Answer
$-\frac{1}{4} $
Work Step by Step
$\frac{1}{2}sin^2(x)+C=-\frac{1}{4}cos(2x) $,
$\frac{1}{2}sin^2(x)+C=-\frac{1}{4}(1-2sin^2(x) $,
$\frac{1}{2}sin^2(x)+C=-\frac{1}{4}+\frac{1}{2}sin^2(x) $,
$ C=-\frac{1}{4} $
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 521: 103
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