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How to find the equations whose roots are equal to the following numbers ?

(a) $\sin^2\frac{\pi}{2n+1}$,$\sin^2\frac{2\pi}{2n+1}$,$\sin^2\frac{3\pi}{2n+1}$,...,$\sin^2\frac{n\pi}{2n+1}$

(b)$\cot^2\frac{\pi}{2n+1}$,$\cot^2\frac{2\pi}{2n+1}$,$\cot^2\frac{3\pi}{2n+1}$,...,$\cot^2\frac{n\pi}{2n+1}$

I got stuck while solving this problem.This was from a complex number chapter practice problem exercise. I tried to fit in Demoivre's theorem considering roots of unity.But no idea how to bring about the squared sine terms. Any suggestions?

  • neglecting this solution $$\prod_{k=1}^{n}\left(x-\sin^2\left(\frac{k\pi}{2n+1}\right)\right)$$ – Chinny84 Sep 14 '15 at 10:22
  • Did'nt get you @Chinny84 –  Sep 14 '15 at 10:27
  • since I know that an equation with roots $a_0,a_1$ and $a_2$ can be given by $$(x-a_0)(x-a_1)(x-a_2)=\prod_{i=0}^2\left(x-a_i\right)$$. This is just a "trivial" solution. – Chinny84 Sep 14 '15 at 10:31
  • Isnt that a bit too trivial :/ –  Sep 14 '15 at 10:33
  • yup, thats why I said neglecting this solution ;). – Chinny84 Sep 14 '15 at 10:33
  • Maybe a more constructive comment would be to know that $$\sin^2 x = \frac{1-\cos 2x}{2}$$ thus you are look for roots, $r_k$ of the form $$r_k = \frac{1-\cos\left(\frac{2k\pi}{2n+1}\right)}{2}$$ – Chinny84 Sep 14 '15 at 12:56

1 Answers1

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For $(b),$ see Trig sum: $\tan ^21^\circ+\tan ^22^\circ+...+\tan^2 89^\circ = ?$

For $(a),$

Method $\#1:$ $$\cot^2y=\dfrac{1-\sin^2y}{\sin^2y}$$

Method $\#2:$

Using De Moivre's formula,

$$\cos(2m+1)x+i\sin(2m+1)x=(\cos y+i\sin y)^{2m+1}=\cdots$$

Equating the imaginary parts (See this),

$$\sin(2m+1)x=(2n+1)\sin x-\cdots+(-1)^{m-1}2^{2m-2}(2m+1)\sin^{2m-1}x+(-1)^m2^{2m}\sin^{2m+1}x$$

Now if $\sin(2m+1)x=0,(2m+1)x=r\pi$ where $r$ is any integer

$x=\dfrac{r\pi}{2m+1}$ where $r\equiv0,\pm1,\pm2,\cdots,\pm m\pmod{2m+1}$

Can you take it from here?

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