# Thread: Very challenging; Simplify the trigonometric complex expression.

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

Simplify

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

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

EDIT;

Stopping at

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

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

Now try it again using the assumption that x is complete, if you should take it.

2. Hello !

Originally Posted by mathwizard
Simplify

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

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 ?

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

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

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

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

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

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

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

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

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Now, remember that :

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

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

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

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

Similarly, we get :

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

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

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

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

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

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

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

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

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

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

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

Yay !

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

Re-edit : new release, mistakes corrected

3. Ok... I have found a quicker method (takin' a shower makes things easier )

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

\displaystyle \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}) \\ &=\ln \left(\sqrt[2i]{\frac{(x-i)(x+i)}{(x-i)^2}}\right) \\ &=\frac{1}{2i} \cdot \ln \left(\frac{x+i}{x-i}\right) \end{aligned}

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\displaystyle \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}} \\ &=\frac 1i \cdot \frac{e^{2iS}-1}{e^{2iS}+1} \end{aligned}

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

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\displaystyle \begin{aligned} \tan S&=\frac 1i \cdot \frac{\frac{x+i}{x-i}-1}{\frac{x+i}{x-i}+1} \\ \\ &=\frac 1i \cdot \frac{x+i-x+i}{x+i+x-i} \\ \\ &=\boxed{\frac 1x} \end{aligned}