Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 231: 46

Answer

$\frac{1}{16}~sin~59^{\circ}$

Work Step by Step

We can use this identity: $sin~2x = 2~sin~x~cos~x$ $\frac{1}{8}~sin~29.5^{\circ}~cos~29.5^{\circ}$ $= \frac{1}{16}~(2~sin~29.5^{\circ}~cos~29.5^{\circ})$ $= \frac{1}{16}~sin~59^{\circ}$
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