## Precalculus (6th Edition) Blitzer

One of the sum-to-product formulas is $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$. Therefore, $\sin x-\sin y$ can be written as shown below: $\sin x-\sin y=2\sin \frac{x-y}{2}\cos \frac{x+y}{2}$ One of the sum-to-product formulas is $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$. Therefore, $\cos x-\cos y$ can be written as shown below: $\cos x-\cos y=-2\sin \frac{x+y}{2}\sin \frac{x-y}{2}$ Now, consider the left side of the provided expression: $\frac{\sin x-\sin y}{\cos x-\cos y}$ The expression can be simplified as provide below: \begin{align} & \frac{\sin x-\sin y}{\cos x-\cos y}=\frac{2\sin \left( \frac{x-y}{2} \right)\cos \left( \frac{x+y}{2} \right)}{-2\sin \left( \frac{x+y}{2} \right)\sin \left( \frac{x-y}{2} \right)} \\ & =-\frac{\sin \left( \frac{x-y}{2} \right)\cos \left( \frac{x+y}{2} \right)}{\sin \left( \frac{x+y}{2} \right)\sin \left( \frac{x-y}{2} \right)} \\ & =-\frac{\sin \left( \frac{x-y}{2} \right)}{\sin \left( \frac{x-y}{2} \right)}\cdot \frac{\cos \left( \frac{x+y}{2} \right)}{\sin \left( \frac{x+y}{2} \right)} \\ & =-\frac{\cos \left( \frac{x+y}{2} \right)}{\sin \left( \frac{x+y}{2} \right)} \end{align} Then, by using one of the quotient identities of trigonometry, which is $\cot x=\frac{\cos x}{\sin x}$, the expression can be further written as: $-\frac{\cos \left( \frac{x+y}{2} \right)}{\sin \left( \frac{x+y}{2} \right)}=-\cot \frac{x+y}{2}$ Thus, the left side of the expression is equal to the right side, which is $\frac{\sin x-\sin y}{\cos x-\cos y}=-\cot \frac{x+y}{2}$.