# Prove that 2\arcsin(x)=\arccos(1-2x^2) for x\in(0,1)

Prove that
$$\displaystyle{2}{\arcsin{{\left({x}\right)}}}={\arccos{{\left({1}-{2}{x}^{{2}}\right)}}}$$ for $$\displaystyle{x}\in{\left({0},{1}\right)}$$

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Tuthornt

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 $$\displaystyle{F}{\left({x}\right)}={G}{\left({x}\right)}\Rightarrow{2}{\arcsin{{\left({x}\right)}}}={\arccos{{\left({1}-{2}{x}^{{2}}\right)}}}$$