Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 6 - The Circular Functions and Their Graphs - 6.2 The Unit Circle and Circular Functions - 6.2 Exercises - Page 588: 31



Work Step by Step

Note that $\dfrac{23\pi}{6}$ is coterminal with: $\dfrac{23\pi}{6} - 2\pi=\dfrac{23\pi}{6} - \dfrac{12\pi}{6}=\dfrac{11\pi}{6}$ 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 $\dfrac{23\pi}{6}$ or $\dfrac{11\pi}{6}$. 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 (iv) above to obtain: $\color{blue}{\sec{(\frac{23\pi}{6})}=\sec{\frac{11\pi}{6}}=\dfrac{1}{\frac{\sqrt3}{2}}=1 \cdot \frac{2}{\sqrt3}=\frac{2}{\sqrt3}=\frac{2\sqrt3}{(\sqrt3)(\sqrt3)}=\frac{2\sqrt3}{3}}$
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