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 3.1 Radian Measure - 3.1 Exercises - Page 99: 77

Answer

$\frac{1}{2}$

Work Step by Step

Convert the angle measure to degrees to obtain: $=\frac{5\pi}{6} \cdot \frac{180^o}{\pi} = 5(30^o)=150^0$ Thus, $\sin{(\frac{5\pi}{6})} \\= \sin{150^o}$ 150 degrees is in Quadrant II so : (i) its reference angle is $180^o-150^o=30^o$ (ii) its sine value is positive From Section 2.1 (page 50) , we learned that: $\sin{30^o} = \frac{1}{2}$ This means that: $\sin{150^o} = \sin{30^o} = \frac{1}{2}$.
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