<s>[INST] Proving a trigonometric identity How can one prove the validity of this trigonometric identity?
\begin{equation}
 2\arccos\sqrt{x} = \frac{\pi }{2}-\arcsin(2x-1)
\end{equation}
 [/INST] EDITED in response to valdo's answer.
Your identity  $$
2\arccos \sqrt{x}=\frac{\pi }{2}-\arcsin (2x-1),\qquad 0\le x\le 1\tag{0},
$$   may be rewritten as $$
\arcsin (2x-1)=\frac{\pi }{2}-2\arccos \sqrt{x},\qquad 0\le x\le 1\tag{1}.
$$
For identity $(1)$ to be valid$^1$ it is enough that  
$$
\sin \left( \arcsin (2x-1)\right) =\sin \left( \frac{\pi }{2}-2\arccos \sqrt{
x}\right).\tag{2} 
$$
The LHS of $(2)$ is $$\sin \left( \arcsin (2x-1)\right) =2x-1,\tag{3}$$ and the RHS,
$$
\sin \left( \frac{\pi }{2}-2\arccos \sqrt{x}\right) =\cos \left( 2\arccos 
\sqrt{x}\right) =2\cos ^{2}\left( \arccos \sqrt{x}\right) -1\tag{4}.
$$
And so, it is enough that we have
$$2x-1 =2\cos ^{2}\left( \arccos \sqrt{x}\right) -1 
=2\left( \sqrt{x}\right) ^{2}-1 
=2x-1,\qquad x\ge 0,\tag{5}
$$
which is indeed an identity. Consequently, all the previous identities are valid e so, also the given identity $(0)$.
--
$^1$ See valdo's detailed explanation in his answer.
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