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Dacă [tex]cos3x=1[/tex], x∈(0,[tex]\frac{\pi }{6}[/tex]) să se calculeze [tex]sin2x+sin4x[/tex].

Răspuns :

[tex]\cos3x=1\Rightarrow\cos^3x=1\\\sin^23x+\cos^23x=1\Rightarrow\sin^23x=1-\cos^23x=0\Rightarrow \sin3x=0\\\sin2x+\sin4x=\sin(3x-x)+\sin(3x+x)=\sin3x\cos x-\cos3x\sin x+\sin3x\cos x+\cos3x\sin x=2\sin3x\cos x=0[/tex]