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 669: 7


a) $\alpha =\frac{5\pi }{12}\text{ and }\beta =\frac{\pi }{12}$ in the expansion $\cos \frac{5\pi }{12}\cos \frac{\pi }{12}+\sin \frac{5\pi }{12}\sin \frac{\pi }{12}$. b) The expression $\cos \frac{5\pi }{12}\cos \frac{\pi }{12}+\sin \frac{5\pi }{12}\sin \frac{\pi }{12}$ is equivalent to $\cos \frac{\pi }{3}$. c) The exact value of $\cos \frac{5\pi }{12}\cos \frac{\pi }{12}+\sin \frac{5\pi }{12}\sin \frac{\pi }{12}$ is $\frac{1}{2}$.

Work Step by Step

(a) From difference formula of cosines, $\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta $ The expansion using the above identity is as follows, $\cos \left( \frac{5\pi }{12}-\frac{\pi }{12} \right)=\cos \frac{5\pi }{12}\cos \frac{\pi }{12}+\sin \frac{5\pi }{12}\sin \frac{\pi }{12}$ Compare the identity with the above expansion to determine the value of $\alpha \text{ and }\beta $. Hence, $\alpha =\frac{5\pi }{12}\text{ and }\beta =\frac{\pi }{12}$. (b) The expansion using the cosine difference formula can be solved as follows, $\begin{align} & \cos \left( \frac{5\pi }{12}-\frac{\pi }{12} \right)=\cos \frac{5\pi }{12}\cos \frac{\pi }{12}+\sin \frac{5\pi }{12}\sin \frac{\pi }{12} \\ & \cos \frac{4\pi }{12}=\cos \frac{5\pi }{12}\cos \frac{\pi }{12}+\sin \frac{5\pi }{12}\sin \frac{\pi }{12} \\ & \cos \frac{\pi }{3}=\cos \frac{5\pi }{12}\cos \frac{\pi }{12}+\sin \frac{5\pi }{12}\sin \frac{\pi }{12} \end{align}$ Hence, the cosine of an angle $\frac{\pi }{3}$ is equivalent to the expression $\cos \frac{5\pi }{12}\cos \frac{\pi }{12}+\sin \frac{5\pi }{12}\sin \frac{\pi }{12}$. (c) Write the expansion using the cosine difference formula and solve as, $\begin{align} & \cos \left( \frac{5\pi }{12}-\frac{\pi }{12} \right)=\cos \frac{5\pi }{12}\cos \frac{\pi }{12}+\sin \frac{5\pi }{12}\sin \frac{\pi }{12} \\ & \cos \frac{\pi }{3}=\cos \frac{5\pi }{12}\cos \frac{\pi }{12}+\sin \frac{5\pi }{12}\sin \frac{\pi }{12} \\ & \frac{1}{2}=\cos \frac{5\pi }{12}\cos \frac{\pi }{12}+\sin \frac{5\pi }{12}\sin \frac{\pi }{12} \end{align}$ Hence, the exact value of $\cos \frac{5\pi }{12}\cos \frac{\pi }{12}+\sin \frac{5\pi }{12}\sin \frac{\pi }{12}$ is $\frac{1}{2}$.
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