<s>[INST] Any hint on how to prove that the two given conditions may not be fulfilled simultaneously? 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?
 [/INST] 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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