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: 71


The result of $\frac{\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)}$ is $\tan \beta $.

Work Step by Step

Let us consider the given expression, $\frac{\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)}$ by using the trigonometric identities, $\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 $ $\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 $ Now, the above expression can be further simplified as: $\begin{align} & \frac{\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)}=\frac{\left( \sin \alpha \cos \beta +\cos \alpha \sin \beta \right)-\left( \sin \alpha \cos \beta -\cos \alpha \sin \beta \right)}{\left( \cos \alpha \cos \beta -\sin \alpha \sin \beta \right)+\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)} \\ & =\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta -\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta -\sin \alpha \sin \beta +\cos \alpha \cos \beta +\sin \alpha \sin \beta } \\ & =\frac{\cos \alpha \sin \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta +\cos \alpha \cos \beta } \\ & =\frac{2\cos \alpha \sin \beta }{2\cos \alpha \cos \beta } \end{align}$ Use, $\frac{\sin x}{\cos x}=\tan x$ $\begin{align} & \frac{2\cos \alpha \sin \beta }{2\cos \alpha \cos \beta }=\frac{\sin \beta }{\cos \beta } \\ & =\tan \beta \end{align}$ Thus, $\frac{\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)}$ can be simplified as $\tan \beta $. Hence, the result of $\frac{\sin \left( \alpha +\beta \right)-\sin \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)}$ is $\tan \beta $.
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