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Let's say we have this-

$$\arctan(\frac{2x}{1-x^2})$$

Now this equals-

$$2\arctan x, x\in[-1,1]$$ $$-\pi+2\arctan x, x\in[1,\infty]$$ $$\pi+2\arctan x, x\in[-\infty,-1]$$

Is there an easy way to find out the definition of such inverse trig functions. I mean what is the way to find out for what values of $x$ which formula to use.(How should I derive the above ranges of $x$ for which I should add $\pi$ for eg in the above example). I could somehow manage to do it by some complicated method with lots of steps that seems too hectic. How to I do it in an easy way.

In general,how should I deal with any inverse trigonometric function such as this or $\arctan\frac{3x-x^3}{1-3x^2}$ etc. and easily find their definition for various values of $x$.

Thanks for any help!

Soham
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1 Answers1

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If $x\in\left(-1,1\right)$, then$$\tan\bigl(2\arctan(x)\bigr)=\frac{2\tan(\arctan x)}{1-\tan^2(\arctan x)}=\frac{2x}{1-x^2}.\tag1$$Since $2\arctan(x)\in\left(-\frac\pi2,\frac\pi2\right)$, it follows from this that$$2\arctan(x)=\arctan\left(\frac{2x}{1-x^2}\right).$$

What if $x>1$? Then $(1)$ still holds, but now $2\arctan x\in\left(\frac\pi2,\frac{3\pi}2\right)$. But $2\arctan(x)-\pi\in\left(-\frac\pi2,\frac\pi2\right)$ and, since $\tan$ is periodc with period $\pi$, it follows that$$2\arctan(x)-\pi=\arctan\left(\frac{2x}{1-x^2}\right).$$

The other case is similar.

José Carlos Santos
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  • How do we know - $2\arctan(x)\in\left(-\frac\pi2,\frac\pi2\right)$? – Soham Aug 26 '18 at 15:23
  • Anyways what you have written is a verification of the statement. In your answer, we start with $x\in(-1,+1)$. How do we know, we have to start with this? I have to derive the formula... not verify it – Soham Aug 26 '18 at 15:25
  • $$x\in(-1,1)\implies\arctan x\in\left(-\frac\pi4,\frac\pi4\right)\implies2\arctan x\in\left(-\frac\pi2,\frac\pi2\right).$$And what I did at first only worked when $x\in(-1,1)$. Therefore, I had to see after what happens in the other cases. – José Carlos Santos Aug 26 '18 at 15:29
  • I understand but I didn't want that actually. For eg. if we want to find the definition of $\arctan\frac{3x-x^3}{1-3x^2}$ we don't know that we have to start with $(-1,1)$ (actually we don't). What to do in that case? So what to do in a general case? I wanted to know that. – Soham Aug 26 '18 at 15:32