## Precalculus (6th Edition) Blitzer

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