Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 237: 57

Answer

$$2\sin58^\circ\cos102^\circ=\sin160^\circ-\sin44^\circ$$

Work Step by Step

$$A=2\sin58^\circ\cos102^\circ$$ The product-to-sum identity that will be applied here is $$\sin X\cos Y=\frac{1}{2}[\sin(X+Y)+\sin(X-Y)]$$ Therefore, A would be $$A=2\times\frac{1}{2}[\sin(58^\circ+102^\circ)+\sin(58^\circ-102^\circ)]$$ $$A=\sin160^\circ+\sin(-44^\circ)$$ We know that $\sin(-X)=-\sin X$. That means $$A=\sin160^\circ-\sin44^\circ$$
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