Prove $ \frac{1}{2} (\arccos(x) - \arccos(-x)) = -\arcsin(x)$ The identity I need to prove is this, and I am very close but I am missing a negative sign which I cannot find.
$$ \frac{1}{2} (\arccos(x) - \arccos(-x)) = -\arcsin(x)$$
I started off by using the $\cos(A-B)$ double angle formula, using $\arccos(x)$ and $\arccos(-x)$ and $A$ and $B$ respectively. Substituting this into the double angle formula, I get that $$\cos(A-B)=1-2x^2$$
Then, from this I say let $x=\sin{y}$, so that I end up with the large expression $$\cos(\arccos(\sin{y})-\arccos(-\sin{y}))=1-2\sin^2(y)$$
Now we can use the cosine double angle formula on the RHS so this becomes $$\cos(\arccos(\sin{y})-\arccos(-\sin{y}))=\cos(2y)$$
Taking the inverse cosine on both sides leads to $$\arccos(\sin{y})-\arccos(-\sin{y})=2y$$
Now reversing the substitution, saying that $y=\arcsin{x}$, we get that $$\arccos(x) - \arccos(-x) = 2\arcsin(x)$$ which is nearly the identity I need to prove except that I have made an error in some step, causing the expression to be wrong. I cannot find this mistake.

Maybe using the following identities will help: $$\arccos a-\arccos b= \arccos (ab +\sqrt {(1-a^2)(1-b^2)}) \tag {1}$$ $$\arcsin a +\arcsin b= \arcsin (a\sqrt {1-b^2} +b\sqrt {1-a^2}) \tag{2}$$ After applying $(1)$, use the triangle formula to convert $\arccos $ to $\arcsin $ component and thus LHS=RHS. Hope it helps. 
