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: 29



Work Step by Step

Note that $-\dfrac{4\pi}{3}$ is coterminal with: $-\dfrac{4\pi}{3} + 2\pi=-\dfrac{4\pi}{3} + \dfrac{6\pi}{3}=\dfrac{2\pi}{3}$ Figure 13 on page 579 of this book shows that the unit circle point $(-\frac{1}{2}, \frac{\sqrt3}{2})$ corresponds to the real number $\dfrac{-4\pi}{3}$ or $\dfrac{2\pi}{3}$. 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 (ii) above to obtain: $\color{blue}{\sin{(-\frac{4\pi}{3})}=\sin{\frac{2\pi}{3}}=\frac{\sqrt3}{2}}$
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