Why is 2 $\cdot$ $\sin(\alpha)$ $\cos(\alpha)$=$\sin(2\alpha)$? I am solving a physics problem involving a 2-dimensional throw. However, I have hit a bump trying to understand
$$2\cdot \sin(\alpha)\cdot\cos(\alpha)=\sin(2\alpha)$$
I have googled and searched on Stack Exhange but found nothing. I hope someone can explain how those two equal each other. 

This follows directly from the well-known angle-addition formula for $\sin$:
$\sin(\alpha + \beta) = (\sin \alpha)(\cos \beta) + (\sin \beta)(\cos \alpha); \tag 1$
by taking $\alpha = \beta$:
$\sin 2 \alpha =  (\sin \alpha)(\cos \alpha) + (\sin \alpha)(\cos \alpha) = 2(\sin \alpha)(\cos \alpha); \tag 2$
see this wikipedia page.
This formula and others may also be derived from the Euler identity
$e^{i\phi} = \cos \phi + i \sin \phi \tag 3$
by taking $\phi = \alpha + \beta$; we obtain
$\cos (\alpha + \beta) + i \sin( \alpha + \beta) = e^{i(\alpha + \beta)} = e^{i\alpha}e^{i\beta} = (\cos \alpha + i \sin \alpha)(\cos \beta + i\sin \beta)$
$= (\cos \alpha)(\cos \beta) - (\sin \alpha)(\sin \beta) + i ((\sin \alpha)(\cos \beta) + (\sin \beta)(\cos \alpha)) , \tag 4$
and look at the real and imaginary parts.
and 
