i've done that and sub in 1-cos^2x but i dont get the correct answer
$\displaystyle \cos (3x)=\cos(x+2x)=\cos x \cos 2x-\sin x \sin 2x$
$\displaystyle =\cos x ( 2 \cos^2 x -1) - \sin x (2 \sin x \cos x)$
$\displaystyle =2 \cos^3 x -\cos x - 2 \sin^2 x \cos x$