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.
[Math] Exact value of $\tan^{-1}(\sqrt3/3)$
algebra-precalculustrigonometry
Related Solutions
Multiply by $\displaystyle 1 = \frac{2-\sqrt 3}{2-\sqrt 3}$. Use $(a-b)(a+b) = a^2-b^2$ on the denominator. It's called "rationalising the denominator" by multiplying the denominator by the "conjugate surd". You should look up the key phrases in quotes.
You can use the same intuition when $\theta$ lies in other quadrants, drawing the same line.
Best Answer
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 $\theta = \frac{\pi}{6}$