第1个回答 2020-03-21
第2个回答 2020-03-21
20. f(x) = 4sin(x-π/6)cosx + 2 = 4[(√3/2)sinx - (1/2)cosx]cosx + 2
= 2√3sinxcosx - 2(cosx)^2 + 2 = √3sin2x - cos2x + 1= 2sin(2x-π/6) + 1
最小正周期 T = 2π/2 = π
f(-π/4) = 2sin(-π/2-π/6) + 1= -2sin(2π/3) + 1 = -√3+1
f(π/12) = 1
f(π/4) = 2sin(π/2-π/6) + 1= 2sin(π/3) + 1 = √3+1
在 [-π/4, π/4]内，最小值 f(-π/4) = -√3+1, 最大值 f(π/4) = √3+1。本回答被网友采纳
= 2√3sinxcosx - 2(cosx)^2 + 2 = √3sin2x - cos2x + 1= 2sin(2x-π/6) + 1
最小正周期 T = 2π/2 = π
f(-π/4) = 2sin(-π/2-π/6) + 1= -2sin(2π/3) + 1 = -√3+1
f(π/12) = 1
f(π/4) = 2sin(π/2-π/6) + 1= 2sin(π/3) + 1 = √3+1
在 [-π/4, π/4]内，最小值 f(-π/4) = -√3+1, 最大值 f(π/4) = √3+1。本回答被网友采纳
第3个回答 2020-03-21
简单，周期T=2pai除以奥秘嘎
相似回答