题目

证明(cosα)的八次方-(sinα)的八次方-cos2α=-1/4sin2αsin4α

提问时间：2020-10-02

答案

证明：
左=(cosα)^8-(sinα)^8-cos2α
=[(cosα)^4+(sinα)^4][(cosα)^4-(sinα)^4]-cos2α
=[(cosα)^4+(sinα)^4][(cosα)^2-(sinα)^2][(cosα)^2+(sinα)^2]-cos2α
=cos2α[(cosα)^4+(sinα)^4]-cos2α
=cos2α{[cosα)^2+(sinα)^2]^2-2(sinα)^2(cosα)^2}-cos2α
=cos2α[1-(sin2α)^2/2]-cos2α
=cos2α-cos2α(sin2α)^2/2-cos2α
=-cos2α(sin2α)^2/2
=-cos2α[(1-cos4α)/2]/2
=-1/4*cos2α(1-cos4α)
=-1/4(cos2α-cos2αcos4α)
=-1/4[cos(4α-2α)-cos2αcos4α]
=-1/4[cos4αcos2α+sin4αsin2α-cos2αcos4α]
=-1/4sin2αsin4α
=右