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Aratati ca (1-sinα)x²-2xcosα+1+sinα≥0 , ∀ x,d ∈ R

Răspuns :

Avem [tex]\sin \alpha=2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}[/tex]
și [tex]\cos\alpha=\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}[/tex]
și [tex]\sin^2\frac{\alpha}{2}+\cos^2\frac{\alpha}{2}=1[/tex]
Atunci expresia se scrie
[tex]\left(\sin\frac{\alpha}{2}-\cos\frac{\alpha}{2}\right)^2x^2-2\left(\cos^\frac{\alpha}{2}-\sin^\frac{\alpha}{2}\right)+\left(\sin\frac{\alpha}{2}+\cos\frac{\alpha}{2}\right)^2[/tex]
care se restrânge la
[tex]\left(\left(\sin\frac{\alpha}{2}-\cos\frac{\alpha}{2}\right)x+\left(\sin\frac{\alpha}{2}+\cos\frac{\alpha}{2}\right)\right)^2\geq 0[/tex]