2

$$ 2\tan^{-1}x=\begin{cases}\tan^{-1}\frac{2x}{1-x^2} &\mbox{ if $|x|<1$}\\ \pi+\tan^{-1}\frac{2x}{1-x^2} &\mbox{ if $|x|>1$} \end{cases} $$

Similarly,

$$ 2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2} &\mbox{ if $|x|\leq1$}\\ \text{____________} &\mbox{ if $|x|>1$} \end{cases} $$

and

$$ 2\tan^{-1}x=\begin{cases}\cos^{-1}\frac{1-x^2}{1+x^2} &\mbox{ if $x\geq0$}\\ \text{____________} &\mbox{ if $x<0$} \end{cases} $$

How do I derive the missing cases ?

My Attempt:

Take $\tan^{-1}x=A\implies x=\tan A$ where $\frac{-\pi}{2}<A<\frac{\pi}{2}\implies -\pi<2A<\pi$.

Case 1 :

$$ \cos2A=\frac{1-\tan^2A}{1+\tan^2A}=\frac{1-x^2}{1+x^2}=\cos\Big(\cos^{-1}\frac{1-x^2}{1+x^2}\Big)\\ \implies \cos^{-1}\frac{1-x^2}{1+x^2}=2n\pi\pm2A=2n\pi\pm2\tan^{-1}x $$ As $0\leq\cos^{-1}\leq\pi$,

If $0\leq2A\leq\pi$ $$ \cos^{-1}\frac{1-x^2}{1+x^2}=2A=2\tan^{-1}x $$

If $-\pi<2A<0$ $$ \cos^{-1}\frac{1-x^2}{1+x^2}= \color{red}{\text{?}} $$

Case 2 :

$$ \sin2A=\frac{2\tan A}{1+\tan^2A}=\frac{2x}{1+x^2}=\sin\Big(\sin^{-1}\frac{2x}{1+x^2}\Big)\\ \implies \sin^{-1}\frac{2x}{1+x^2}=n\pi+(-1)^n.2A=n\pi+(-1)^n.2\tan^{-1}x $$ As $\frac{-\pi}{2}\leq\sin^{-1}\leq\frac{\pi}{2}$,

If $\frac{\pi}{2}\leq2A\leq{\pi}$ $$ \sin^{-1}\frac{2x}{1+x^2}= \color{red}{\text{?}} $$

what value should $\cos^{-1}\frac{1-x^2}{1+x^2}$ and $\sin^{-1}\frac{2x}{1+x^2}$ take and how do I proceed further ?

My Understanding:

For, $$ \tan2A=\tan\big(\tan^{-1}\frac{2x}{1-x^2}\big)\implies\tan^{-1}\frac{2x}{1-x^2}=n\pi+2A $$ If $\frac{\pi}{2}<2A<{\pi}$ I would take $$ \tan^{-1}\frac{2x}{1-x^2}=-\pi+2A=-\pi+2\tan^{-1}x $$

Sooraj S
  • 7,573

2 Answers2

2

As $\arctan(-x)=-\arctan x,$

Set $x=-y$ with $x\ge0,$ in $$\arccos\dfrac{1-x^2}{1+x^2}=2\arctan x$$

to find $$\arccos\dfrac{1-(-y)^2}{1+(-y)^2}=2\arctan(-y)=?$$

In $$\arcsin\dfrac{2x}{1+x^2}=2\arctan x$$ with $x\le1$

set $x=\dfrac1y$ to find $$\arcsin\dfrac{2/y}{1+(/y)^2}=2\arctan(1/y)$$

Now use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?

lab bhattacharjee
  • 274,582
  • thanx. i think it is the easiest way to derive other cases. could u pls confirm $2\tan^{-1}x=\pi-\sin^{-1}\frac{2x}{1+x^2}$ when $|x|>1$ and $2\tan^{-1}x=-\cos^{-1}\frac{1-x^2}{1+x^2}$ when $x<0$ – Sooraj S Jan 27 '18 at 18:26
  • @ss1729, What if $x=-\sqrt3?$ Cosine is fine – lab bhattacharjee Jan 27 '18 at 18:28
  • thanks. it is $2\tan^{-1}x=\pi-\sin^{-1}\frac{2x}{1+x^2}$ for $|x|>1, x>0$ and $2\tan^{-1}x=-\pi-\sin^{-1}\frac{2x}{1+x^2}$ for $|x|>1, x<0$. – Sooraj S Jan 28 '18 at 16:46
0

hint

If $A=\tan (\frac {x}{2}) $ then

$$\cos (x)=\frac {1-A^2}{1+A^2} $$ and $$\sin (x)=\frac {2A}{1+A^2} $$

hamam_Abdallah
  • 62,951