Question

已知sin(α+π/3）+sina=-4√3/5,-π/2＜α＜0，则cos(α+2π/3）等于

Answer 1

sin(α+π/3）+sina
=（1/2）sinα+（√3/2）cosα+sina
=（3/2）sinα+（√3/2）cosα
=√3[（√3/2）sinα+（1/2）cosα]
=√3[sin(α+π/6)]
=-4√3/5
所以sin(α+π/6)=-4/5
cos(α+2π/3）
=cos(α+π-π/3）
=cos(π+α-π/3）
=-cos(α-π/3）
=-sin（π/2+α-π/3）
=-sin（α+π/6)
=-[-4/5]
=4/5

Answer 2

sin(α+π/3) + sina
= sinαcosπ/3+cosαsinπ/3+sinα
= 1/2sinα+√3/2cosα+sinα
= 3/2sinα+√3/2cosα
= √3(1/2sinα+1/2cosα)
= √3(sinαcosπ/6+cosαsinπ/6）
= √3sin(α+π/6) = -4√3/5
sin(α+π/6) = -4/5
cos(α+2π/3) = cos(α+π/6+π/2) = -sin(α+π/6) = 4/5

Answer 3