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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.

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    Divide by 2 on both sides and it is correct – 2017-01-03
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    There is a minus sign missing. – 2017-01-03

2 Answers 2

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Here's one approach $$\arccos(x) - \arccos(-x)$$ $$=\frac{\pi}{2}-\arcsin(x) - \pi+\arccos(x)$$ $$=\frac{\pi}{2}-\arcsin(x) - \pi+\frac{\pi}{2}-\arcsin(x)$$ $$=-2\arcsin(x)$$

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    That's a very neat way of showing it! – 2017-01-03
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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.