已知向量a=(√3,cosωx),b=(sinωx,1)(ω>0)函数f(x)=aXb,且最小正周期为4π.
1.设α,β∈[π/2,π],f(2α-π/3)=6/5,f(2β+2π/3)=-24/13,求sin(α+β)的值
解析：∵向量a=(√3,cosωx),b=(sinωx,1)(ω>0)
∴函数f(x)=a·b=√3sinωx+ cosωx=2sin(ωx+π/6)
∵最小正周期为4π
∴f(x)=2sin(1/2x+π/6)
设α,β∈[π/2,π],
f(2α-π/3)= 2sin(α)=6/5==>sinα=3/5==>cosα=-4/5,
f(2β+2π/3)=2sin(β+π/2)=-24/13==>cosβ=-12/13==>sinβ=5/13
sin(α+β)=sinαcosβ+cosαsinβ=-36/65-20/65=-56/65