Show that arccos(...) = ...

**Twig** · June 11th 2009, 01:55 AM

Hi

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

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

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

But what do I do with my right-side?

Kinda stuck here.

**Amer** · June 11th 2009, 02:37 AM

Originally Posted by Twig

Hi

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

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

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

But what do I do with my right-side?

Kinda stuck here.

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

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

as you can see from the pic

$tan(u) = x$

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

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

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

**TheAbstractionist** · June 11th 2009, 02:40 AM

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

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

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

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

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

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

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

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

**Twig** · June 11th 2009, 04:32 AM

Thanks guys, always nice with a few different approaches.

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

**Soroban** · June 11th 2009, 05:24 AM

Hello, Twig!

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

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

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

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

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

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

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

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