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I have these two conditions for $0<a<2\pi$ and $b>0$ and real. $$ \sin \left(\frac{\pi a}{2 (\pi -a)}\right)=\frac{a }{2 \pi -a}\;\sin \left(\frac{\pi (a-2 \pi )}{2 (a-\pi )}\right)+\frac{4 b }{2 \pi -a}\;\sin \left(\frac{\pi ^2}{2 (\pi -a)}\right) $$ and $$ \cos \left(\frac{\pi a}{2 (\pi -a)}\right)=\frac{a}{2 \pi -a} \;\cos \left(\frac{\pi (a-2 \pi )}{2 (a-\pi )}\right)\qquad\qquad\qquad \qquad\quad\qquad$$

As I check them numerically, I see that these two conditions may not be fulfilled simultaneously; Are there any hopes to prove this analytically?

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    $\begingroup$ Hint: $$\frac{\pi a}{2(\pi - a)} + \frac\pi2 \;=\; \frac{\pi^2}{2(\pi - a)}\;=\;\frac{\pi (a - 2 \pi)}{2(a - \pi)} - \frac\pi2$$ $\endgroup$
    – Blue
    Commented Apr 7, 2022 at 11:28
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    $\begingroup$ @Blue Thank you. $\endgroup$
    – MsMath
    Commented Apr 7, 2022 at 12:11
  • $\begingroup$ @Blue Thank you again, it worked for this problem, I could prove it using your hint :). $\endgroup$
    – MsMath
    Commented Apr 7, 2022 at 14:47
  • $\begingroup$ Congratulations! :) Now, post your solution as an answer (and accept it), so that we can all up-vote your success. (Plus, doing so will get your question out of the Unanswered queue.) $\endgroup$
    – Blue
    Commented Apr 7, 2022 at 15:21
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    $\begingroup$ @Blue Thank you for your comment. Then, I will do it (add my answer) as soon as I find some time. $\endgroup$
    – MsMath
    Commented Apr 7, 2022 at 21:18

1 Answer 1

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By the hint given by @Blue , considering $x:=\frac{\pi ^2}{2 (\pi -a)}$ my equations will be $$ \sin \left( x-\frac\pi2\right)=\frac{a }{2 \pi -a}\;\sin \left(x+\frac\pi2\right)+\frac{4 b }{2 \pi -a}\;\sin \left(x\right) $$ $$ \cos \left(x-\frac\pi2\right)=\frac{a}{2 \pi -a} \;\cos \left(x+\frac\pi2\right)\qquad\qquad\qquad $$

Further, they can be simplified as

$$ -\cos x=\frac{a }{2 \pi -a}\;\cos x+\frac{4 b }{2 \pi -a}\;\sin \left(x\right) \qquad (1) $$ $$ \sin x=\frac{-a}{2 \pi -a} \;\sin x\qquad\qquad\qquad \qquad (2) $$

The equation $(2)$ may be fulfilled either by $\sin x=0$ or $\frac{-a}{2 \pi -a} =1$ where the latter is not valid. Investigating then $(1)$ for $\sin x=0$ in which $\cos x=\pm1$, we get $$ \mp1=\frac{a }{2 \pi -a} ,$$ which never holds which completes the proof.

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    $\begingroup$ +1. Looks good to me! (Thanks for the sense of satisfaction. :) ... Interestingly enough, I initially did the same case analysis as you (either $\sin x = 0$ or we get something invalid) and then realized that I could/should have simplified things further. Eg, moving things in equation $(2)$ to one side and clearing fractions gives $2\pi\sin x=0$, so that $\sin x=0$ without having to consider the alternative. Likewise, the first equation reduces to, say, $-2\pi\cos x=4b\sin x$, so that if $\sin x=0$, then $\cos x=0$ as well, giving the contradiction. But no matter; the argument works as-is. $\endgroup$
    – Blue
    Commented Apr 8, 2022 at 13:12

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