Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 670: 74


The result is $\frac{\sqrt{3}}{2}$.

Work Step by Step

Let us consider the expression: $\sin \left( \frac{\pi }{3}-\alpha \right)\cos \left( \frac{\pi }{3}+\alpha \right)+\cos \left( \frac{\pi }{3}-\alpha \right)\sin \left( \frac{\pi }{3}+\alpha \right)$ By using the trigonometric identity, $\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta $ Now, the above expression can be further simplified as: $\begin{align} & \sin \left( \frac{\pi }{3}-\alpha \right)\cos \left( \frac{\pi }{3}+\alpha \right)+\cos \left( \frac{\pi }{3}-\alpha \right)\sin \left( \frac{\pi }{3}+\alpha \right)=\sin \left[ \left( \frac{\pi }{3}-\alpha \right)+\left( \frac{\pi }{3}+\alpha \right) \right] \\ & =\sin \left( \frac{\pi }{3}-\alpha +\frac{\pi }{3}+\alpha \right) \\ & =\sin \left( \frac{\pi }{3}+\frac{\pi }{3} \right) \\ & =\sin \left( \frac{2\pi }{3} \right) \end{align}$ Use, $\sin \frac{2\pi }{3}=\frac{\sqrt{3}}{2}$. $\sin \left( \frac{2\pi }{3} \right)=\frac{\sqrt{3}}{2}$ Thus, $\sin \left( \frac{\pi }{3}-\alpha \right)\cos \left( \frac{\pi }{3}+\alpha \right)+\cos \left( \frac{\pi }{3}-\alpha \right)\sin \left( \frac{\pi }{3}+\alpha \right)$ can be simplified as $\frac{\sqrt{3}}{2}$.
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