Answer
See below.
Work Step by Step
$RHS=\frac{1}{2}(ln|1+2cos^2\theta-1|-ln2)=\frac{1}{2}(ln|2cos^2\theta|-ln2)=\frac{1}{2}(ln2+2ln|cos\theta|-ln2)=ln|cos\theta|=LHS$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 7 - Analytic Trigonometry - 7.6 Double-angle and Half-angle Formulas - 7.6 Assess Your Understanding - Page 498: 70
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