Inegalitate sin pi/2n>=1/n

[Tudorel Lupu]

Sa se arate ca daca \( n\in\mathbb{N}^{*} \) atunci are loc:

\( \sin\frac{\pi}{2n}\geq\frac{1}{n} \).

Marius Cavachi, Olimpiada judeteana Constanta, 1993

[Wizzy]

Sa luam numarul complex :

\( \omega=\cos\frac {\pi}{n}+i\sin\frac{\pi}{n} \) unde \( \omega ^n=-1 \) adica \( \omega ^n-1=-2 \)

Aplicand modul la ultima relatie,avem :

\( 2=|\omega^n-1|=|\omega-1| |\omega^{n-1}+\omega^{n-2}+...+\omega+1| \)

Din inegalitatea modulelor obtinem :

\( 2\leq n|\omega-1| \) , deci \( |\omega-1|\geq \frac{2}{n} \)

Dar \( \omega-1=\cos\frac {\pi}{n}-1+i\sin\frac{\pi}{n}=-2\sin^2 \frac{\pi}{2n}+2i \sin\frac{\pi}{2n}\cos\frac{\pi}{2n}=2i \sin\frac{\pi}{2n}(\cos\frac{\pi}{2n}+i\sin\frac{\pi}{2n}) \)

Asadar : \( |\omega-1|=2\sin\frac{\pi}{2n}\geq \frac{2}{n} \)


\( \sin\frac{\pi}{2n}\geq \frac{1}{n} \)