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 668: 5

Answer

a) $\alpha =50{}^\circ \text{ and }\beta =20{}^\circ $ in the expansion $\cos 50{}^\circ \cos 20{}^\circ +\sin 50{}^\circ \sin 20{}^\circ $. b) The expression $\cos 50{}^\circ \cos 20{}^\circ +\sin 50{}^\circ \sin 20{}^\circ $ is equivalent to $\cos 30{}^\circ $. c) The exact value of $\cos 50{}^\circ \cos 20{}^\circ +\sin 50{}^\circ \sin 20{}^\circ $ is $\frac{\sqrt{3}}{2}$.

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 can be written as, $\cos \left( 50{}^\circ -20{}^\circ \right)=\cos 50{}^\circ \cos 20{}^\circ +\sin 50{}^\circ \sin 20{}^\circ $ Compare the identity with the above expansion to determine the value of $\alpha \text{ and }\beta $. Hence, $\alpha =50{}^\circ \text{ and }\beta =20{}^\circ $. (b) The expansion using the cosine difference formula can be solved as, $\begin{align} & \cos \left( 50{}^\circ -20{}^\circ \right)=\cos 50{}^\circ \cos 20{}^\circ +\sin 50{}^\circ \sin 20{}^\circ \\ & \cos 30{}^\circ =\cos 50{}^\circ \cos 20{}^\circ +\sin 50{}^\circ \sin 20{}^\circ \end{align}$ Hence, the cosine of an angle $30{}^\circ $ is equivalent to the expression $\cos 50{}^\circ \cos 20{}^\circ +\sin 50{}^\circ \sin 20{}^\circ $. (c) The expansion using the cosine difference formula can be solved as, $\begin{align} & \cos \left( 50{}^\circ -20{}^\circ \right)=\cos 50{}^\circ \cos 20{}^\circ +\sin 50{}^\circ \sin 20{}^\circ \\ & \cos 30{}^\circ =\cos 50{}^\circ \cos 20{}^\circ +\sin 50{}^\circ \sin 20{}^\circ \\ & \frac{\sqrt{3}}{2}=\cos 50{}^\circ \cos 20{}^\circ +\sin 50{}^\circ \sin 20{}^\circ \end{align}$ Hence, the exact value of $\cos 50{}^\circ \cos 20{}^\circ +\sin 50{}^\circ \sin 20{}^\circ $ is $\frac{\sqrt{3}}{2}$.
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