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

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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
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

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
①cos(7π-α)=-cosα,cos(-α)=cosα,sin(2π+α)=sinα,所以整个式子化简得-sinα
②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(θ+π/3)=sinθ*1/2+cosθ*根3/2=(4-3倍根3)/10

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