I want to prove that2arctan⁡x=arcsin⁡x−1x+1+π2

Question

Haiden Frazier · Accepted Answer

One way would be to notice that both functions have the same derivative and then find out what the "constant" is by plugging in

x = 0

.
Here's another way. Look at

\frac{π}{2} - 2 \arctan \sqrt{x} = 2 (\frac{π}{4} - \arctan \sqrt{x})

= 2 (\arctan 1 - \arctan \sqrt{x})

\arctan u - \arctan v = \arctan \frac{u - v}{1 + u v}

(This follows from the usual identity for the tangent of a sum.) The left side above becomes

2 \arctan \frac{1 - \sqrt{x}}{1 + \sqrt{x}}

The double-angle formula for the sine says

\sin (2 u) = 2 \sin u \cos u

. Apply that:

\sin (2 \arctan \frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = 2 \sin (\arctan \frac{1 - \sqrt{x}}{1 + \sqrt{x}}) \cos (\arctan \frac{1 - \sqrt{x}}{1 + \sqrt{x}})

\sin (2 \arctan \frac{1 - \sqrt{x}}{1 + \sqrt{x}}) = 2 \sin (\arctan \frac{1 - \sqrt{x}}{1 + \sqrt{x}}) \cos (\arctan \frac{1 - \sqrt{x}}{1 + \sqrt{x}})

Now remember that

\sin (\arctan u) = \frac{u}{\sqrt{1 + u^{2}}}

and

\cos (\arctan u) = \frac{1}{\sqrt{1 + u^{2}}}

Then use algebra:

2 \cdot \frac{(\frac{1 - \sqrt{x}}{1 + \sqrt{x}})}{\sqrt{1 + {(\frac{1 - \sqrt{x}}{1 + \sqrt{x}})}^{2}}} \cdot \frac{1}{{\sqrt{1 + (\frac{1 - \sqrt{x}}{1 + \sqrt{x}}}}}^{2}} = \frac{1 - x}{1 + x}

I want to prove that 2\arctan\sqrt{x}=\arcsin\frac{x-1}{x+1}+\frac{\pi}{2}