Answer
See below.
Work Step by Step
$RHS=sin^2\theta cos^2\theta=\frac{1}{4}(2sin\theta cos\theta)^2=
\frac{1}{4}sin^2(2\theta)=\frac{1}{4}(\frac{1-cos(4\theta)}{2})
=\frac{1-cos(4\theta)}{8}=LHS$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 6 - Analytic Trigonometry - Section 6.6 Double-angle and Half-angle Formulas - 6.6 Assess Your Understanding - Page 519: 56
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