## Precalculus (6th Edition) Blitzer

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 $\frac{\sin \,\left( x+h \right)-\sin \,x}{h}=\cos \,x\frac{\sin \,h}{h}+\sin \,x\frac{\cos \,h-1}{h}$.