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Question Number 192758 by vishal1234 last updated on 26/May/23

w h a t i s t h e v a l u e o f d y / d x i f

y = {c o t}^{- 1} [\frac{\sqrt{1 + s i n x} + \sqrt{1 - s i n x}}{\sqrt{1 + s i n x} - \sqrt{1 - s i n x}}]; x \in (0, \frac{π}{2})

p l e a s e g i v e c o m p l e t e e x p l a n a t i o n t h a t

w h y c a n w e n o t t a k e

{[s i n x / 2 - c o s x / 2]}^{2} = 1 - s i n x

Answered by a.lgnaoui last updated on 26/May/23

for the second question

{(\sin \frac{x}{2} - \cos \frac{x}{2})}^{2} = {(\sin \frac{x}{2})}^{2} + {(\cos \frac{x}{2})}^{2}

- 2 (\sin \frac{x}{2}) (\cos \frac{x}{2})

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

posons u = \sin \frac{x}{2} z = {(\sin \frac{x}{2} - \cos \frac{x}{2})}^{2}

{\begin{cases} z = 1 + 2 u \sqrt{1 - u^{2}} (1) \\ z = 1 - 2 u \sqrt{1 - u^{2}} (2) \end{cases}

1 + 2 u \sqrt{1 - u^{2}} = 1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}

= 1 + \sin x

1 - 2 u \sqrt{1 - u^{2}} = 1 - \sin x

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

∙ - \frac{π}{2} + 2 k π < x < + \frac{π}{2} + 2 k π

\sin x > 0 z = 1 + \sin x

∙ \frac{π}{2} + 2 k π < x < \frac{3 π}{2} + 2 k π

\sin x < 0 z = 1 - \sin x

donc z n est pas toujours egale (1 - sinx)

Commented by a.lgnaoui last updated on 26/May/23

- \frac{π}{4} < \frac{x}{2} < \frac{π}{4} cas 1

\frac{π}{4} < \frac{x}{2} < π + \frac{π}{4} cas 2

∙ \sin \frac{x}{2} \cos \frac{x}{2} > 0 cas 1 \Rightarrow z = 1 - \sin x

∙ \sin \frac{x}{2} \cos \frac{x}{2} < 0 cas 2 \Rightarrow z = 1 + \sin x

Answered by mahdipoor last updated on 26/May/23

w h y c a n n o t t a k e ?!

t a k e t h a t, b o t p a y a t t e n t i o n :

\sqrt{1 - s i n x} = \sqrt{{(s i n \frac{x}{2} - c o s \frac{x}{2})}^{2}} =

∣ s i n \frac{x}{2} - c o s \frac{x}{2} ∣= c o s \frac{x}{2} - s i n \frac{x}{2} a t 0 < x < \frac{π}{2}

a n d

\sqrt{1 + s i n x} = c o s \frac{x}{2} + s i n \frac{x}{2}

y = {c o t}^{- 1} (\frac{c o s \frac{x}{2}}{s i n \frac{x}{2}}) = {c o t}^{- 1} (c o t (\frac{x}{2})) = \frac{x}{2} \Rightarrow \frac{d y}{d x} = \frac{1}{2}

Answered by a.lgnaoui last updated on 26/May/23

First question

\cot y = \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}

= \frac{{[\sqrt{1 + \sin x} + \sqrt{1 - \sin x}]}^{2}}{2 \sin x}

= \frac{1 + ∣ \cos x ∣}{\sin x}

0 < x < \frac{π}{2} \Rightarrow \cos x > 0

\Rightarrow \cot y = \frac{1 + \cos x}{\sin x}

\frac{d (\cot y)}{dx} = \frac{d (\cot y)}{dy} \times \frac{dy}{dx}

- \frac{\cos x + 1}{\sin^{2} x} = - \frac{1}{\sin^{2} y} \times \frac{dy}{dx}

\frac{1 + \cos x}{\sin^{2} x} \sin^{2} y = \frac{dy}{dx}

\frac{dy}{dx} = [\frac{1 + \cos x}{\sin^{2} x}] (1 + \cot^{2} y)

= \frac{1 + \cos x}{\sin^{2} x} (1 + \frac{1 + \cos x}{\sin x})

= \frac{1 + \cos x}{\sin^{2} x} + \frac{{(1 + \cos x)}^{2}}{\sin^{3} x}

\frac{d y}{dx} = \frac{(1 + \cos x) (\sin x + \cos x + 1)}{\sin^{3} x}