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√3sin(a+π/6)=-4/5cos(a+π/6)=√[1-(-4/5)^2]=3/5sin(a+π/6)=sinacos(π/6)+cosasin(π/6)=√3/2sina+1/2co...2sin[(a+π/3+a)/2]*cos[(a+π/3-a)/2]怎么出來的根據(jù)和差化積公式：sinθ+sinφ = 2sin[(θ+φ)/2] cos[(θ-φ)/2]=√3sin(a+π/6)=-4/5√3是不是應該是√3sin(a+π/6)=-4√3/5