高一数学，帮忙解答

Question

如题所述

Answer 1

(1)f(x)=-cos(2ωx-π/6)+√3sin(2ω-π/6)
=2sin(2ωx-π/6-π/6)=2sin(2ωx-π/3)
∵T=2π/2ω=π
∴ω=1，f(x)=2sin(2x-π/3)
由-π/2+2kπ≤2x-π/3≤π/2+2kπ，k∈Z
得-π/12+kπ≤x≤5π/12+kπ，k∈Z
即f(x)的单调递增区间是[-π/12+kπ，5π/12+kπ],(k∈Z)。
(2)∵f(x0)=2sin(2x0-π/3)=6/5
∴f(x0)=sin(2x0-π/3)=3/5
∵x0∈[5π/12,11π/12]
∴2x0-π/3∈[π/2，3π/2]，cos(2x0-π/3)<0
cos(2x0-π/3)=-√[1-(3/5)^2]=-4/5
sin2x0=sin[(2x0-π/3)+π/3]=sin(2x0-π/3)cosπ/3+cos(2x0-π/3)sinπ/3
=(3/5)×(1/2)+(-4/5)×(√3/2)=(3-4√3)/10。

Answer 2

把sin给拆成1/2sin cos
再合并就可以了

Answer 3