# Piecewise equivalence of trigonometric inverses

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?

• Note that $\tan$ is not bijective. The tangent also is $\pi$-periodic. – vitamin d Feb 20 at 19:31
• Could you explain what you mean by pi-periodic? @vitamind – Buraian Feb 20 at 19:34
• $\tan(x)\equiv\tan(x+\pi)\equiv\tan(x-\pi)$ – vitamin d Feb 20 at 19:35
• Teresa Lisbon, CBI, MSE reporting. I will take a look. Over – Teresa Lisbon Feb 22 at 15:33
• 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. – 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.

• "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 – Buraian 2 days ago
• Ok I will expand on that. Just a minute. – 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.

• 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 – Buraian Feb 20 at 19:55
• 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 – Buraian Feb 20 at 19:57
• Here it is :desmos.com/calculator/bszdq7feca – Buraian Feb 20 at 19:59
• it is (-pi/2 , pi/2) – Buraian Feb 20 at 20:00
• Yeah I agree with that – Buraian Feb 20 at 20:02