## Precalculus (6th Edition) Blitzer

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