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: 50


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Work Step by Step

Let us consider the left side of the given expression, $\frac{\sin \,\left( x+h \right)-\sin \,x}{h}$ By using the identity of trigonometry, $\sin \,\left( \alpha +\beta \right)=\sin \,\alpha \,\cos \beta +\cos \,\alpha \,\sin \,\beta $ , the above expression can be further simplified as, $\begin{align} & \frac{\sin \,\left( x+h \right)-\sin \,x}{h}=\frac{\sin \,x\,\cos \,h+\cos \,x\,\sin \,h-\sin \,x}{h} \\ & =\frac{\cos \,x\,\sin \,h+\sin \,x\,\cos \,h-\sin \,x}{h} \\ & =\frac{\cos \,x\,\sin \,h+\sin \,x\,\left( \cos \,h-1 \right)}{h} \\ & =\cos \,x\,\frac{\sin \,h}{h}+\sin \,x\frac{\left( \,\cos \,h-1 \right)}{h} \end{align}$ Thus, the left side of the expression is equal to the right side, which is $\frac{\sin \,\left( x+h \right)-\sin \,x}{h}=\cos \,x\frac{\sin \,h}{h}+\sin \,x\frac{\cos \,h-1}{h}$. Hence, it is proved that $\frac{\sin \,\left( x+h \right)-\sin \,x}{h}=\cos \,x\frac{\sin \,h}{h}+\sin \,x\frac{\cos \,h-1}{h}$.
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