原式=(sin²x+cos²x)(sin^4x-sin²xcos²x+cos^4x)+2sin²xcos²x
=(sin^4x-sin²xcos²x+cos^4x)+2sin²xcos²x
=sin^4x+sin²xcos²x+cos^x
=(sin²x+cos²x)²-sin²xcos²x
=1-sin²xcos²x
=1-(1/4)sin²2x
=1-(1/4)[(1/2)(1-cos4x)]
=(7/8)+(1/8)cos4x
原式=(sin²x+cos²x)(sin^4x-sin²xcos²x+cos^4x)+2sin²xcos²x
=(sin^4x-sin²xcos²x+cos^4x)+2sin²xcos²x
=sin^4x+sin²xcos²x+cos^x
=(sin²x+cos²x)²-sin²xcos²x
=1-sin²xcos²x
=1-(1/4)sin²2x
=1-(1/4)[(1/2)(1-cos4x)]
=(7/8)+(1/8)cos4x