Biết rằng sin^4x+cos^4x = mcos4x + n(m,n thuộc Q). Tính tổng S=m + n

Ta có:

${\sin }^{4}x+{\cos }^{4}x={\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x$

$=1-2{\left(\sin x\cos x\right)}^2$

$=1-2{\left(\frac{1}{2}\sin 2x\right)}^2$

$=1-2.\frac{1}{4}{\sin }^{2}2x$

$=1-\frac{1}{2}{\sin }^{2}2x$

$=1-2.\frac{1}{4}{\sin }^{2}2x$

$=1-\frac{1}{2}.\frac{1-\cos 4x}{2}$

$=1-\frac{1}{4}.\left(1-\cos 4x\right)$

$=1-\frac{1}{4}+\frac{1}{4}\cos 4x$

$=\frac{1}{4}\cos 4x+\frac{3}{4}$