Ta có: \[{\sin ^4}x + {\cos ^4}x = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x = 1 - \frac{{{{\sin }^2}2x}}{2}\]
\[ = 1 - \frac{{\frac{{1 - \cos 4x}}{2}}}{2} = 1 - \frac{{1 - \cos 4x}}{4} = \frac{1}{4}\cos 4x + \frac{3}{4}\,\,\,\,\left( \text{đpcm} \right)\]
#crystal