# Thread: Show that arccos(...) = ...

1. ## Show that arccos(...) = ...

Hi

Another "show that" problem involving inverse trig. functions.

Show that $\displaystyle arccos(\frac{1}{\sqrt{1+x^{2}}}) = arctan(|x|)$

Since $\displaystyle | \frac{1}{\sqrt{1+x^{2}}} | \leq 1$ , I assume I can take $\displaystyle cos(arccos(\frac{1}{\sqrt{1+x^{2}}})) = \frac{1}{\sqrt{1+x^{2}}}$

But what do I do with my right-side?

Kinda stuck here.

2. Originally Posted by Twig
Hi

Another "show that" problem involving inverse trig. functions.

Show that $\displaystyle arccos(\frac{1}{\sqrt{1+x^{2}}}) = arctan(|x|)$

Since $\displaystyle | \frac{1}{\sqrt{1+x^{2}}} | \leq 1$ , I assume I can take $\displaystyle cos(arccos(\frac{1}{\sqrt{1+x^{2}}})) = \frac{1}{\sqrt{1+x^{2}}}$

But what do I do with my right-side?

Kinda stuck here.
let $\displaystyle u = arccos(\frac{1}{\sqrt{1+x^{2}}})$

$\displaystyle cos(u)=(\frac{1}{\sqrt{1+x^{2}}})$

as you can see from the pic

$\displaystyle tan(u) = x$

$\displaystyle arctan(tan(u))=arctan(x)$

$\displaystyle u=arctan(x) = arccos(\frac{1}{\sqrt{1+x^{2}}})$ since

we let
$\displaystyle u=arccos(\frac{1}{\sqrt{1+x^{2}}})$

3. Let $\displaystyle \theta=\arccos\left(\frac1{\sqrt{1+x^2}}\right).$ Then

$\displaystyle \cos\theta\ =\ \frac1{\sqrt{1+x^2}}$

$\displaystyle \implies\ \cos^2\theta\ =\ \frac1{1+x^2}\quad\ldots\,\fbox1$

$\displaystyle \therefore\ \sin^2\theta\ =\ 1-\cos^2\theta\ =\ 1-\frac1{1+x^2}\ =\ \frac{x^2}{1+x^2}\quad\ldots\,\fbox2$

$\displaystyle \therefore\ \tan^2\theta\ =\ \frac{\sin^2\theta}{\cos^2\theta}\ =\ x^2$

$\displaystyle \implies\ \tan\theta\ =\ \pm|x|$

$\displaystyle \implies\ \theta\ =\ \arctan(\pm|x|)\ =\ \pm\arctan|x|$

But since $\displaystyle \arccos y\ge0$ for all $\displaystyle 0<y\le1,$ $\displaystyle \theta\ge0.$ Hence $\displaystyle \arccos\left(\frac1{\sqrt{1+x^2}}\right)=\arctan|x |.$

4. Thanks guys, always nice with a few different approaches.

I think the drawing a triangle solution is something I need to remember.

5. Hello, Twig!

Show that: .$\displaystyle \arccos\left(\frac{1}{\sqrt{1+x^{2}}}\right) \:=\: \arctan(|x|)$
Let: .$\displaystyle y \;=\;\arccos\left(\frac{1}{\sqrt{1+x^2}}\right) \quad\Rightarrow\quad \cos y \:=\:\frac{1}{\sqrt{1+x^2}}$

Then: .$\displaystyle \cos^2\!y \:=\:\frac{1}{1+x^2} \quad\Rightarrow\quad 1-\cos^2\!y \:=\:1 - \frac{1}{1+x^2} \:=\:\frac{x^2}{1+x^2}$

. . Hence: .$\displaystyle \sin^2\!y \:=\:\frac{x^2}{1+x^2} \quad\Rightarrow\quad \sin y \:=\:\frac{\pm x}{\sqrt{1+x^2}}$

Then: .$\displaystyle \tan y \:=\:\frac{\sin y}{\cos y} \;=\;\frac{\dfrac{\pm x}{\sqrt{1+x^2}}}{\dfrac{1}{\sqrt{1+x^2}}}\quad\Ri ghtarrow\quad \tan y \;=\; \pm x$

. . Hence: .$\displaystyle y \:=\:\arctan(\pm x) \;=\;\arctan|x|$

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