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
Gold Member
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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})?##
 
Last edited:
  • #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
 

FAQ: Proof of an inverse trigonometric identity

What is an inverse trigonometric identity?

An inverse trigonometric identity is an equation that relates the inverse trigonometric functions (arcsine, arccosine, and arctangent) to the trigonometric functions (sine, cosine, and tangent).

Why is it important to prove an inverse trigonometric identity?

Proving an inverse trigonometric identity allows us to verify the relationship between the inverse trigonometric functions and the trigonometric functions. It also helps us to better understand the properties and behaviors of these functions.

What is the process for proving an inverse trigonometric identity?

The process for proving an inverse trigonometric identity involves using algebraic manipulation, trigonometric identities, and properties of inverse functions to show that the two sides of the equation are equal.

Are there any common techniques or strategies for proving inverse trigonometric identities?

Yes, there are some common techniques and strategies that can be used when proving inverse trigonometric identities. These include using double angle formulas, using trigonometric identities involving the sum or difference of angles, and using the Pythagorean identities.

Can inverse trigonometric identities be used in real-world applications?

Yes, inverse trigonometric identities have many real-world applications, particularly in fields such as physics, engineering, and navigation. They are used to solve problems involving angles, distances, and trajectories.

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