In this website, it is a theorem that:

$$ 2 \tan^{-1}(x)= \begin{cases} \sin^{-1} \frac{2x}{1+x^2}, x \in [-1,1] \cr \pi-\sin^{-1} \frac{2x}{1+x^2}, x >1 \cr -\pi - \sin^{-1} \frac{2x}{1+x^2}, x<1 \end{cases}$$

Why is there such a piecewise definition? I considered proving equivalence of tan inverse and sine inverse and I was only able to achieve the first definition.

My proof:

We begin with $ \sin^{-1} \frac{2x}{1+x^2}$ , substitute $ x = \tan \phi$ , the previous result simplfies as $ \sin^{-1} ( 2 \tan \phi \cos^2 \phi) = \sin^{-1} ( 2 \sin \phi \cos \phi) = 2 \phi = 2\tan^{-1} (x)$. Where is my mistake?

  • 1
    $\begingroup$ Note that $\tan$ is not bijective. The tangent also is $\pi$-periodic. $\endgroup$ – vitamin d Feb 20 at 19:31
  • $\begingroup$ Could you explain what you mean by pi-periodic? @vitamind $\endgroup$ – Buraian Feb 20 at 19:34
  • 1
    $\begingroup$ $\tan(x)\equiv\tan(x+\pi)\equiv\tan(x-\pi)$ $\endgroup$ – vitamin d Feb 20 at 19:35
  • $\begingroup$ Teresa Lisbon, CBI, MSE reporting. I will take a look. Over $\endgroup$ – Teresa Lisbon Feb 22 at 15:33
  • $\begingroup$ Let's take an example, say $x =2$. Then, $\frac{2x}{1+x^2} = \frac 45$. According to Wolfram , $\arctan 2 = 63.3^\circ$, so double that is $126.6$ degrees. The arcsin is $53.13$ degrees, and there is the problem : it is to ensure that the codomain of $\arcsin$ is between $-\frac \pi 2$ and $\frac \pi 2$ that we do the transformation. So in your working , up till $\arcsin(\sin 2 \phi)$ everything is fine. But this is not equal to $2 \phi$, because the codomain of $\arcsin$ is to be between $-\frac \pi 2$ and $\frac \pi 2$, so you have to transform some side. $\endgroup$ – Teresa Lisbon Feb 22 at 16:30

The arcsine function, $\arcsin : [-1,1] \to [-\frac \pi 2, \frac \pi 2]$ ($\color{red}{\mathit{note\ the \ codomain}}$ , and the function is often written as $\sin^{-1}$ but I prefer $\arcsin$) is defined as follows : if $a \in [-1,1]$ we can find a unique $x \in [-\frac \pi 2, \frac \pi 2]$ such that $\sin x =a$. We let $x = \arcsin a$.

Since there may be many $x$ such that $\sin x = a$, it is important to specify the codomain of the arcsine, so that we need not worry about multiple values.

Similarly, $\arctan : \mathbb R \to (-\frac \pi 2, \frac \pi 2)$ is defined as : for $a \in \mathbb R$ there is a unique $-\frac \pi 2 < x <\frac \pi 2$ with $\tan x =a$. We let $x = \arctan a$.

Now, we must prove the given statement. First of all, note that $-1\leq \frac{2x}{1+x^2} \leq 1$ for all values of $x$, so certainly $\arcsin \frac{2x}{1+x^2}$ is defined.

Now let $x \in \mathbb R$ and $\phi = \arctan x$ so that $x = \tan \phi$. Note that $-\frac \pi 2 < \phi < \frac \pi 2$. The argument you make shows that $$ \frac{2x}{1+x^2} = \sin 2 \phi $$

and therefore, $$ \arcsin\left(\frac{2x}{1+x^2}\right) = \arcsin(\sin 2 \phi) $$

Now, we would like to simplify the RHS. For this, we cannot simply cancel out the two functions, because of codomain issues. To be precise : $\arcsin(\sin 2 \phi) = \theta$ where $\theta$ is the unique angle in $[-\frac \pi 2, \frac \pi 2]$ such that $\sin 2 \phi = \sin \theta$. Therefore, $\arcsin(\frac{2x}{1+x^2}) = \theta$.

