## Precalculus (6th Edition) Blitzer

Let us consider the right-hand side of the given expression: $2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ One of the product-to-sum formulas is $\cos \alpha \cos \beta =\frac{1}{2}\left[ \cos \left( \alpha -\beta \right)+\cos \left( \alpha +\beta \right) \right]$. Therefore, here $\alpha$ is $\frac{\alpha +\beta }{2}$ and $\beta$ is $\frac{\alpha -\beta }{2}$. Now, the expression can be calculated as shown below: \begin{align} & 2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=2\cdot \frac{1}{2}\left[ \cos \left( \frac{\alpha +\beta }{2}-\frac{\alpha -\beta }{2} \right)+\cos \left( \frac{\alpha +\beta }{2}+\frac{\alpha -\beta }{2} \right) \right] \\ & =2\cdot \frac{1}{2}\left[ \cos \left( \frac{\alpha +\beta -\alpha +\beta }{2} \right)+\cos \left( \frac{\alpha +\beta +\alpha -\beta }{2} \right) \right] \\ & =\cos \left( \frac{2\beta }{2} \right)+\cos \left( \frac{2\alpha }{2} \right) \\ & =\cos \beta +\cos \alpha \end{align} Then, rearrange the terms: $\cos \beta +\cos \alpha =\cos \alpha +\cos \beta$ Thus, the right-hand side of the expression is equal to the left-hand side: $\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$.