Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 694: 78

Answer

$\displaystyle \sin\frac{3\theta}{10}$

Work Step by Step

Comparing with the Half-Angle Identity, $\displaystyle \sin\frac{A}{2}=\pm\sqrt{\frac{1-\cos A}{2}}$ if we replace $A$ with $\displaystyle \frac{3\theta}{5}$ in the identity, the RHS equals the given expression. So, the LHS $=\displaystyle \sin\frac{A}{2}$= $\displaystyle \sin\frac{\frac{3\theta}{5}}{2}=\sin\frac{3\theta}{10}$
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