求证:cos^8x-sin^8x-cos2x = 1/8(cos6x - cos2x)

cos^8x-sin^8x-cos2x
=(cos^4x+sin^4x)(cos^2x+sin^2x)(cos^2x-sin^2x)-cos2x
=(cos^4x+sin^4x)*1*cos2x-cos2x
=[(cos^2x+sin^2x)^2-2(sinxcosx)^2]cos2x-cos2x
=[1-2(sinxcosx)^2]cos2x-cos2x
=[1-2(sinxcosx)^2-1]cos2x
=-(2sinxcosx)^2/2*cos2x
=-sin^2 2x/2*co2x
=-(1+cos4x)*cos2x/4
1/8(cos6x - cos2x)
=1/8*(-2)*sin(8x/2)sin(4x/2)
=-1/4sin4xsin2x
=-1/4*(2sin2xcos2x)sin2x
=-1/4*2(sin2x)^2cos2x
=-1/4(1+cos4x)cos2x
所以cos^8x-sin^8x-cos2x = 1/8(cos6x - cos2x)