## Precalculus (6th Edition) Blitzer

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