若sin(π/2﹢α﹚=﹣4/5,α∈﹙π/2,π﹚,则cos﹙π/3﹣α﹚=?

问题描述:

若sin(π/2﹢α﹚=﹣4/5,α∈﹙π/2,π﹚,则cos﹙π/3﹣α﹚=?

sin(π/2+a)=cosa=-4/5,因a∈(π/2,π),则sina=3/5
cos(π/3-a)=cos(π/3)cosa+sin(π/3)sina=(1/2)×(-4/5)+(√3/2)×(3/5)=[-4+3√3]/10