arctan and arcsin equation

Question

How can I prove that :

$2 \arctan(x) + \arcsin \Big( \frac{ 2x }{ 1 + x^2 }\Big) = \pi$ , $x > 1$

What is the best way to do this ?

Answer 1

You can prove it without differentiating: set $t=2\arctan x$. This means $$\tan\frac t2=x\quad\text{and}\quad -\pi <t<\pi.$$ Furthermore, as $x>1$, we have $\;\dfrac\pi2<t<\pi$.

Now, by the half-angle formulae (actually this is where the formula we seek to prove comes from), $$\sin t=\frac{2x}{1+x^2}, \quad\text{whence}\quad t\equiv \begin{cases}\arcsin\frac{2x}{1+x^2}\quad\text{or} \\ \pi -\arcsin\frac{2x}{1+x^2}\end{cases}\mod2\pi.$$ Knowing $\;\dfrac\pi2<t<\pi$, we necessarily have $$t=2\arctan x= \pi -\arcsin\frac{2x}{1+x^2}.$$

Answer 2

An easy way is to differentiate the function $$f(x)=2\arctan(x)+\arcsin\left(\dfrac{2x}{1+x^2}\right)$$ and show that $f'(x)=0$ so that $f$ is a constant. Then calculate the constant by letting $x$ be a well-chosen number.

Answer 3

Let put $x=tan(\frac{t}{2})$ with

$t\in(\frac{\pi}{2},\pi)$ cause $x>1$

then

$1+x^2=\frac{1}{\cos^2(\frac{t}{2})}$

and

$f(x)=t+arcsin(\sin(t))$

$=t-arcsin(\sin(t-\pi))$.

but $-\frac{\pi}{2}<t-\pi<0$, thus

$f(x)=t-t+\pi=\pi$.