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For $k=0,\cdots, m$ and $l=0,\cdots 2m+1$ let us put $$ \alpha_{kl}=\frac{4k-4l+1}{4(m+1)}\pi\quad \beta_{kl}=\frac{4k+4l+3}{4(m+1)}\pi $$

and

$$x_{kl}=\frac{1}{4}\Big(\frac{1}{\sin\alpha_{kl}}+\frac{1}{\sin\beta_{kl}} \Big)$$

$$y_{kl}=-\frac{\sqrt{2}}{4}\Big(\frac{\sin(\frac{\pi}{4}-\alpha_{kl})}{\sin\alpha_{kl}}+\frac{\sin(\frac{3\pi}{4}-\beta_{kl})}{\sin(\beta_{kl})} \Big) $$

$$z_{kl}=\frac{1}{2}\Big(\frac{\sin\frac{\alpha_{kl}}{2}}{\sin\alpha_{kl}}-\frac{\sin\frac{\beta_{kl}}{2}}{\sin\beta_{kl}}\Big)$$

How can we show that $x_{kl}+y_{kl}+z_{kl}>0$ ?

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  • $\begingroup$ Having been on Math.SE for a while, you should know that the community prefers/expects a question to include something of what the asker knows (What have you tried? Where did you get stuck? etc.) This helps answerers avoid wasting time explaining things you already understand. ... That said, if my quick Mathematica calculation is correct, the sum simplifies rather nicely. $\endgroup$
    – Blue
    Commented Jan 31, 2023 at 14:47
  • $\begingroup$ To prove the invertibility of a principal submatrix of discrete sine transform matrix of type 1, I faced to this problem. I checked it by MATLAB but could not find any clear approach to get that. $\endgroup$
    – ABB
    Commented Jan 31, 2023 at 15:09
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    $\begingroup$ Well, I can tell you that the sum reduces to (ignoring subscripts) $$\frac{\sin\frac14(\alpha+\beta)}{4\sin\frac14(\pi-\alpha)\;\sin\frac14(\pi+\beta)}$$ Can you take it from there? $\endgroup$
    – Blue
    Commented Feb 1, 2023 at 2:50
  • $\begingroup$ Great, it is helpful. Please write details. $\endgroup$
    – ABB
    Commented Feb 1, 2023 at 6:21

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