Question

sin（α+π/3）+sinα= -4根号3分之5，-π/2＜α＜0，则cosα=?

Answer 1

sin(a+π/3)+sina=-4√3/5
1/2sina+√3/2cosa+sina=-4√3/5
3/2sina+√3/2cosa=-4√3/5
√3/2sina+1/2cosa=-4/5
cos(a-π/3)=-4/5
-π/2<a<0
-5π/6<a-π/3<-π/3
sin(a-π/3)=-3/5
cosa=cos(a-π/3+π/3)=cos(a-π/3)cos(π/3)-sin(a-π/3)sin(π/3)
=(-4/5)×(1/2)+(3/5)×(√3/2)
=(3√3-4)/10

Answer 2

sin（α+π/3）+sinα= -4根号3分之5
sin(a+π/3)+sina=-4√3/5
1/2sina+√3/2cosa+sina=-4√3/5
3/2sina+√3/2cosa=-4√3/5
√3/2sina+1/2cosa=-4/5
cos(a-π/3)=-4/5
-π/2<a<0
-5π/6<a-π/3<-π/3
sin(a-π/3)=-3/5
cosa=cos(a-π/3+π/3)=cos(a-π/3)cos(π/3)-sin(a-π/3)sin(π/3)
=(-4/5)×(1/2)+(3/5)×(√3/2)
=(3√3-4)/10