I'm trying to figure out how to find the exact value of $\tan^{-1}(\sqrt3/3)$. I tried doing the following because I couldn't see any points where $\sqrt3/3$ lies on the Unit Circle $$\tan^{-1}(\sqrt3/3)=x$$ $$\tan(x)=\sqrt3/3$$ but from here I'm lost, I thought about maybe multiplying top and bottom by $1/\sqrt3$ but I just ended up with the same thing. Any suggestions.
$\endgroup$ 15 Answers
$\begingroup$Obviously, the best solution is to rationalize the numerator $$\frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{3}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{3}{3\sqrt{3}}=\frac{1}{\sqrt{3}}$$
But lets say that completely went by us. One solution may be to try to build right triangle with an angle $\theta$ such that $$\tan\theta = \frac{\sqrt{3}}{3}$$
Setting $opposite = \sqrt{3}$, and $adjacent=3$, by Pythagoras we have $$hypotenuse = \sqrt{opposite ^2 + adjacent^2}=\sqrt{12}=2\sqrt{3}$$
Notice that the hypotenuse is exactly twice that of the of the opposite side.
By reflecting this triangle onto itself, we can build an equilateral triangle, giving us an angle of $\theta = \frac{\pi}{6}$
$\endgroup$ 1 $\begingroup$$$\tan^{-1}\frac{\sqrt{3}}{3}=\tan^{-1}\frac{1}{\sqrt{3}}=\frac{\pi}6$$
$\endgroup$ 5 $\begingroup$By Pythagoras, the height of an equilateral triangle of unit side length is $\frac{\sqrt 3}{2}$. Conclude from this that $\tan 30^\circ = \frac{1/2}{\sqrt 3 2}=\frac{\sqrt 3}3$.
$\endgroup$ $\begingroup$The range of the arctangent (inverse tangent) function is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Since $$\frac{\sqrt{3}}{3} > 0$$ we may conclude that $$\arctan\left(\frac{\sqrt{3}}{3}\right) \in \left(0, \frac{\pi}{2}\right)$$ so we are looking for an answer in the first quadrant.
Since
$$\frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$$
we are looking for a first quadrant angle $\theta$ with
\begin{align*}
\sin\theta & = \frac{1}{2}\\
\cos\theta & = \frac{\sqrt{3}}{2}
\end{align*}
The only such angle is
$$\theta = \frac{\pi}{6}$$
Hence,
$$\arctan\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}$$
Notice, when $z\in\mathbb{C}$ and $\Re[z]>0$:
- $$\arg[z]=\arctan\left(\frac{\Im[z]}{\Re[z]}\right)$$
In your example we can say:
- $$\Re[z]=3$$
- $$\Im[z]=\sqrt{3}$$
So:
$$z=\Re[z]+\Im[z]i=3+\sqrt{3}i$$
And we now that:
$$\arg[3+\sqrt{3}i]=\frac{\pi}{6}$$
So:
$$\arg[3+\sqrt{3}i]=\arctan\left(\frac{\sqrt{3}}{3}\right)=\frac{\pi}{6}$$
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