Answer
$\frac{1}{16}~sin~59^{\circ}$
Work Step by Step
We can use this identity: $sin~2x = 2~sin~x~cos~x$
$\frac{1}{8}~sin~29.5^{\circ}~cos~29.5^{\circ}$
$= \frac{1}{16}~(2~sin~29.5^{\circ}~cos~29.5^{\circ})$
$= \frac{1}{16}~sin~59^{\circ}$
Trigonometry (11th Edition) Clone
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
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
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