\(\begin{aligned} \sin ^{4} x &=\left(\sin ^{2} x\right)^{2} \\ &=\left(\frac{1-\cos 2 x}{2}\right)^{2} \\ &=\frac{1-2 \cos 2 x+\cos ^{2} 2 x}{4} \\ &=\frac{1}{4}\left(1-2 \cos 2 x+\frac{1+\cos 4 x}{2...sin4x=(sin2x)2=(1−cos2x2)2=1−2cos2x+cos22x4=14(1−2cos2x+1+cos4x2) &=2\ sin x\ cos x+\ izquierda (\ cos ^ {2} x-\ sin ^ {2} x\ sin ^ {2} x\ derecha)\ cos x- (2\ sin x\ cos x)\ sin x\\ &=2\ sin x\ cos x+\ cos ^ {3} x-\ sin ^ {2} x\ cos x-2\ sin ^ {2} x\ cos x\
