sin (a+兀/3)+sina=-4根号3/5,求cos a

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sin (a+兀/3)+sina=-4根号3/5,求cos a
-兀/2

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√3
sin(a+π/6)=-4/5
cos(a+π/6)=√[1-(-4/5)^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=3/5 (2)
(1)+√3* (2),得
cosa=(3√3-4)/10
再问： 2sin[(a+π/3+a)/2]*cos[(a+π/3-a)/2] 怎么出来的
再答：根据和差化积公式：sinθ+sinφ = 2sin[(θ+φ)/2] cos[(θ-φ)/2]
再问： =√3sin(a+π/6)=-4/5√3 是不是应该是√3sin(a+π/6)=-4√3/5

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