Inverse trignometric functions $\arcsin\left(2x\sqrt{1 - x^2}\right)$

Question

I recently learnt about inverse trigonometric functions. While studying, I came across a formula

$$\arcsin\left(2x\sqrt{1-x^2}\right)=2\arcsin x$$

The proof of this was given by substituting $x$ by $\sin\phi$. However if we substitute $x$ by $\cos\phi$, we can prove that the above expression equals $2\arccos x$. And indeed while going through another book I found that the above expression equals $2\arcsin x$ for $x$ between $-1/\sqrt{2}$ and $1/\sqrt{2}$ and $2\arccos x$ for $x$ between $1/\sqrt{2}$ and $1$. I graphed the function in GeoGebra and wanted to post the images but I can't. Links: arcsin(2x√1-x²) 2arcsin(x) 2arccos(x)

But it was clear that $\arcsin\left(2x\sqrt{1-x^2}\right)$ is not equal to $\arcsin x$ throughout its domain.

So the question is:why isn't $\arcsin\left(2x\sqrt{1-x^2}\right)=2\arcsin x$ throughout its domain? Why is there another definition for it?

Also as a bonus question, what can be a definition for the expression for $x$ between $-1/\sqrt{2}$ and $-1$?

Please explain in a simple way.

I'm new to high level mathematics that is common here in mathstackexchange :)

What is $\phi$? You could post a link to the images you mentioned. Somebody with sufficient reputation could then edit your question to post the images. — N. F. Taussig, Sep 16 '18 at 17:25
I believe $\phi$ is a random name for a variable that satisfies $x=\sin\phi$ — , Sep 16 '18 at 17:26
@GarryHost, See https://math.stackexchange.com/questions/1764431/solving-arcsin-left2x-sqrt1-x2-right-2-arcsin-x — lab bhattacharjee, Sep 17 '18 at 10:03

score 2 · Answer 1 · answered Sep 16 '18 at 18:32

Substituting $x = \sin \phi$ gives on the LHS, by trig addition theorem:

$\arcsin(2x\sqrt{1-x^2}) = \arcsin(2\sin \phi \cos \phi) = \arcsin(\sin ( 2\phi))$

Note that $\sin ( 2\phi)$ is increasing with $\phi$ only for $0 \le \phi \le \pi/4$ so

$\arcsin(\sin ( 2\phi)) = 2 \phi = 2 \arcsin(x)$ is only valid for the corresponding $0\le x \le \sin \pi/4 = \frac{1}{\sqrt 2}$ .

However, for $\pi/4 \le \phi \le \pi/2$, $\sin ( 2\phi)$ is decreasing, so substituting, $x = \sin \phi$ gives in this range $\frac{1}{\sqrt 2} \le x \le 1$ and

$\arcsin(2x\sqrt{1-x^2}) = \arcsin(2\sin \phi \cos \phi) = \arcsin(\sin ( 2\phi)) = \pi - 2 \phi = \pi - 2 \arcsin x = 2 \arccos x$

So the answer in your "other book" is accurate.

score 0 · Answer 2 · answered Sep 16 '18 at 17:28

0

There are two issues. One is the $\pm$ sign in $x=\sin \phi\implies \sqrt{1-x^2}=\pm\cos\phi$. But even if the $\pm$ sign is a $+$ so that $\arcsin (2x\sqrt{1-x^2})=\arcsin\sin 2\phi$, that doesn't necessarily simplify to $2\phi$.

answered Sep 16 '18 at 17:28

J.G.

115,835

for arcsin(sin2ø) to simplify to 2ø, 2ø has to lie between -π/2 to π/2 right? – Garry Host Sep 16 '18 at 17:42
@GarryHost Right. – J.G. Sep 16 '18 at 18:08

Inverse trignometric functions $\arcsin\left(2x\sqrt{1 - x^2}\right)$

2 Answers2