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Math Help - Very challenging; Simplify the trigonometric complex expression.

  1. #1
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    Very challenging; Simplify the trigonometric complex expression.

    Simplify


    <br />
tan[ln(\sqrt[2i]{x^2 + 1}) - ln(\sqrt[i]{x-i})]


    http://i29.tinypic.com/2ijroqq.jpg

    EDIT;

    Stopping at

     \frac{1}{2i} ln(x+i) - \frac{1}{2i} ln(x-i) ,

    note that ln(x+i) = ln(x^2+1) + i\ arctan(1/x) and
    ln(x-i) = ln(x^2+1) - i\ arctan(1/x).
    The rest follows.

    Now try it again using the assumption that x is complete, if you should take it.
    Last edited by mathwizard; June 17th 2008 at 03:31 PM.
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  2. #2
    Moo
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    Hello !

    Your problems are really challenging

    Quote Originally Posted by mathwizard View Post
    Simplify


    <br />
tan[ln(\sqrt[2i]{x^2 + 1}) - ln(\sqrt[i]{x-i})]


    http://i29.tinypic.com/2ijroqq.jpg
    I don't know if I'm going to do it correctly...but well, what happens if one doesn't try anything ?

    \tan \left(\ln(\sqrt[2i]{x^2+1})-\ln(\sqrt[i]{x-i})\right)

    I'll study first \left(\ln(\sqrt[2i]{x^2+1})-\ln(\sqrt[i]{x-i})\right)=S

    S=\ln\left((x^2+1)^{\frac{1}{2i}}\right)-\ln\left((x-i)^{\frac 1i}\right)

    S=\frac{1}{2i} \ln [(x-i)(x+i)]-\frac 1i \ln[(x-i)]

    S=\frac{1}{2i} \ln(x-i)+\frac{1}{2i} \ln(x+i)-\frac 1i \ln(x-i)

    S=\frac{1}{2i} \ln(x+i)-\frac{1}{2i}\ln(x-i)

    S=\frac{1}{2i} \ln \left(\frac{x+i}{x-i}\right)


    (I know we can do it in a more direct way, but it doesn't matter)

    --------------------------

    Now, remember that :

    \cos x=\frac{e^{ix}+e^{-ix}}{2}

    \sin x=\frac{e^{ix}-e^{-ix}}{2i}


    \tan x=\frac{\sin x}{\cos x} \implies \tan S=\frac 1i \cdot \frac{e^{iS}-e^{-iS}}{e^{iS}+e^{-iS}}

    ---------------------------

    e^{iS}=e^{i \cdot \frac{1}{2i} \ln \left(\frac{x+i}{x-i}\right)}=e^{\frac 12 \ln \left(\frac{x+i}{x-i}\right)}=\sqrt{\frac{x+i}{x-i}}

    Similarly, we get :

    e^{-iS}=\sqrt{\frac{x-i}{x+i}}


    ----------------------------

    \tan S=\frac 1i \cdot \left(\sqrt{\frac{x+i}{x-i}}-\sqrt{\frac{x-i}{x+i}}\right) \cdot \frac{1}{\sqrt{\frac{x+i}{x-i}}+\sqrt{\frac{x-i}{x+i}}}

    Multiply by 1=\frac{\sqrt{\frac{x+i}{x-i}}-\sqrt{\frac{x-i}{x+i}}}{\sqrt{\frac{x+i}{x-i}}-\sqrt{\frac{x-i}{x+i}}} :

    \tan S=\frac 1i \cdot {\color{blue}\left(\sqrt{\frac{x+i}{x-i}}-\sqrt{\frac{x-i}{x+i}}\right)^2} \cdot \frac{1}{\color{red}\frac{x+i}{x-i}-\frac{x-i}{x+i}}

    ------------------------------

    {\color{red}\frac{x+i}{x-i}-\frac{x-i}{x+i}}=\frac{(x+i)^2-(x-i)^2}{x^2+1}=\frac{4ix}{x^2+1}


    {\color{blue}\left(\sqrt{\frac{x+i}{x-i}}-\sqrt{\frac{x-i}{x+i}}\right)^2}=\frac{x+i}{x-i}+\frac{x-i}{x+i} -2 \underbrace{\sqrt{\frac{x+i}{x-i}} \cdot \sqrt{\frac{x-i}{x+i}}}_{=1}

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

    ------------------------------

    \tan S=\frac 1{\color{red}i} \cdot \left(\frac{2(x^2-1)}{x^2+1}-2\right) \cdot \frac{x^2+1}{4{\color{red}i}x}

    \tan S={\color{red}-} \frac{1}{4x} \cdot \left(2(x^2-1)-2(x^2+1)\right)


    \tan S=-\frac{1}{4x} \cdot (-4)



    \boxed{\tan S=\frac{1}{x}}



    Yay !

    Edit : and this was done assuming that x had values that didn't impeach anything in it...

    Re-edit : new release, mistakes corrected
    Last edited by Moo; June 16th 2008 at 03:33 PM.
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  3. #3
    Moo
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    Ok... I have found a quicker method (takin' a shower makes things easier )

    \tan \left(\ln(\sqrt[2i]{x^2+1})-\ln(\sqrt[i]{x-i})\right)

    \begin{aligned} S=\ln(\sqrt[2i]{x^2+1})-\ln(\sqrt[i]{x-i}) &=\ln(\sqrt[2i]{(x-i)(x+i)})-\ln(\sqrt[2i]{(x-i)^2}) \\<br />
&=\ln \left(\sqrt[2i]{\frac{(x-i)(x+i)}{(x-i)^2}}\right) \\<br />
&=\frac{1}{2i} \cdot \ln \left(\frac{x+i}{x-i}\right) \end{aligned}

    -------------------

    \begin{aligned} \tan S &=\frac 1i \cdot \frac{e^{iS}-e^{-iS}}{e^{iS}+e^{-iS}} \quad \leftarrow \quad \text{multiply by } \frac{e^{iS}}{e^{iS}} \\<br />
&=\frac 1i \cdot \frac{e^{2iS}-1}{e^{2iS}+1} \end{aligned}

    -------------------

    e^{2iS}=e^{2i \cdot \frac{1}{2i} \cdot \ln \left(\frac{x+i}{x-i}\right)}=\frac{x+i}{x-i}

    -------------------

    \begin{aligned} \tan S&=\frac 1i \cdot \frac{\frac{x+i}{x-i}-1}{\frac{x+i}{x-i}+1} \\ \\<br />
&=\frac 1i \cdot \frac{x+i-x+i}{x+i+x-i} \\ \\<br />
&=\boxed{\frac 1x} \end{aligned}
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