Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 3 - Radian Measure and the Unit Circle - Section 3.3 The Unit Circle and Circular Functions - 3.3 Exercises - Page 117: 21

Answer

$-\frac{\sqrt3}{3}$

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 , page 111 of the textbook) This angle 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, $\tan{\frac{5\pi}{6}} \\= \dfrac{y}{x} \\=\dfrac{\frac{1}{2}}{-\frac{\sqrt3}{2}} \\=\frac{1}{2} \cdot (-\frac{2}{\sqrt3}) \\=-\frac{1}{\sqrt3}$ Rationalize the denominator by multiplying $\sqrt3$ to both the numerator and the denominator to obtain: $=-1 \cdot \frac{\sqrt3}{\sqrt3} \\=-\frac{\sqrt3}{3}$
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