Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 25

Answer

See the explanation below.

Work Step by Step

${{\left( \sin \theta +\cos \theta \right)}^{2}}=1+\sin 2\theta $ Recall the algebraic formula. ${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$ Consider the left side of the given expression and apply the algebraic formula. ${{\left( \sin \theta +\cos \theta \right)}^{2}}={{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta $ Recall the Trigonometric Identities. $\begin{align} & \sin 2\theta =2\sin \theta \cos \theta \\ & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\ \end{align}$ Apply the Trigonometric Identities. ${{\left( \sin \theta +\cos \theta \right)}^{2}}=1+\sin 2\theta $ Hence, it is proved that the given identity ${{\left( \sin \theta +\cos \theta \right)}^{2}}=1+\sin 2\theta $ holds true.
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