Question

已知sin( α+ π/3 )+ sinα =- 4√3 / 5,(- π/2 )<α<0,求cosα的值

Answer 1

- π/2<α<0

sin( α+ π/3 )+ sinα =- 4√3 / 5
sinαcosπ/3+cosαsinπ/3+ sinα =- 4√3 / 5
1/2sinα+√3/2*cosα+ sinα =- 4√3 / 5
3/2sinα+√3/2*cosα=- 4√3 / 5
√3/2sinα+1/2*cosα=- 4 / 5
√3/2sinα+1/2*cosα=- 4 / 5
sinπ/3sinα+cosπ/3cosα=- 4 / 5
cos(α-π/3)=- 4 / 5
- π/2<α<0
- 5π/6<α-π/3<-π/3
所以- 5π/6<α-π/3<-π/2
sin(α-π/3)=- 3/ 5

cosα
=cos[(α-π/3)+π/3]
=cos(α-π/3)cosπ/3-sin(α-π/3)sinπ/3
=cos(α-π/3)cosπ/3-sin(α-π/3)sinπ/3
=-4/5*1/2-(-3/5)*√3/2
=-2/5+3√3/10

Answer 2

sin(x+π/3)+sinx
=sinxcosπ/3+cosxsinπ/3+sinx
=1/2sinx+√3/2cosx+sinx
=3/2sinx+√3/2cosx=(-4√3)/5
∴√3sinx+cosx=-8/5
∴sinx=(-8/5 -cosx)/√3
∵(sinx)^2+(cosx)^2=1
∴[(-8/5 -cosx)/√3]^2+(cosx)^2=1
整理后：100(cosx)^2+80cosx-11=0
∴cosx=(-4+3√3)/10 或cosx=(-4-3√3)/10
∵(-π/2)<x<0
∴cosx=(-4+3√3)/10