①化简[sin(2π+α)*cos(7π-α)]/cos(-α)②已知cosθ=-3/5,θ∈(π/2,π),求sin(θ+π/3)的值
问题描述：
①化简[sin(2π+α)*cos(7π-α)]/cos(-α)②已知cosθ=-3/5,θ∈(π/2,π),求sin(θ+π/3)的值
答
1.
= -sin(alpha)*cos(alpha)/cos(alpha)
= -sin(alpha)
2.
sin(θ+π/3)
= sin(θ)*cos(π/3)+cos(θ)*sin(π/3)
= 4/5*1/2-3/5*sqrt(3)/2
= (4-3*sqrt(3))/10
= -0,1196152
答
①cos(7π-α)=-cosα,cos(-α)=cosα,sin(2π+α)=sinα,所以整个式子化简得-sinα②根据cos²θ+sin²θ=1,cosθ=-3/5,得sinθ=±4/5；又θ∈(π/2,π),得sinθ=4/5；又sin(A+B)=sinAcosB+cosAsinB,得sin(...