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 6.6 Double-angle and Half-angle Formulas - 6.6 Assess Your Understanding - Page 519: 36

Answer

$\frac{\sqrt {15}}{8}$

Work Step by Step

1. $f(2\alpha)=sin(2\alpha)=2sin\alpha cos\alpha$ 2. Use the given figures, we have $cos\alpha=-\frac{1}{4}, sin\alpha=-\frac{\sqrt {15}}{4},$ 3. $f(2\alpha)=sin(2\alpha)=2(-\frac{\sqrt {15}}{4})(-\frac{1}{4})=\frac{\sqrt {15}}{8}$
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