Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 237: 46



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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