f(x)=\frac{1}{2}\sin{(2x)}(\cos{2x}-cos{4x})=\frac{1}{4}\sin{4x}-\frac{1}{2}\sin{(2x)}\cos{(4x)}=\\
\frac{1}{4}\sin{4x}-\frac{1}{4}(\sin{6x}+\sin{2x})=\frac{1}{4}\sin{4x}-\frac{1}{4}\sin{6x}-\frac{1}{4}\sin{2x}