# Thread: [SOLVED] Equation involving arctan

1. ## [SOLVED] Equation involving arctan

Show that $\forall x \in R$, that

$arctan(x+1) - arctanx = arctan\frac{1}{x^2 + x + 1}$

tan is a continuous function, so (can anyone tell me why that is important when doing these kinda proofs?)

$tan(arctan(x+1) - arctanx) = tan\left(arctan\frac{1}{x^2 + x + 1}\right)$

$u = arctan(x+1), v = arctanx \implies$

$tan(u-v) = \frac{tanu - tanv}{1 + tanu \cdot tanv} = tan\left(arctan\frac{1}{x^2 + x + 1}\right)$

Now all I need to prove is that

$\frac{-\pi}{2} < arctan(x+1) - arctanx < \frac{\pi}{2}, \forall x \in R$

Suggestions?

(How do you write the symbol for real numbers in latex by the way?)

2. Hello, Spec!

You started off okay . . .

Show that $\forall x \in R$, that:

. . $\arctan(x+1) - \arctan(x) \:= \:\arctan\left(\frac{1}{x^2 + x + 1}\right)$
Let $\theta \:=\:\arctan(x+1) - \arctan(x)$

Take the tangent:

. . $\tan\theta \;=\;\tan\left[\arctan(x+1) - \arctan(x)\right]$

. . . . . . $=\;\frac{\tan[\arctan(x+1)] - \tan[\arctan(x)]}{1 + \tan[\arctan(x+1)]\tan[\arctan(x)]}$

. . . . . . $= \;\frac{x + 1 - x}{1 + (x+1)x}$

. . . . . . $=\;\frac{1}{x^2+x+1}$

We have: . $\tan\theta \;=\;\frac{1}{x^2+x+1}$

. . Hence: . $\theta \;=\;\arctan\left(\frac{1}{x^2+x+1}\right)$

Therefore: . $\arctan(x+1) - \arctan(x) \;=\;\arctan\left(\frac{1}{x^2+x+1}\right)$

3. That much I've already figured out, but according to my math professor, I also need to be able to show that

$\frac{-\pi}{2} < arctan(x+1) - arctanx < \frac{\pi}{2}, \forall x \in R$

4. Originally Posted by Spec
$\frac{-\pi}{2} < arctan(x+1) - arctanx < \frac{\pi}{2}, \forall x \in R$
$\frac{d}{dx}\text{Arctan}(x+1) - \text{Arctan}x = 0$

$\frac{1}{(x+1)^2+1} - \frac{1}{x^2+1} = 0$

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

$(\text{ For }(x+1)^2+1 \neq 0, x^2+1 \neq 0)\text{ }$

$(x+1)^2+1 = x^2+1$
... (some easy work)
$x = -\frac{1}{2}$
(I leave you the proof of this point is a maxima)

Now we know that this function has only one extrema. But we must still check it for global maximums/minimums.

$\lim _{x\to +\infty }\text{Arctan}(x+1)-\text{Arctan}(x)=0$

$\lim _{x\to -\infty }\text{Arctan}(x+1)-\text{Arctan}(x)=0$

So, the functions largest value is $-\frac{1}{2}$ and least value is $0$.

5. Originally Posted by Spec
That much I've already figured out, but according to my math professor, I also need to be able to show that

$\frac{-\pi}{2} < arctan(x+1) - arctanx < \frac{\pi}{2}, \forall x \in R$
Are you allowed to use $\arctan(x+1) - \arctan(x) \;=\;\arctan\left(\frac{1}{x^2+x+1}\right) \,$ ?

Note that $\frac{3}{4} \leq x^2 + x + 1 = \left( x + \frac{1}{2} \right)^2 + \frac{3}{4} < \infty$

Therefore $0 < \frac{1}{x^2+x+1} \leq \frac{4}{3}$.

What does that tell you ......?