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: 73


The result of $\cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)$ is $\frac{1}{2}$.

Work Step by Step

Let us consider the given expression, $\cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)$ By using the trigonometric identity, $\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta $ Now, the above expression can be further simplified as, $\begin{align} & \cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)=\cos \left[ \left( \frac{\pi }{6}+\alpha \right)+\left( \frac{\pi }{6}-\alpha \right) \right] \\ & =\cos \left( \frac{\pi }{6}+\alpha +\frac{\pi }{6}-\alpha \right) \\ & =\cos \left( \frac{\pi }{6}+\frac{\pi }{6} \right) \\ & =\cos \left( \frac{2\pi }{6} \right) \end{align}$ Use, $\cos \frac{\pi }{3}=\frac{1}{2}$. $\begin{align} & \cos \left( \frac{2\pi }{6} \right)=\cos \left( \frac{\pi }{3} \right) \\ & =\frac{1}{2} \end{align}$ Thus, $\cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)$ can be simplified as $\frac{1}{2}$. Hence, the result of $\cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)$ is $\frac{1}{2}$.
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