Resolved: "Prove that2sin−1x−sin−1(2x−1)=π2"

Adolfo Hebert

Answered question

2022-04-01

Prove that

2 \sin^{- 1} \sqrt{x} - \sin^{- 1} (2 x - 1) = \frac{π}{2}

Answer & Explanation

Cailyn Hanson

Beginner2022-04-02Added 11 answers

Proceeding, we let $f (x, y) = 2 \arcsin (x) - \arcsin (y)$ and note that
$\sin (f (x, y)) = \sin (2 \arcsin (x)) \cos (\arcsin (y)) - \sin (\arcsin (y)) \cos (2 \arcsin (x))$
$= 2 x \sqrt{1 - x^{2}} \sqrt{1 - y^{2}} - y (1 - 2 x^{2})$ (1)
Substituting $x \to \sqrt{x}$ and $y \to 2 x - 1$ into (1) reveals
$\sin (2 \arcsin (\sqrt{x}) + \arcsin (2 x + 1))$

$= 2 \sqrt{x} \sqrt{1 - x} 2 \sqrt{x} \sqrt{1 - x} - (2 x - 1) (1 - 2 x)$

$= 1$
Therefore, $2 \arcsin (\sqrt{x}) - \arcsin (2 x - 1) = \frac{π}{2} + 2 n π$ . Inasmuch as the arcsine is bounded in absolute value by $\frac{π}{2}$ , then we conclude immediately that $n = 0$ and
$2 \arcsin (\sqrt{x}) - \arcsin (2 x - 1) = \frac{π}{2}$

Mason Knight

Beginner2022-04-03Added 11 answers

alternative to differentiating, let

ϕ = 2 \sin^{- 1} \sqrt{x}

\Rightarrow x = \sin^{2} (\frac{ϕ}{2}) = \frac{12}{1 - \cos ϕ}

\Rightarrow \cos ϕ = 1 - 2 x

\Rightarrow ϕ = \cos^{- 1} (1 - 2 x) = \frac{π}{2} - \sin^{- 1} (1 - 2 x) = \frac{π}{2} + \sin^{- 1} (2 x - 1)

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