MykolasD PRO +2154
Apskaičiuokite :[tex]\frac{3\sin ^{2}\beta +\sin ^{2}\alpha +\sin ^{2}\gamma }{2\sin ^{2}\beta -\sin \alpha \sin \gamma }[/tex][tex],[/tex] kai [tex]\sin \alpha +\sin \beta +\sin \gamma = 0[/tex].
Tomas PRO +4529
Apskaičiuokite [tex]:\frac{3\sin^2β+\sin^2α+\sin^2γ}{2\sin^2β−\sinα\sinγ},[/tex] kai [tex]\sinα+\sinβ+\sinγ=0.[/tex]
Sprendimas:
[tex]\sinα+\sinβ+\sinγ=0\implies \sinα+\sinγ=-\sinβ\implies \\(\sinα+\sinγ)^2=(-\sinβ)^2\implies \sin^2α+2\sinα\sinγ+\sin^2γ=\sin^2β[/tex]
[tex]\frac{3\sin^2β+\sin^2α+\sin^2γ}{2\sin^2β−\sinα\sinγ}=\frac{3(\sin^2α+2\sinα\sinγ+\sin^2γ)+\sin^2α+\sin^2γ}{2(\sin^2α+2\sinα\sinγ+\sin^2γ)−\sinα\sinγ}=\frac{3\sin^2α+6\sinα\sinγ+3\sin^2γ+\sin^2α+\sin^2γ}{2\sin^2α+4\sinα\sinγ+2\sin^2γ−\sinα\sinγ}=\frac{4\sin^2α+6\sinα\sinγ+4\sin^2γ}{2\sin^2α+3\sinα\sinγ+2\sin^2γ}=\frac{2(2\sin^2α+3\sinα\sinγ+2\sin^2γ)}{2\sin^2α+3\sinα\sinγ+2\sin^2γ}=2.[/tex]