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: 48


See the full explanation below.

Work Step by Step

Let us consider the left side of the given expression: $\frac{\cos \left( \alpha +\beta \right)}{\cos \left( \alpha -\beta \right)}$ By using the identities of trigonometry, $\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta $ $\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +sin\alpha \sin \beta $ , Now above provided expression can be further simplified as: $\frac{\cos \left( \alpha +\beta \right)}{\cos \left( \alpha -\beta \right)}=\frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta +sin\alpha \sin \beta }$ By dividing the numerator and denominator by the $\cos \,\alpha \cos \,\beta $, we get: $\begin{align} & \frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta +sin\alpha \sin \beta }=\frac{\frac{\cos \alpha \cos \beta }{\cos \alpha \cos \beta }-\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}{\frac{\cos \alpha \cos \beta }{\cos \alpha \cos \beta }+\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }} \\ & =\frac{1-\frac{\sin \alpha }{\cos \alpha }\frac{\sin \beta }{\cos \beta }}{1+\frac{\sin \alpha }{\cos \alpha }\frac{\sin \beta }{\cos \beta }} \\ & =\frac{1-\tan \alpha \tan \beta }{1+\tan \alpha \tan \beta } \end{align}$ Hence, the left side of the given expression is equal to the right side, which is $\frac{\cos \left( \alpha +\beta \right)}{\cos \left( \alpha -\beta \right)}=\frac{1-\tan \alpha \tan \beta }{1+\tan \alpha \tan \beta }$.
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