Show an $\arctan$ and $\arcsin$ function is constant

Question

Show that for every $x\geq1$ the following is true: $2\arctan x + \arcsin \frac{2x}{1+x^2} = \pi$

One way (mentioned in the link at the bottom) would be to calculate the derivative of the left side, show that it is always $0$ then show that for $x=1$ the equation is true. I'm trying for some time to find a cleaner way to prove the equality, without so much algebra. Does anyone have any idea where to start?

Someone already mentioned the same problem here.

Answer 1

Outline: We can do it with just trigonometric identities. Note first that $$\tan(2\arctan x)=\frac{2x}{1-x^2}.\tag{1}$$ Then note that $$\tan\left(\arcsin\left(\frac{2x}{1+x^2}\right)\right)=\frac{2x}{x^2-1} \tag{2}$$ because $x\ge 1$. So the right-hand side of (2) is the negative of the right-hand side of (1).

No derivatives!

Answer 2

We can show $\frac {2x}{1-x^2}=\tan 2 \arctan x = \tan (\pi-\arcsin \frac {2x}{1+x^2}=\tan \arcsin \frac {2x}{1+x^2})$ The left equality comes from $\tan 2y = \frac {2 \tan y}{1-\tan^2 y}$ and the second from drawing a right triangle with one angle $\arcsin \frac {2x}{1+x^2}$ As $\arctan$ is a bijection over this range, we are done.

Answer 3

Let $f$ be the function

$$f(x)=2 \arctan x+\arcsin\left(\frac{2x}{1+x^2}\right)$$

Note that the arcsine function can be expressed alternatively as the arctangent function using $\arcsin y=\arctan \frac{y}{\sqrt{1-y^2}}$ for $|y|<1$.

Here, we have $y= \frac{2x}{1+x^2}$ for $|x|>1$. Thus,

$$\arcsin \frac{2x}{1+x^2}=\arctan \frac{2x}{|1-x^2|}=\arctan \frac{2x}{x^2-1}$$

We also note that for $x>1$, $\arctan x>0$. Therefore, we have

$$2 \arctan x=\pi+\arctan \frac{2x}{1-x^2}$$

since $\arctan \frac{2x}{1-x^2}<0$.

Finally, we obtain the coveted result

$2 \arctan x+\arcsin\left(\frac{2x}{1+x^2}\right)=\pi$.

Answer 4

This can be solved using basic Inverse Trigonometric Identities

$$2\arctan(x) + \arcsin{\dfrac{2x}{1+x^2}} $$

$$\Rightarrow \arctan\dfrac{2x}{1-x^2} + \arctan\dfrac{2x}{x^2-1}$$

$$\Rightarrow \arctan\dfrac{2x}{1-x^2} + \arctan\dfrac{-2x}{1-x^2}$$

$$\Rightarrow \arctan\dfrac{0}{1+\Big(\dfrac{2x}{1-x^2}\Big)^2}$$

Now since $x\geq 1$

$$ \Rightarrow \arctan(0) = \pi$$

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I guess you can also use Andre Nicolas' answer as a hint!

Answer 5

if $x\geq 1$, then $x=\tan\frac{\theta}{2}$ for some $\theta\in\left[\frac{\pi}{2},\pi\right)$, then: $$2\arctan x + \arcsin\frac{2x}{1+x^2} = 2\arctan\tan\frac{\theta}{2}+\arcsin\sin\theta=\theta+(\pi-\theta)=\pi. $$

Answer 6

i will use the letter $m$ so that i can interpret it as the slope of the radius in the init circle. let $$t = \tan^{-1}(m), m \ge 1.\tag 1$$ this implies $$\pi/4 \le t < \infty, \tan t = m, \cos t = \frac 1{\sqrt {1+m^2}}, \sin t = \frac m{\sqrt{1+m^2}}. $$ we also know from the double angle formula $$ \sin 2t = 2\sin t \cos t = \frac{2m}{1+m^2}$$ since $\pi/2 \le 2t < p,$ we have $$\sin^{-1}(\sin 2t)) = \pi - 2t. \tag 2$$

putting $(1)$ and $(2)$ gives $$ \text{ if } 1\le m \text{ then } 2 \tan^{-1}m + \sin^{-1}\left(\frac{2m}{1+m^2}\right)= \pi.$$

Answer 7

\begin{align} 2\arctan(x)+\arcsin\left(\frac{2x}{1+x^2}\right) &=\pi \end{align}

Let's $\chi=\arctan(x)\ge\pi/4$ since $x\ge1$.

Consider a unit circle

Then in a $\triangle OAB$ we have $|OA|=|OB|=1$, $\angle OAB=\angle OBA=\chi$, $\angle AOB=\alpha=\pi-2\chi$, $|AH|=\sin(\alpha)=a$.

In the $\triangle ABH$ \begin{align} |AH|&=|BH|\tan(\chi) \\ \sin(\alpha) &= (1-\cos(\alpha))\tan(\chi) \\ \sin(\alpha) &= \left(1-\sqrt{1-\sin^2(\alpha)}\right)\tan(\chi) \\ a&= \left(1-\sqrt{1-a^2}\right)x \\ \left(1-\frac{a}{x}\right)^2&= 1-a^2 \\ \frac{a^2}{x^2}+a^2 &= 2\frac{a}{x} \\ a\left(\frac{1}{x^2}+1\right) &= \frac{2}{x} \\ a&=\sin(\alpha)= \frac{2x}{1+x^2}. \end{align}