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: 45



Work Step by Step

We can use this identity: $cos~2x = 1-2~sin^2~x$ $\frac{1}{4}-\frac{1}{2}sin^2~47.1^{\circ}$ $= \frac{1}{4}~(1-2~sin^2~47.1^{\circ})$ $= \frac{1}{4}~cos~94.2^{\circ}$
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