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

  1. #1
    Senior Member Twig's Avatar
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    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.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Twig View Post
    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

    Show that arccos(...) = ...-23334.jpg

    $\displaystyle tan(u) = x $

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

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

    we let
    $\displaystyle u=arccos(\frac{1}{\sqrt{1+x^{2}}})$
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    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 $\displaystyle \arccos\left(\frac1{\sqrt{1+x^2}}\right)=\arctan|x |.$
    Last edited by TheAbstractionist; Jun 11th 2009 at 03:31 AM.
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  4. #4
    Senior Member Twig's Avatar
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    Thanks guys, always nice with a few different approaches.

    I think the drawing a triangle solution is something I need to remember.
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  5. #5
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    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: .$\displaystyle \arccos\left(\frac{1}{\sqrt{1+x^2}}\right) \;=\;\arctan|x|$

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