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: 15

Answer

$\frac{1}{2}$

Work Step by Step

Using the Double Angle Formula in reverse, $2sin\theta cos\theta = sin2\theta$ we have $2sin15^\circ cos15^\circ = sin2(15^\circ)=sin30^\circ=\frac{1}{2}$
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