# Prove $\sqrt{\arctan(x)} = (1/2) \arccos((1-x)/(1+x))$

$$\sqrt{\arctan(x)} = \dfrac{1}{2} \arccos\left(\dfrac{1-x}{1+x}\right)$$

I have been trying to solve this problem for the past hour, but I'm not able to solve it as I have just started solving difficult trigonometric problems. I'm not able to get any logic to solve this problem. I'm not able to put any trigonometric formula to solve this please help.

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## migrated from mathematica.stackexchange.comAug 27 '12 at 16:00

This question came from our site for users of Mathematica.

This looks like a math problem, nothing to do with Mathematica. Migrate? – stevenvh Aug 27 '12 at 15:17
This statement is not true: Sqrt[ArcTan[x]] === 1/2 ArcCos[(1 - x)/(1 + x)] --> False. Can anybody check if my TeX-ification is correct? – stevenvh Aug 27 '12 at 15:35
@stevenvh the original 1-x/1+x could be interpreted variously as (1 - x)/(1 + x), 1 - x/(1 + x), or 1. However, none of these make the statement true (except for $x=0$). So I see no problem with your interpretation but still don't really know how to proceed. – Oleksandr R. Aug 27 '12 at 15:43
This user already has several questions like this on Math.SE, so it's a bit strange to post this here now, having found the right place in the past. Even though it was purely accidental, hopefully they'll be glad that we saved them from a low-quality question... – Oleksandr R. Aug 27 '12 at 15:54
I think you mean $\arctan \sqrt{x}$, not $\sqrt{\arctan x}$. – Robert Israel Aug 27 '12 at 19:48

The statement isn't true. For instance you can plot the difference or compute an asymptotic expansion of the difference in $0$ :

$$\frac{2\,{x}^{\frac{3}{2}}}{3}-\frac{11\,{x}^{\frac{5}{2}}}{15}+\frac{2\,{x}^{\frac{7}{2}}}{7}+\cdots$$

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Here is 2-steps plan:

1. Salvage your accept rate from its current appalling value.
2. Prove that $$\arctan(\sqrt{x}) = \frac{1}{2} \arccos\left(\dfrac{1-x}{1+x}\right).$$
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I seem to recall first proving this identity as follows:

• Compute the derivatives of the left and right sides and show that those are the same, so the left and right sides differ by a constant; then
• Show that they're equal at $x=0$, so that constant must be $0$.

It should also be possible to prove it as a corollary of other trigonometric identities and/or basic geometry.

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Here is short demonstration, using trigonometric identities. Suppose $\sqrt{x} = \tan(\phi)$ for some $0\leqslant \phi < \frac{\pi}{2}$. Then: $$\frac{1-x}{1+x} = \frac{1-\tan^2(\phi)}{1+\tan^2(\phi)} = \frac{\cos^2(\phi)-\sin^2(\phi)}{\cos^2(\phi)+\sin^2(\phi)} = \frac{\cos^2(\phi)-\sin^2(\phi)}{1}= \cos(2\phi)$$ Hence: $$\arctan\left(\sqrt{x}\right) = \phi = \frac{1}{2}\arccos\left(\frac{1-x}{1+x}\right)$$ – Sasha Aug 27 '12 at 19:23
Pleaaaase, no derivatives here... – Did Aug 27 '12 at 20:14
I thought something was wrong with the way the identity was stated, but I was too rushed earlier to think about it. – Michael Hardy Aug 27 '12 at 21:20