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

Question

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

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.

Answer 1

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

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

Answer 2

Here is 2-steps plan:

1. Salvage your accept rate from its current appalling value.
2. Prove that $$ \arctan(\sqrt{x}) = \frac{1}{2} \arccos\left(\dfrac{1-x}{1+x}\right). $$

Answer 3

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.