Question

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

Answer 1

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