We can prove these by two ways . First is by using derivatives of functions. And second by using identities of trigonometry.

Answer: To prove \(\displaystyle{2}{\arcsin{{\left({x}\right)}}}={\arccos{{\left({1}-{2}{x}^{{2}}\right)}}},{x}\in{\left({0},{1}\right)}\)

Let \(\displaystyle{F}{\left({x}\right)}={2}{\arcsin{{\left({x}\right)}}},{\quad\text{and}\quad}{G}{\left({x}\right)}={\arccos{{\left({1}-{2}{x}^{{2}}\right)}}}\)

we start by showing that \(\displaystyle{F}'{\left({x}\right)}={G}'{\left({x}\right)}\), we have

\(\displaystyle{F}'{\left({x}\right)}={\frac{{{2}}}{{\sqrt{{{1}-{x}^{{2}}}}}}}\)

\(\displaystyle{G}'{\left({x}\right)}={\frac{{-{1}}}{{\sqrt{{{1}-{\left({1}-{2}{x}^{{2}}\right)}^{{2}}}}}}}\times{\left(-{4}{x}\right)}={\frac{{{4}{x}}}{{\sqrt{{{1}-{\left({1}-{2}{x}^{{2}}\right)}^{{2}}}}}}}={\frac{{{4}{x}}}{{\sqrt{{{4}{x}^{{2}}-{4}{x}^{{4}}}}}}}={\frac{{{4}{x}}}{{{2}{x}\sqrt{{{1}-{x}^{{2}}}}}}}\)

\(\displaystyle{\frac{{{2}}}{{\sqrt{{{1}-{x}^{{2}}}}}}}\), therefore \(\displaystyle{F}'{\left({x}\right)}={G}{\left({x}\right)}\)

This means that F(x) and G(x) are differ by a constant to verify that a constant is '0'. Substitute in any value of x test you choise such as 0. We have \(\displaystyle{F}{\left({0}\right)}={2}{\arcsin{{\left({0}\right)}}}={0}\) while \(\displaystyle{G}{\left({0}\right)}={\arccos{{\left({2}\right)}}}={0}\) constant is zero. Therefore \(\displaystyle{F}{\left({x}\right)}={G}{\left({x}\right)}\Rightarrow{2}{\arcsin{{\left({x}\right)}}}={\arccos{{\left({1}-{2}{x}^{{2}}\right)}}}\)