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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.