Question

find the exact value of s in the given interval that has the given circular function value. do not use a calculator. (left\frac{pi}{2},pi
ight); (\tan s = -\frac{sqrt{3}}{3}) (s = square) radians (simplify your answer. type an exact answer, using (pi) as needed. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Recall tangent values

We know that \(\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}\) and \(\tan(\pi - \theta)=-\tan\theta\), \(\tan(2\pi - \theta)=-\tan\theta\). The interval is \([\frac{\pi}{2},\pi]\), which is the second quadrant where tangent is negative.

Step2: Find the angle in the interval

We need an angle \(s\) in \([\frac{\pi}{2},\pi]\) such that \(\tan s = -\frac{\sqrt{3}}{3}\). Since \(\tan\frac{5\pi}{6}=\tan(\pi - \frac{\pi}{6})=-\tan\frac{\pi}{6}=-\frac{\sqrt{3}}{3}\) and \(\frac{5\pi}{6}\) is in \([\frac{\pi}{2},\pi]\) (because \(\frac{\pi}{2}=\frac{3\pi}{6}\) and \(\pi=\frac{6\pi}{6}\), so \(\frac{3\pi}{6}\leq\frac{5\pi}{6}\leq\frac{6\pi}{6}\)).

Answer:

\(\frac{5\pi}{6}\)