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: 47

Answer

$-cos~\frac{4\pi}{5}$

Work Step by Step

We can use this identity: $cos~2x = cos^2~x - sin^2~x$ $sin^2~\frac{2\pi}{5}-cos^2~\frac{2\pi}{5}$ $= -(cos^2~\frac{2\pi}{5}-sin^2~\frac{2\pi}{5})$ $= -cos~\frac{4\pi}{5}$
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