# How to prove that $2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}$

I want to prove that$$2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}, x\geq 0$$

I have started from the arcsin part and I tried to end to the arctan one but I failed.

Can anyone help me solve it?

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Define $\displaystyle g(x)=2\arctan(\sqrt{x})-\arcsin\left(\frac{x-1}{x+1}\right)-\frac{\pi}{2}$. Differentiate to find that $g'(x)=0$. Since this was true on the connected set $[0,\infty)$ you can conclude that $g$ is contant. Note then that $g(0)=0$ to finish.

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+1. A nice way to look at it but unnecessarily complicated. – user17762 Dec 9 '11 at 0:33

Let $\arctan{\sqrt{x}} = \theta$. Then we have $x = \tan^2 (\theta)$. Hence, $$\frac{x-1}{x+1} = \frac{\tan^2 (\theta)-1}{\tan^2 (\theta)+1} = \sin^2(\theta) - \cos^2(\theta) = - \cos(2 \theta)= \sin \left(2 \theta - \frac{\pi}{2} \right)$$ Hence, $$2 \arctan{\sqrt{x}} = 2 \theta = \arcsin \left(\frac{x-1}{x+1} \right) + \frac{\pi}{2}$$

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Very nice.. One short comment: $\frac{\tan^2 (\theta)-1}{\tan^2 (\theta)+1} = - \cos(2 \theta)$ is just the standard $t =\tan(\frac{x}{2})$ substitution :) – N. S. Dec 9 '11 at 0:41

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 $$\frac\pi2 - 2\arctan\sqrt{x} = 2\left( \frac\pi4 - \arctan\sqrt{x} \right) = 2\left( \arctan1-\arctan\sqrt{x} \right).$$ Now remember the identity for the difference of two arctangents: $$\arctan u - \arctan v = \arctan\frac{u-v}{1+uv}.$$ (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 $\sin(2u)=2\sin u\cos u$. Apply that: $$\sin\left(2\arctan\frac{1-\sqrt{x}}{1+\sqrt{x}}\right) = 2 \sin\left(\arctan\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)\cos\left(\arctan\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)$$

Now remember that $\sin(\arctan u) = \dfrac{u}{\sqrt{1+u^2}}$ and $\cos(\arctan u) = \dfrac{1}{\sqrt{1+u^2}}$

Then use algebra: $$2\cdot\frac{\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)}{\sqrt{1+\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)^2}}\cdot \frac{1}{\sqrt{1+\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)^2}} = \frac{1-x}{1+x}.$$

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OK, I hope I've now got everything except any remaining minor typos into shape in this posting. – Michael Hardy Dec 9 '11 at 0:20