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: 47


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Work Step by Step

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