Giải thích các bước giải:
$\sin^8x+\cos^8x$
$=(\sin^4x)^2+2\sin^4x\cos^4x+(\cos^4x)^2-2\sin^4x\cos^4x$
$=(\sin^4x+\cos^4x)^2-\dfrac{\sin^42x}{8}$
$=(\sin^4x+2\sin^2x\cos^2x+\cos^4x-2\sin^2x\cos^2x)^2-\dfrac{\sin^42x}{8}$
$=\bigg{[}(\sin^2x+\cos^2x)^2-\dfrac{\sin^22x}{2}\bigg{]}^2-\dfrac{\left(\dfrac{1-\cos4x}{2}\right)^2}{8}$
$=\left(1-\dfrac{1-\cos4x}{4}\right)^2-\dfrac{(1-\cos4x)^2}{32}$
$=\left(\dfrac{3+\cos4x}{4}\right)^2-\dfrac{(1-\cos4x)^2}{32}$
$=\dfrac{(3+\cos4x)^2}{16}-\dfrac{(1-\cos4x)^2}{32}$
$=\dfrac{9+6\cos4x+\cos^24x}{16}-\dfrac{1-2\cos4x+\cos^24x}{32}$
$=\dfrac{18+12\cos4x+2\cos^24x-1+2\cos4x-\cos^24x}{32}$
$=\dfrac{17+14\cos4x+\cos^24x}{32}$
$=\dfrac{17+14\cos4x+\dfrac{1+\cos8x}{2}}{32}$
$=\dfrac{\dfrac{35}{2}+14\cos4x+\dfrac{\cos8x}{2}}{32}$
$=\dfrac{35}{64}+\dfrac{7}{16}\cos4x+\dfrac{1}{64}\cos8x$
Vậy $\sin^8x+\cos^8x=\dfrac{35}{64}+\dfrac{7}{16}\cos4x+\dfrac{1}{64}\cos8x$ (Đpcm).
