函数y=sin（θ－π/2)cos(θ+π/2),θ∈[0,2π/3]的最小值

Question

函数y=sin（θ－π/2)cos(θ+π/2),θ∈[0,2π/3]的最小值

Answer 1

y=sin(θ－π/2)cos(θ+π/2)
=－sin(π/2-θ)cos(π/2+θ)
=cosθsinθ
=(sin2θ)/2
θ∈[0,2π/3],2θ∈[0,4π/3],－√2/2≤sin2θ≤1
所以 －√2/4≤y≤1/2
则y最小值为－√2/4