Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.4 Trigonometric Identities - 6.4 Assess Your Understanding - Page 497: 33


The left hand side is equivalent to $2$ therefore the given equation is an identity. Refer to the solution below.

Work Step by Step

$\text{ LHS } = (\sin{\theta}+\cos{\theta})^2+(\sin{\theta}-\cos{\theta})^2$ $\text{By Expanding:}$ \begin{align} \text{LHS } &= (\sin^2{\theta}+2 \sin{\theta} \cos{\theta} + \cos^2{\theta})+(\sin^2{\theta}-2 \sin{\theta} \cos{\theta} + \cos^2{\theta}) \\[2mm] &= \sin^2{\theta}+\cos^2{\theta}+\sin^2{\theta}+\cos^2{\theta} \\[2mm] &= 2\sin^2{\theta}+2 \cos^2{\theta} \\[2mm] & = 2(\sin^2{\theta}+\cos^2{\theta}) \\[2mm] &= 2(1)\\[2mm] &= 2\\[2mm] &= \text{ RHS} \end{align}
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