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: 20

Answer

$\frac{2\sqrt3}{3}$

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) Note that $\frac{13\pi}{3} = 4\frac{1}{3}\pi$. This is $\frac{\pi}{3}$ radians more than $4\pi$, which is 2 full rotations from the positive x-axis. The terminal side of this angle is the same as the terminal side of $\frac{\pi}{3}$. This angle intersects the unit circle at the point $(\frac{1}{2}, \frac{\sqrt3}{2})$. This point has: $x= \frac{1}{2}$ $y=\frac{\sqrt3}{2}$ Thus, $\csc{\frac{13\pi}{3}} \\= \csc{\frac{\pi}{3}} \\=\frac{1}{y} \\=\frac{1}{\frac{\sqrt3}{2}} \\= 1 \cdot \frac{2}{\sqrt3} \\=\frac{2}{\sqrt3}$ Rationalize the denominator by multiplying $\sqrt3$ to both the numerator and the denominator to obtain: $=\frac{2}{\sqrt3} \cdot \frac{\sqrt3}{\sqrt3} \\=\frac{2\sqrt3}{3}$
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