Show-that-arcsin-x-pi-4-1-2-arcsin-2x-1-

Question Number 114812 by Ar Brandon last updated on 21/Sep/20

Show that

\arcsin (\sqrt{x}) = \frac{π}{4} + \frac{1}{2} \arcsin (2 x - 1)

Answered by mr W last updated on 21/Sep/20

s a y t = \sin^{- 1} \sqrt{x}

\Rightarrow \sqrt{x} = \sin t

\Rightarrow x = \sin^{2} t

\Rightarrow 2 x - 1 = 2 \sin^{2} t - 1 = - \cos 2 t

= - \sin (\frac{π}{2} - 2 t) = \sin (2 t - \frac{π}{2})

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

\Rightarrow \frac{1}{2} \sin^{- 1} (2 x - 1) = t - \frac{π}{4}

\Rightarrow t = \frac{π}{4} + \frac{1}{2} \sin^{- 1} (2 x - 1)

\Rightarrow \sin^{- 1} \sqrt{x} = \frac{π}{4} + \frac{1}{2} \sin^{- 1} (2 x - 1)

Commented by Ar Brandon last updated on 21/Sep/20

�� Thanks Mr W.

Answered by 1549442205PVT last updated on 21/Sep/20

Put \arcsin (2 x - 1) = φ (φ \in [- \frac{π}{2}; \frac{π}{2}]) .

. We will prove thar

P = \sin (\frac{π}{4} + \frac{φ}{2}) = \sqrt{x} . Indeed,

\sin φ = 2 x - 1 (1), so P = \sin (\frac{π}{4} + \frac{φ}{2}) =

\sin \frac{π}{4} \cos \frac{φ}{2} + \cos \frac{π}{4} \sin \frac{φ}{2} =

\frac{\sqrt{2}}{2} (\cos \frac{φ}{2} + \sin \frac{φ}{2}) \Rightarrow \sqrt{2} P = \cos \frac{φ}{2} + \sin \frac{φ}{2}

\Rightarrow {2 P}^{2} = \cos^{2} \frac{φ}{2} + \sin^{2} \frac{φ}{2} + 2 \sin \frac{φ}{2} \cos \frac{φ}{2}

\Leftrightarrow {2 P}^{2} = 1 + \sin φ = 1 + (2 x - 1) = 2 x (by

(1)) Since φ \in [\frac{- π}{2}; \frac{π}{2}], \frac{π}{4} + \frac{φ}{2} \in [0; π]

\Rightarrow P = \sin (\frac{π}{4} + \frac{φ}{2}) ⩾ 0 . Therefore,

{2 P}^{2} = 2 x \Leftrightarrow P^{2} = x \Leftrightarrow P = \sin (\frac{π}{4} + \frac{φ}{2}) = \sqrt{x}

\Rightarrow \arcsin (\sqrt{x}) = \frac{π}{4} + \frac{φ}{2} .

Thus, \arcsin (\sqrt{x}) = \frac{π}{4} + \frac{1}{2} \arcsin (2 x - 1)

(due to \arcsin (2 x - 1) = φ) (Q . E . D

Answered by mathmax by abdo last updated on 22/Sep/20

let f (x) = \arcsin (\sqrt{x}) - \frac{1}{2} \arcsin (2 x - 1) - \frac{π}{4}

we have f^{^{'}} (x) = \frac{1}{2 \sqrt{x} \sqrt{1 - x}} - \frac{1}{2} \times \frac{2}{\sqrt{1 - {(2 x - 1)}^{2}}}

= \frac{1}{2 \sqrt{x - x^{2}}} - \frac{1}{\sqrt{1 - {4 x}^{2} + 4 x - 1}} = \frac{1}{2 \sqrt{x - x^{2}}} - \frac{1}{2 \sqrt{x - x^{2}}} = 0 \Rightarrow

f (x) = constante C

f (0) = C = - \frac{1}{2} \arcsin (- 1) - \frac{π}{4} = \frac{π}{4} - \frac{π}{4} = 0 \Rightarrow

f (x) = 0 = \arcsin (\sqrt{x}) - \frac{1}{2} \arcsin (2 x - 1) - \frac{π}{4} \Rightarrow

\arcsin (\sqrt{x}) = \frac{π}{4} + \frac{1}{2} \arcsin (2 x - 1)