✀已知函数f(x)=(sinx+cosx)^2-2cos^2x 求f(x)最小正周期和单调递增区间

2024-10-31 16:28:41
推荐回答(1个)
回答1:

f(x)=(sinx+cosx)²-2cos²x=sin²x+cos²x+2sinxcosx-2cos²x=sin2x+1-2cos²x=sin2x-cos2x=√2sin(2x-π/4)
∴f(x)最小正周期T=2π/2=π
∵y=sinx的单调递增区间为[-π/2+2kπ,π/2+2kπ],(K∈Z)
∴-π/2+2kπ≤2x-π/4≤π/2+2kπ
∴-π/8+kπ≤x≤3π/8+kπ
∴此函数的单调递增区间为[-π/8+kπ,3π/8+kπ],(K∈Z)