Reasoning that $ \sin2x=2 \sin x \cos x$

$$\color{red}{\sin 2x}=\color{blue}{2\sin x}\cos x$$

$$\sin(2x)=\mathrm{Im}(e^{2ix})=\mathrm{Im}(e^{ix}e^{ix})=\mathrm{Im}((\cos x+i\sin x)(\cos x+i\sin x))=2\sin x\cos x$$

Assuming that you already know the sum of angles formula, this is pretty easy to get:

$\sin(x + y) = \sin(x)\cos(y) + \cos(x)\sin(y)$, so

$\sin(2x) = \sin(x + x) = \sin(x)\cos(x) + \cos(x)\sin(x) = 2\sin(x)\cos(x)$.

Edit: If you don't want to take the sum of angles formula for sines as a given, this page on Wikipedia can explain it better than I can, especially since it has a diagram.