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. 
 A: 
Here is one visualization.
The area of the red triangle is $\frac 12 bh$
the base is $1,$ the height is $\sin 2a$
$A = \frac 12 \sin 2a$
But is can also be decomposed into 2 congruent right triangles with legs $\cos a, \sin a$
$A = \cos a\sin a$
$\frac 12 \sin 2a = \cos a\sin a\\
\sin 2a = 2\cos a\sin a$
Here is another.

The hypotenuse of the triangle with vertexes $(-1,0), (\cos 2a, \sin 2a)$ is $2\cos a$
$\sin a = \frac {\text{opposite}}{\text{hypotenuse}} = \frac {\sin 2a}{2\cos a}\\
2\sin a\cos a = \sin 2a$
Or you can be really boring and say that is falls out of the angle addition rules
$\sin 2a = \sin (a+a) = \sin a\cos a + \cos a\sin a$
A:  
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 $$\sin(\angle BEC)=\sin(2\alpha)=\frac{[CD]}{[EC]}=[CD] \Rightarrow \sin(2\alpha)=2·\sin(\alpha)·\cos(\alpha)$$
A: 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 
