Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 692: 4

Answer

$4\rightarrow A$

Work Step by Step

$\cos ^{2}\alpha -\sin ^{2}\alpha =\cos 2\alpha \Rightarrow \cos ^{2}\dfrac {\pi }{6}-\sin ^{2}\dfrac {\pi }{6}=\cos \left( 2\times \dfrac {\pi }{6}\right) =\cos \dfrac {\pi }{3}=\dfrac {1}{2}\Rightarrow $ $4\rightarrow A$
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