Question

已知ω〉0且函数f(x)=cos^2ωx-sin^2ωx的最小正周期为π,则f(x)在[π/3,5π/6]上的最大值为_

Answer 1

f(x)=cos^2ωx-sin^2ωx

=cos2wx,

T=2π/2w=π的哦w=1,

所以f(x)=cos2x,

x∈[π/3,5π/6],

2x∈[2π/3,5π/3],

f（x）max=cos5π/3=cosπ/3=1/2.