The following question is from a previous post. The reason that I am posting this again is because it had two questions which were not really related and hence, one of the questions was not answered. The question is from a book, "Mathematical Analysis - 2nd Edition" by Apostol. The question has three parts and is as follows:

  1. By equating imaginary parts in DeMoivre's formula, prove that $$\sin{\left(n\theta\right)} = \sin^n\theta \left\lbrace \binom{n}{1}\cot^{n - 1}\theta - \binom{n}{3} \cot^{n - 3}\theta + \binom{n}{5} \cot^{n - 5}\theta - + \cdots \right\rbrace$$

  1. If $0 < \theta < \dfrac{\pi}{2}$, prove that $$\sin{\left(\left( 2m + 1 \right)\theta\right)} = \sin^{2m + 1}\theta P_m(\cot^2\theta)$$ where $P_m$ is the polynomial of degree $m$ given by $$P_m\left( x \right) = \binom{2m + 1}{1}x^m - \binom{2m + 1}{3}x^{m-1} + \binom{2m + 1}{5}x^{m-2} - + \cdots$$ Use this to show that $P_m$ has zeros at $m$ distinct points $x_k = \cot^2\left( \dfrac{\pi k}{2m + 1} \right)$ for $k = 1, 2, \dots, m$.

  1. Show that the sum of zeros of $P_m$ is given by $$\sum\limits_{k=1}^{m} \cot^2\dfrac{\pi k}{2m + 1} = \dfrac{m \left( 2m - 1 \right)}{3}$$ and that the sum of theie squares is given by $$\sum\limits_{k=1}^{m} \cot^4\dfrac{\pi k}{2m + 1} = \dfrac{m \left( 2m - 1 \right) \left( 4m^2 + 10m - 9 \right)}{45}$$

As far as the solution is concerned, I have been able to prove the first part and upto proving the existence of the polynomial in the second part. For later parts, I do not have any insights in proceeding towards the solution.

Help will be appreciated!



For part 2, since you already showed the existence of the polynomial s.t. $$\sin \left( 2m + 1 \right)\theta = \left(\sin^{2m + 1}\theta \right)P_m(\cot^2\theta)$$ replace $\theta$ with $\frac{k\pi}{2m+1}$ and $$\left(\sin^{2m + 1}\frac{k\pi}{2m+1} \right)\color{red}{P_m\left(\cot^2\frac{k\pi}{2m+1}\right)}=\sin \left(\left( 2m + 1 \right)\frac{k\pi}{2m+1}\right)=\sin{k\pi}=\color{red}{0}$$

For part 3, have you tried using Vieta's formulas? It is a simple application of $$\color{red}{\sum\limits_{k=1}^{m}x_k}=\sum\limits_{k=1}^{m}\cot^2\frac{\pi k}{2m+1}=\color{red}{-\frac{a_{m-1}}{a_m}}=-\frac{-\binom{2m+1}{3}}{\binom{2m+1}{1}}=\\ \frac{(2m+1)2m(2m-1)}{1\cdot2\cdot3}\cdot \frac{1}{2m+1}=\frac{m(2m-1)}{3}$$ For the 2nd part of $3$ use the fact that $$\sum\limits_{k=1}^{m}x_k^2=\left(\sum\limits_{k=1}^{m}x_k\right)^2-2\sum\limits_{k<t}x_kx_t=\left(-\frac{a_{m-1}}{a_m}\right)^2-2\left(\frac{a_{m-2}}{a_m}\right)$$

  • $\begingroup$ What about the second part? How will I show that the roots are what we have to show? $\endgroup$ – Aniruddha Deshmukh Feb 18 '18 at 10:52
  • $\begingroup$ @AniruddhaDeshmukh is it better now? $\endgroup$ – rtybase Feb 18 '18 at 11:18
  • 1
    $\begingroup$ Yes! It was pretty simple. I could not see it, though! $\endgroup$ – Aniruddha Deshmukh Feb 18 '18 at 11:54

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