Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 231: 65

Answer

$$\sin25^\circ+\cos(-48^\circ)=-2\sin\frac{23^\circ}{2}\cos\frac{73^\circ}{2}$$

Work Step by Step

$$A=\sin25^\circ+\cos(-48^\circ)$$ Here you can choose whether to deal with the negative $-48^\circ$ right away or to leave it alone. In this case, I will leave it alone. The sum-to-product identity that will be applied here is $$\sin X+\sin Y=2\sin(\frac{X+Y}{2})\cos(\frac{X-Y}{2})$$ Therefore, A would be $$A=2\sin[\frac{25^\circ+(-48^\circ)}{2}]\cos[\frac{25^\circ-(-48^\circ)}{2}]$$ $$A=2\sin[\frac{25^\circ-48^\circ}{2}]\cos[\frac{25^\circ+48^\circ}{2}]$$ $$A=2\sin(\frac{-23^\circ}{2})\cos(\frac{73^\circ}{2})$$ As we know $\sin(-X)=-\sin X$, therefore, $$A=-2\sin\frac{23^\circ}{2}\cos\frac{73^\circ}{2}$$
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