Why is arctan of $-\frac{\sqrt{3}}{3} = -\frac{1}{6}\pi$?
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!
$\endgroup$ 92 Answers
$\begingroup$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:
$$ \frac{1}{\sqrt{3}} = \frac{1\cdot\sqrt{3}}{\sqrt{3}\sqrt{3}} = \frac{\sqrt{3}}{3}. $$
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 5
