Proof of an inverse trigonometric identity

In summary, when solving the equation ##\arcsin(2x\sqrt{1-x^2}) = 2\arccos(x)## for 1/√2 < x < 1, it is not possible to substitute x=siny and obtain the answer as 2arcsinx. This is because the substitution results in the equation ##\arcsin(2sinycosy) = 2y##, which is only true for -π/2 ≤ 2y ≤ π/2. However, for 1/√2 < x < 1, we have 0 < 2y < π, which means that the substitution does not hold. Therefore
  • #1

Krushnaraj Pandya

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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
 
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  • #2
Perhaps because you still get ##2y=2\arccos x\ ##?
 
  • #3
BvU said:
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
 
  • #4
Krushnaraj Pandya said:

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})?##
 
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  • #5
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.
 
  • #6
Ray Vickson said:
arcsin(2x√1−x2)?
very sorry for the ambifuity, the entire thing is in a bracket
 
  • #7
vela said:
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
 

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