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
 A: If you let $\theta = \cos^{-1}(x)$, you can use the fact that $\sin(2\theta) = 2 \sin \theta \cos \theta$. You then need to write $\sin \theta$ and $\cos \theta$ in terms of $x$ in order to get the result you are looking for (and you can certainly use a triangle to help with that part).
A: $\sin^2y+\cos^2y =1 \implies \sin y= \pm \sqrt{1-\cos^22y} \tag{1}$
$\sin 2y=2\sin y\cos y \tag{2}$  .
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   
$\pm\sqrt{1-\cos^2(\arccos x)}\times\cos(\arccos x)=\pm\dfrac{1}{2} x\sqrt{1-x^2}$
A: Do you know duplication rule? Then ot's really trivial to prove the identity...
$$\frac14\sin(2\arccos(x))\\=\frac14 2\sin(\arccos(x))\cos(\arccos(x))\\=\frac12 x\sqrt{1-[\cos(\arccos(x))]^2}=\frac{x}{2}\sqrt{1-x^2}$$
A: If $\arccos x=y\implies x=\cos y$ and from this, $0\le y\le\pi\implies\sin y\ge0$
$\displaystyle\implies\sin y=+\sqrt{1-x^2}$
and subsequently   $\displaystyle\sin(2\arccos x)=\sin2y=2\sin y\cos y=\cdots$ 
A: your stategy is 2-fold.
1) eliminate the 2arccos(x)s by using a double angle identity
2) re-express any tan(arccos(x))s or sin(arccos(x))s as cos(arccos(x))s
as you go through some of the other answers, keep these strategies in mind.
