Suman Cole

2021-08-20

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

Tuthornt

We have two ways to demonstrate this. The first is by employing function derivatives. Secondly, by applying trigonometric

$2\mathrm{arcsin}\left(x\right)=\mathrm{arccos}\left(1-2{x}^{2}\right),x\in \left(0,1\right)$
Let
${F}^{\prime }\left(x\right)={G}^{\prime }\left(x\right)$, so
${F}^{\prime }\left(x\right)=\frac{2}{\sqrt{1-{x}^{2}}}$
${G}^{\prime }\left(x\right)=\frac{-1}{\sqrt{1-{\left(1-2{x}^{2}\right)}^{2}}}×\left(-4x\right)=\frac{4x}{\sqrt{1-{\left(1-2{x}^{2}\right)}^{2}}}=\frac{4x}{\sqrt{4{x}^{2}-4{x}^{4}}}=\frac{4x}{2x\sqrt{1-{x}^{2}}}$
$\frac{2}{\sqrt{1-{x}^{2}}}$, therefore ${F}^{\prime }\left(x\right)=G\left(x\right)$
In order to confirm that a constant is '0,' this means that F(x) and G(x) must differ by a constant. Replace with any value of the x test that you like, such as 0. $F\left(0\right)=2\mathrm{arcsin}\left(0\right)=0$ while $G\left(0\right)=\mathrm{arccos}\left(2\right)=0$ constant is zero. So, $F\left(x\right)=G\left(x\right)⇒2\mathrm{arcsin}\left(x\right)=\mathrm{arccos}\left(1-2{x}^{2}\right)$

Do you have a similar question?