    $\begingroup$ The arctangent is odd, so $\arctan(-x)=-\arctan(x)$. Now, would you happen to have a drawing of the 30-60-90 triangle lying around? $\endgroup$ Commented Dec 8, 2011 at 0:00
  • $\begingroup$ It's not. You want $-\pi/6$, not $-1/(6 \pi)$. $\endgroup$ Commented Dec 8, 2011 at 0:09
  • $\begingroup$ 1. I think you're writing $-1/(6\pi)$ when you mean $-(\pi/6)$. 2. If what you are reporting accurately represents what's on Khan Academy then Khan Academy is wrong. It happens to the best of us. $\endgroup$ Commented Dec 8, 2011 at 0:11
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    $\begingroup$ You have shown that $\tan(-\pi/6)=-1/\sqrt{3}=-\sqrt{3}/3$. Since $\arctan x$ is the angle in $(-\pi/2,\pi/2)$ whose $\tan$ is $x$, you have shown that $\arctan(-\sqrt{3}/3)=-\pi/6$. $\endgroup$ Commented Dec 8, 2011 at 0:23
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    $\begingroup$ It is slightly unfortunate that Khan Academy says Note: please answer in terms of radians (ex: "3/4 pi"), partly because whether $\pi$ is part of the denominator or not is unclear, and partly because it would not accept "5/6 pi" in answer to this particular question, instead insisting on "-1/6 pi" as the principal value. $\endgroup$
    – Henry
    Commented Dec 8, 2011 at 0:51

2 Answers 2

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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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  • $\begingroup$ Wow. Yes I completely missed that. That seems to have been my error. Thank you Michael. $\endgroup$
    – drc
    Commented Dec 8, 2011 at 1:02
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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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  • $\begingroup$ What happens if y is outside of those bounds you specify? $\endgroup$
    – drc
    Commented Dec 8, 2011 at 2:00
  • $\begingroup$ Also your use of ][, I'm assuming it indicates an open interval? I'm wondering why you used that instead of (-pi/2 < y < pi/2) $\endgroup$
    – drc
    Commented Dec 8, 2011 at 2:06
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    $\begingroup$ @drc: It's not uncommon notation; I've seen it many times. $\endgroup$ Commented Dec 8, 2011 at 2:26
  • $\begingroup$ @drc: I usually use the reverse square brakets for the open interval just to avoid confusion with other symbols; in fact, for example, $(a,b)$ is also used to denote the ordered pair with first entry $a$ and second entry $b$. $\endgroup$
    – Pacciu
    Commented Dec 8, 2011 at 12:41
  • $\begingroup$ @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 $]-\pi/2, \pi/2[$. $\endgroup$
    – Pacciu
    Commented Dec 8, 2011 at 12:44

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