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. 

 
Consider the semicircle over the diameter $[AB]$ of radius 1 and center $E$. Let $C$ be an arbitrary point on the semicircle and $D$ the foot of the altitude of the side $[AB]$ in the triangle $\Delta ABC$. Denote furthermore by $\alpha$ the angle $\angle BAC$. It follows that $\angle BEC=2\alpha$.
By the definition of the sine and cosine $$\sin(\alpha)=\frac{[CB]}{[AB]}$$
$$\cos(\alpha)=\frac{[AC]}{[AB]}$$ Thus $$\sin(\alpha)·\cos(\alpha)=\frac{[CB]·[AC]}{[AB]^2}$$
Now, since $[CD]·[AB]=[AC]·[CB]$ (different ways to get the area of $\Delta ABC$)
$$\sin(\alpha)·\cos(\alpha)=\frac{[CB]·[AC]}{[AB]^2}=\frac{[CD]}{[AB]}=\frac{[CD]}{2}$$ $$\Rightarrow 2·\sin(\alpha)·\cos(\alpha)=[CD]$$
Finally $$\sin(\angle BEC)=\sin(2\alpha)=\frac{[CD]}{[EC]}=[CD] \Rightarrow \sin(2\alpha)=2·\sin(\alpha)·\cos(\alpha)$$
