## Precalculus (6th Edition) Blitzer

Let us consider the left side of the given expression: $\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$ By using the identities of trigonometry, $\cos \,\left( \alpha +\beta \right)=\cos \,\alpha \,\cos \beta -\sin \,\alpha \,\sin \,\beta$ $\cos \,\left( \alpha -\beta \right)=\cos \,\alpha \,\cos \beta +\sin \,\alpha \,\sin \,\beta$ , the above expression can be further simplified as: \begin{align} & \cos \,\left( \alpha +\beta \right)+\cos \,\left( \alpha -\beta \right)=\left( \cos \,\alpha \,\cos \beta -\sin \,\alpha \,\sin \,\beta \right)+\left( \cos \,\alpha \,\cos \beta +\sin \,\alpha \,\sin \,\beta \right) \\ & =\cos \,\alpha \,\cos \beta +\cos \,\alpha \,\cos \beta -\sin \,\alpha \,\sin \,\beta +\sin \,\alpha \,\sin \,\beta \\ & =2\cos \,\alpha \,\cos \beta \end{align} Hence, the left side of the given expression is equal to the right side, which is $\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)=2\cos \,\alpha \cos \,\beta$.