Answer
See the full explanation below.
Work Step by Step
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 $.
