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Subiectul III Gheorghe Lazar 2009 Sibiu
Posted: Sun Mar 22, 2009 8:58 pm
by DrAGos Calinescu
Sa se demonstreze egalitatea:
\( \frac{1}{\cos a}+2\frac{\cos a}{\cos 2a}+4\frac{\cos 3a}{\cos 4a}+8\frac{\cos 7a}{\cos 8a}+16\frac{\cos 15a}{\cos 16a}=0 \) , unde \( a=\frac{\pi}{31} \)
GM 12/2008
Posted: Tue Mar 24, 2009 4:50 pm
by maxim bogdan
Sa enuntam cazul general al problemei:
\( \bullet \) Daca \( a=\frac{\pi}{2^{n+1} -1} \) atunci are loc identitatea:
\(
\frac{1}{\cos a}+2\cdot\frac{\cos a}{\cos 2a}+4\cdot\frac{\cos 3a}{\cos 4a}+\dots +2^{n}\cdot\frac{\cos \left (2^{n}-1\right )a}{\cos 2^{n}a}=0.
\)
Se foloseste lema:
Lema. Identitatea:
\(
\frac{1}{\cos x}+2\cdot\frac{\cos x}{\cos 2x}+4\cdot\frac{\cos 3x}{\cos 4x}+\dots+2^{n}\cdot\frac{\cos \left (2^{n}-1\right )x}{\cos 2^n x}=2^{n+1}\cdot\frac{\sin\left (2^{n+1}-1\right )x}{\sin 2^{n+1}x},
\)
are loc pentru orice \( x \) (care respecta conditiile de existenta) si \( n\in\mathbb{N}^*. \)
Demonstratia e imediata prin inductie dupa \( n \) folosind formule de manual.
Aplicand lema (pentru \( x:=a \)) ramane de aratat ca:
\( 2^{n+1}\cdot\frac{\sin(2^{n+1}-1)\cdot\frac{\pi}{2^{n+1}-1}}{\sin 2^{n+1}x}=0 \), care este evident adevarata deoarece \( \sin\pi=0. \)