So we can break into cases :

  • If $2 \phi \in [-\frac \pi 2 , \frac \pi 2]$ then of course $\theta = 2 \phi$.

  • If $2\phi \in (\frac \pi 2 , \pi]$ then $\sin(\pi - x) = \sin x$ gives $\sin(\pi - 2\phi) = \sin 2 \phi$ so $\theta = \pi - 2\phi$ (which lies in the codomain).

  • If $2 \phi \in [-\pi , -\frac \pi 2)$ then $\sin(-\pi-x) = -\sin(\pi + x) = \sin x$ so $\sin(-\pi - 2\phi) = \sin(2 \phi)$, and hence $\theta = -\pi - 2 \phi$.

Finally, we know that $x>1, x<1,x \in [-1,1]$ if and only if $2 \phi > \frac \pi 2 , 2\phi < - \frac \pi 2, 2\phi \in [-\frac \pi 2, \frac \pi 2]$ respectively.

Therefore, we get :

  • If $-1 \leq x \leq 1$ then $\arcsin\left(\frac{2x}{1+x^2}\right) = 2 \phi$.

  • If $x>1$ then $\arcsin\left(\frac{2x}{1+x^2}\right) =\pi- 2 \phi$.

  • If $x<1$ then $\arcsin\left(\frac{2x}{1+x^2}\right) =-\pi- 2 \phi$

Bring $2\phi$ to one side in each of these equations, and you have your result.

  • $\begingroup$ "Now, we would like to simplify the RHS. For this, we cannot simply cancel out the two functions, because of codomain issues. To be precise : arcsin(sin2ϕ)=θ where θ is the unique angle in [−π2,π2] such that sin2ϕ=sinθ. Therefore, arcsin(2x1+x2)=θ." this part I can't understand the codomain issue thing like how did the final conclusion follow from the previous statement $\endgroup$ – Buraian 2 days ago
  • $\begingroup$ Ok I will expand on that. Just a minute. $\endgroup$ – Teresa Lisbon 2 days ago

The identity $\arcsin\sin x= x$ is given by definition. So we have to prove that $$\sin(2\arctan x) = \frac{2x}{x^2+1}.$$ Making use of the double angle identity: $$\sin(2\arctan x)=2\sin(\arctan x)\cos(\arctan x).$$ We have two beautiful relations now: $$\sin(\arctan x) = \frac{x}{\sqrt{x^2+1}}, \quad \cos(\arctan x) = \frac{1}{\sqrt{x^2+1}}.$$ Putting all of this together gives us our desired result. $$2\frac{x}{\sqrt{x^2+1}}\frac{1}{\sqrt{x^2+1}}=\frac{2x}{x^2+1}$$

Remark: The $\arctan$ has different values if $\lvert x\rvert$ is smaller or greater than one. But they have a nice connection, because of the period of the function.

  • $\begingroup$ hmm, this is fine but it doesn't answer my question of why there is a piecewise relation between tan inverse and sine inverse/ how to derive to that piecewise relation $\endgroup$ – Buraian Feb 20 at 19:55
  • $\begingroup$ I sort of get what you are trying to say, but I don't get how to use that idea to get the piecewise representation between sine and tan inverses $\endgroup$ – Buraian Feb 20 at 19:57
  • $\begingroup$ Here it is :desmos.com/calculator/bszdq7feca $\endgroup$ – Buraian Feb 20 at 19:59
  • $\begingroup$ it is (-pi/2 , pi/2) $\endgroup$ – Buraian Feb 20 at 20:00
  • $\begingroup$ Yeah I agree with that $\endgroup$ – Buraian Feb 20 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.