Question

化简(1-cos^4x-sin^4x)/(1-cos^6x-sin^6x)

Answer 1

已知sin^2x+cos^2x=1 分子配方 1-sin^4x-cos^4x=1+2sin^2xcos^2x-(sin^2x+cos^2x)^2=1+2sin^2xcos^2x-1=2sin^2xcos^2x 分母先用立方和公式,再配方 1-sin^6-cos^6x=1-(sin^2x+cos^2x)*(sin^4x-sin^2xcos^2x+cos^4x)=1-(sin^4x-sin^2xcos^2x+cos^4x)=1+3sin^2xcos^2x-(sin^2x+cos^2x)^2=3sin^2xcos^2x 因此原式=2/3