Question

-\frac{pi}{2}<\theta<\frac{pi}{2}. find the value of \theta in radians.
\tan(\theta)=\frac{sqrt{3}}{3}
write your answer in simplified, rationalized form. do not round.
\theta =

Explanation:

Step1: Recall inverse - tangent property

We know that if $\tan(\theta)=x$, then $\theta = \arctan(x)$. Here $x = \frac{\sqrt{3}}{3}$, so $\theta=\arctan(\frac{\sqrt{3}}{3})$.

Step2: Use the unit - circle values

The tangent function $y = \tan\theta=\frac{\sin\theta}{\cos\theta}$. We know that $\tan(\frac{\pi}{6})=\frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6})}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$, and $\frac{\pi}{6}\in(-\frac{\pi}{2},\frac{\pi}{2})$.

Answer:

$\frac{\pi}{6}$