Giải thích các bước giải:
$\cos^6x-\sin^6x$
$=(\cos^2x)^3-(\sin^2x)^3$
$=(\cos^2x-\sin^2x)(\cos^4x+\cos^2x\sin^2x+\sin^4x)$
$=\cos2x.(\cos^4x+2\cos^2x\sin^2x+\sin^4x-\cos^2x\sin^2x)$
$=\cos2x.[(\cos^x+\sin^2x)^2-\cos^2x\sin^2x]$
$=\cos2x.(1-\cos^2x\sin^2x)$
$=\cos2x.\left(1-\dfrac{\sin^22x}{4}\right))$
$=\cos2x.\left((1-\dfrac{1-\cos4x}{8}\right)$
$=\cos2x-\dfrac{\cos2x(1-\cos4x)}{8}$
$=\cos2x-\dfrac{\cos2x-\cos2x.\cos4x}{8}$
$=\cos2x-\dfrac{\cos2x-\dfrac{1}{2}(\cos6x+\cos2x)}{8}$
$=\cos2x-\dfrac{\dfrac{1}{2}\cos2x-\dfrac{1}{2}\cos6x}{8}$
$=\cos2x-\dfrac{1}{16}\cos2x+\dfrac{1}{16}\cos6x$
$=\dfrac{15}{16}\cos2x+\dfrac{1}{16}\cos6x$
Vậy $\cos^6x-\sin^6x=\dfrac{15}{16}\cos2x+\dfrac{1}{16}\cos6x$ (Đpcm).
