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The ratio is as follows:

$$1 - \cos2x + i\sin2x \over 1 + \cos2x - i\sin 2x$$

I am unsure how to simplify this, as the numerator poses a problem as I try to multiply this equation by $\operatorname{cis}(2x)$ to get a real denominator.

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    $\begingroup$ You can say $\frac{a+ib}{c-id} = \frac{(a+ib)(c+id)}{(c-id)(c+id)} = \frac{ac-bd }{c^2+d^2}+ i\frac{ad+bd}{c^2+d^2}$ and here $a=1-\cos 2x$, $b=\sin 2x$, $c=1+\cos2x$, $d=\sin 2x$. It looks to me as if the numerators simplify nicely $\endgroup$
    – Henry
    Sep 29, 2019 at 9:19
  • $\begingroup$ I’m one of the people who leaves $\rm e\ i\ π$ upright, just like the other universal numbers like $1\ 2\ 3$, but I left it as $i$ like a variable since that’s what everyone else here has done. $\endgroup$ Sep 29, 2019 at 9:44

3 Answers 3

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HINT

Recall that

  • $\cos t = \frac{e^{it}+e^{-it}}{2}$

  • $\sin t = \frac{e^{it}-e^{-it}}{2i}$

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    $\begingroup$ Thank you for allowing me to work it out myself! You are a legend! $\endgroup$ Sep 29, 2019 at 9:24
  • $\begingroup$ @DonkeyKong You are welcome! That's the best way to learn. Bye $\endgroup$
    – user
    Sep 29, 2019 at 9:28
  • $\begingroup$ @DrZafarAhmedDSc Yes of course! Thanks to have pointed the typos out! $\endgroup$
    – user
    Sep 29, 2019 at 9:37
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$$F=\frac{1-\cos 2x +i \sin 2x}{1+\cos 2x-i\sin 2x} = \frac{2 \sin ^2 x+ 2i \sin x \cos x}{2 \cos^2 x-2 i \sin x \cos x} =i \tan x \frac{\cos x -i \sin x}{\cos x -i \sin x}= i \tan x.$$

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Using Intuition behind euler's formula

$$\dfrac{1-e^{-2ix}}{1+e^{2ix}}=\dfrac{e^{-ix}}{e^{ix}}\dfrac{e^{ix}-e^{-ix}}{?}=e^{2ix}\cdot\dfrac{2i\sin x}{2\cos x}=?$$

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  • $\begingroup$ Surely that would not simplify the denominator? $\endgroup$ Sep 29, 2019 at 9:36
  • $\begingroup$ @Donkey, I have left the calculation of the denominator for you $\endgroup$ Sep 29, 2019 at 9:40
  • $\begingroup$ Multiplying the denominator by e^(ix) will give you e^(ix) + e^(3ix) won't it? How will that help? $\endgroup$ Sep 29, 2019 at 10:12
  • $\begingroup$ @Donkey, Please find the updated post $\endgroup$ Sep 29, 2019 at 10:50

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