Prove that 2arcsin⁡(x)=arccos⁡(1−2x2) for x∈(0,1)

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Prove that   2arcsin⁡(x)=arccos⁡(1−2x2) for x∈(0,1)

Tuthornt · Accepted Answer

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

$2 \arcsin (x) = \arccos (1 - 2 x^{2}), x \in (0, 1)$
Let $F (x) = 2 \arcsin (x), and G (x) = \arccos (1 - 2 x^{2})$
$F^{'} (x) = G^{'} (x)$ , so
$F^{'} (x) = \frac{2}{\sqrt{1 - x^{2}}}$
$G^{'} (x) = \frac{- 1}{\sqrt{1 - {(1 - 2 x^{2})}^{2}}} \times (- 4 x) = \frac{4 x}{\sqrt{1 - {(1 - 2 x^{2})}^{2}}} = \frac{4 x}{\sqrt{4 x^{2} - 4 x^{4}}} = \frac{4 x}{2 x \sqrt{1 - x^{2}}}$
$\frac{2}{\sqrt{1 - x^{2}}}$ , therefore $F^{'} (x) = G (x)$
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. $F (0) = 2 \arcsin (0) = 0$ while $G (0) = \arccos (2) = 0$ constant is zero. So, $F (x) = G (x) \Rightarrow 2 \arcsin (x) = \arccos (1 - 2 x^{2})$

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

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