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: 19

Answer

$\color{blue}{\sin{(\frac{7\pi}{6})}=-\frac{1}{2}}$

Work Step by Step

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 $\dfrac{7\pi}{6}$. RECALL: (i) $\cos{s} = x$ (ii) $\sin{s} = y$ (iii) $\tan{s}=\dfrac{y}{x}$ Use the coordinates of the unit circle point above and the formula in (ii) above to obtain: $\color{blue}{\sin{(\frac{7\pi}{6})}=-\frac{1}{2}}$
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