Proof of an inverse trigonometric identity

[Krushnaraj Pandya]

Homework Statement

Show that $\arcsin 2x \sqrt{1-x^2} = 2 \arccos{x}$ when 1/√2 < x < 1

Homework Equations

All trigonometric and inverse trigonometric identities, special usage of double angle identities here

The Attempt at a Solution

I can get the answer by puting x=cosy, the term inside arcsin becomes 2cosysiny=sin2y, so the answer is 2y=2arccosx
BUT
why can't I put x=siny and obtain the answer as 2arcsinx as in the case when 1/√2< x <1/√2

 

[BvU]

Perhaps because you still get $2y=2\arccos x\$?

 

[Krushnaraj Pandya]

Perhaps because you still get $2y=2\arccos x\$?

When we put x=siny, we get {arcsin(2sinycosy=sin2y)}=2y, now arcsinx=y, so we get 2arcsinx not arccosx

 

[Ray Vickson]

Homework Statement

Show that $\arcsin 2x \sqrt{1-x^2} = 2 \arccos{x}$ when 1/√2 < x < 1

Homework Equations

All trigonometric and inverse trigonometric identities, special usage of double angle identities here

The Attempt at a Solution

I can get the answer by puting x=cosy, the term inside arcsin becomes 2cosysiny=sin2y, so the answer is 2y=2arccosx
BUT
why can't I put x=siny and obtain the answer as 2arcsinx as in the case when 1/√2< x <1/√2

Do you mean $\arcsin(2x) \, \cdot \sqrt{1-x^2} = \sqrt{1-x^2} \arcsin(2x)$ or do you mean $\arcsin (2x \sqrt{1-x^2})?$

 

[vela]

Note that you can only say $\arcsin \sin 2y = 2y$ only if $-\pi/2 \le 2y \le \pi/2$.

Try plotting the two sides of the equation for $0 \le x \le 1$ to get some insight into what's going on.

 

[Krushnaraj Pandya]

arcsin(2x√1−x2)?

very sorry for the ambifuity, the entire thing is in a bracket

 

[Krushnaraj Pandya]

Note that you can only say $\arcsin \sin 2y = 2y$ only if $-\pi/2 \le 2y \le \pi/2$.

Try plotting the two sides of the equation for $0 \le x \le 1$ to get some insight into what's going on.

I'll do that and get back with what I understand, thank you