Question

sin (a+兀/3)+sina=-4根号3/5,求cos a

-兀/2<a<0

Answer 1

sin (a+π/3)+sina
=2sin[(a+π/3+a)/2]*cos[(a+π/3-a)/2]
=2sin(a+π/6)cos(π/6)
=√3sin(a+π/6)=-4/5√樱圆3
sin(a+π/6)=-4/5
cos(a+π/6)=√[1-(-4/高茄5)^2]=3/5
sin(a+π/6)=sinacos(π/6)+cosasin(π/6)=√3/2sina+1/2cosa=-4/5 (1)
cos(a+π/6)=cosacos(π/6)-sinasin(π/6)=√3/2cosa-1/2sina=3/脊念塌5 (2)
(1)+√3* (2),得
cosa=(3√3-4)/10

追问

2sin[(a+π/3+a)/2]*cos[(a+π/3-a)/2]
怎么出来的

追答

根据和差化积公式：sinθ+sinφ = 2sin[(θ+φ)/2] cos[(θ-φ)/2]

追问

=√3sin(a+π/6)=-4/5√3
是不是应该是√3sin(a+π/6)=-4√3/5