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$ \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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    Satisfied with an answer below?2012-09-19

4 Answers 4

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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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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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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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    I _thought_ something was wrong with the way the identity was stated, but I was too rushed earlier to think about it.2012-08-27
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Arctan . Jpeg

Click on that blue thingy above - and no, I swear this isn't a spam. The math formula programming was simply too tiring to learn in 3min. flat, so I've provided a written vers instead. Don't judge my handwriting, perhaps? Hope it helped! :)