y=sin^4x+cos^4x
=sin^4x+2sin^2xcos^2x+cos^4x-2sin^2xcos^2x
=(sin^2 2x+cos^2 2x)^2-(1/2)(2sinxcosx)^2
=1-(1/2)(sin2x)^2
=1-(1/2)(1-cos^2 2x)
=1-1/2*(1-2/9)
=1-7/18
=11/18sin^4x+2sin^2xcos^2x+cos^4x-2sin^2xcos^2x怎么得到=(sin^2 2x+cos^2 2x)^2-(1/2)(2sinxcosx)^2怎么X变成2X??sin^4x+2sin^2xcos^2x+cos^4x=[(sin^2（2x)+cos^2( 2x)]^2(1/2)(2sinxcosx)^2=1/2*4*sin^2xcos^2x=2sin^2xcos^2x明白了吗