## Trigonometry 7th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.3 - Definition III: Circular Functions - 3.3 Problem Set - Page 143: 18

#### Answer

$\dfrac{\sqrt3}{3}$

#### Work Step by Step

RECALL: In a unit circle $\cos {t}=x \\\sin{t}=y \\\tan{t} = \dfrac{y}{x}, x\ne0 \\\cot{t} = \dfrac{x}{y}, y \ne 0 \\\sec{t} = \dfrac{1}{x}, x\ne0 \\\csc{t} = \dfrac{1}{y}, y \ne0$ Using the definition above and the unit circle in Figure 5 on page 137 of this book, then: Point on the unit circle: $\left(-\dfrac{1}{2}, -\dfrac{\sqrt3}{2}\right)$ Thus, $\cot{(\frac{4\pi}{3})} = \dfrac{x}{y}=\dfrac{-\frac{1}{2}}{-\frac{\sqrt3}{2}}=-\dfrac{1}{2} \cdot \dfrac{-2}{\sqrt3}=\dfrac{1}{\sqrt3}$ Rationalize the denominator to obtain: $\cot{(\frac{4\pi}{3})}=\dfrac{1}{\sqrt3} \cdot \dfrac{\sqrt3}{\sqrt3} \\\cot{(\frac{4\pi}{3})}=\dfrac{\sqrt3}{3}$

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