## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 3 - Radian Measure and the Unit Circle - Section 3.3 The Unit Circle and Circular Functions - 3.3 Exercises - Page 123: 22

#### Answer

$-\sqrt3$

#### Work Step by Step

RECALL: $\sin{s} = y \\\cos{s} = x \\\tan{s} = \frac{y}{x} \\\cot{s} = \frac{x}{y} \\\sec{s} = \frac{1}{x} \\\csc{s}=\frac{1}{y}$ (refer to Figure 11 on page 111 of the textbook) The angle $\frac{5\pi}{6}$ intersects the unit circle at the point $(-\frac{\sqrt3}{2}, \frac{1}{2})$. This point has: $x= -\frac{\sqrt3}{2}$ $y=\frac{1}{2}$ Thus, $\cot{\frac{5\pi}{6}} \\= \frac{x}{y} \\=\dfrac{-\frac{\sqrt3}{2}}{\frac{1}{2}} \\=-\frac{\sqrt3}{2} \cdot \frac{2}{1} \\=-\sqrt3$

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