# Proof of an inverse trigonometric identity

• Krushnaraj Pandya
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

Gold Member

## 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

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

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

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:
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.

Ray Vickson said:
arcsin(2x√1−x2)?
very sorry for the ambifuity, the entire thing is in a bracket

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