Trigonometry from integration (simplifying) How do I simplify $$\frac14\sin(2\arccos(x))$$
to $$\frac12x\sqrt{1-x^2}$$
Note, it used to more complex I simplified it to this level, I think you have to use a triangle but im not sure...
Thank you,
Good Luck

$\sin^2y+\cos^2y =1 \implies \sin y= \pm \sqrt{1-\cos^22y} \tag{1}$
$\sin 2y=2\sin y\cos y \tag{2}$  .
Using $1$ and $2$:  
$\sin(2(\arccos(x))=2\sin(\arccos x)\cos(\arccos x)=\pm2\sqrt{1-\cos^2(\arccos x)}\times\cos(\arccos x)$.  
You know $\cos(\arccos(x))=x$    
So   
$\pm\sqrt{1-\cos^2(\arccos x)}\times\cos(\arccos x)=\pm\dfrac{1}{2} x\sqrt{1-x^2}$
