(cos2x-sin2x)/[(1-cos2x)(1-tan2x)] =cos2x/(1-cos2x)=[cosx)^2-(sin_数学解答

(cos2x-sin2x)/[(1-cos2x)(1-tan2x)] =cos2x/(1-cos2x)=[cosx)^2-(sinx)^2]/2(sinx)^2 这步是怎么换算的?

人气：205 ℃　时间：2019-10-10 05:57:51

解答

(cos2x-sin2x)/[(1-cos2x)(1-tan2x)]=cos2x[1-(sin2x/cos2x)]/[(1-cos2x)(1-tan2x)] （分母部分提出cos2x）=cos2x(1-tan2x)/[(1-cos2x)(1-tan2x)]=cos2x/(1-cos2x) （分子分母约去(1-tan2x)）=[(cosx)^2-(sinx)^2]/2...