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I am familiar with solving trigonometric identities using De Moivre's Theorem, where only $\sin(x)$ and $\cos(x)$ terms are involved. But could not use it to solve identities involving other ratios. For example,

(1) $\tan\left(\frac{\theta}{2}\right)\sec(x)+\tan\left(\frac{\theta}{2^2}\right)\sec\left(\frac{x}{2}\right)+...+\tan\left(\frac{\theta}{2^n}\right)\sec\left(\frac{x}{2^{n-1}}\right)$

(2) $\csc(x)+\csc(2x)+...+\csc(2^nx)$

Is there any way to simplify this kind of problems and express them in smaller terms using De Moivre's Theorem?

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  • $\begingroup$ What are the problems? Do you seek a shorter way to express those sums? Or do you want to show if they converge? $\endgroup$ – ajotatxe Jun 5 at 6:37
  • $\begingroup$ No I am not trying to show they converge. I am trying to find a simplified form. The way they have done it here -math.stackexchange.com/a/1538186/854039 $\endgroup$ – Abhinandan Saha Jun 5 at 6:45
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    $\begingroup$ You have mixed $x$s and $\theta$s in your expressions. Can you edit your post so as to make it clear exactly what you mean? $\endgroup$ – Prime Mover Jun 5 at 6:47
  • $\begingroup$ The question did have mixed expressions of $\theta$ and $x$. $\endgroup$ – Abhinandan Saha Jun 5 at 6:50
  • $\begingroup$ In the second one, use the complex definition of cosecant function and then try to find a geometric series. Probably you will get a double series. $\endgroup$ – Nikhil Kumar Singh Jun 5 at 6:54
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We have $$\sin x=\frac{ e^{ix}-e^{-ix}}{2i}$$

And therefore it's reciprocal as $$\csc x=\frac{2i}{ e^{ix}-e^{-ix}}$$ which can also be written as $$\csc x=\frac{2i.e^{ix}}{ e^{2ix}-1}$$ therefore your series becomes

$\displaystyle{\frac{2i.e^{ix}}{ e^{2ix}-1}+\frac{2i.e^{2ix}}{ e^{4ix}-1}+\frac{2i.e^{4ix}}{ e^{8ix}-1}\cdots\frac{2i.e^{2^{n}ix}}{ e^{2^{n+1}ix}-1}}$

$2i\displaystyle{(\frac{e^{ix}}{ e^{2ix}-1}+\frac{e^{2ix}}{ e^{4ix}-1}+\frac{e^{4ix}}{ e^{8ix}-1}\cdots\frac{e^{2^{n}ix}}{ e^{2^{n+1}ix}-1})}$

$2i\displaystyle{(\frac{e^{ix}+1-1}{ (e^{ix}-1)(e^{ix}+1)}+\frac{e^{2ix}+1-1}{ (e^{2ix}-1)(e^{2ix}+1)}+\frac{e^{4ix}-1+1}{ (e^{4ix}-1)(e^{4ix}+1)}\cdots\frac{e^{2^{n}ix}+1-1}{ (e^{2^{n}ix}-1)(e^{2^{n}ix}+1)})}$

$2i((\frac{1}{e^{ix}-1}-\frac{1}{e^{2ix}-1})+(\frac{1}{e^{2ix}-1}-\frac{1}{e^{4ix}-1})+(\frac{1}{e^{4ix}-1}-\frac{1}{e^{8ix}-1})\cdots (\frac{1}{e^{2^{n}ix}-1}-\frac{1}{e^{2^{n+1}ix}-1}))$

which at last simplifies to $$2i((\frac{1}{e^{ix}-1}-\frac{1}{e^{2^{n+1}ix}-1}))$$

Now you may take it forward

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