## Precalculus (6th Edition)

Published by Pearson

# Chapter 6 - The Circular Functions and Their Graphs - 6.2 The Unit Circle and Circular Functions - 6.2 Exercises - Page 588: 33

#### Answer

$\color{blue}{\tan{(\frac{5\pi}{6})}=-\dfrac{\sqrt3}{3}}$

#### Work Step by Step

Figure 13 on page 579 of this book shows that the unit circle point $(-\frac{\sqrt3}{2}, \frac{1}{2})$ corresponds to the real number $\dfrac{5\pi}{6}$. RECALL: (i) $\cos{s} = x$ (ii) $\sin{s} = y$ (iii) $\tan{s}=\dfrac{y}{x}$ (iv) $\sec{s} =\dfrac{1}{x}$ (v) $\csc{s} = \dfrac{1}{y}$ (vi) $\cot{s} = \dfrac{x}{y}$ Use the coordinates of the unit circle point above and the formula in (iii) above to obtain: $\color{blue}{\tan{(\frac{5\pi}{6})}=\dfrac{\frac{1}{2}}{-\frac{\sqrt3}{2}}=\frac{1}{2} \cdot \frac{-2}{\sqrt3}=-\frac{1}{\sqrt3}=-\frac{1\sqrt3}{(\sqrt3)(\sqrt3)}=-\frac{\sqrt3}{3}}$

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