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: 47



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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