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: 28



Work Step by Step

Figure 13 on page 579 of this book shows that the unit circle point $(\frac{-\sqrt2}{2}, -\frac{\sqrt2}{2})$ corresponds to the real number $\dfrac{5\pi}{4}$. 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: $\color{blue}{\sec{\frac{5\pi}{4}}=\dfrac{1}{\frac{-\sqrt2}{2}}=1 \cdot \frac{-2}{\sqrt2}=-\frac{2}{\sqrt2}=-\frac{2(\sqrt2)}{(\sqrt2)(\sqrt2)}=-\sqrt2}$
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