题目

f(x)=2sinx(sinx-cosx)+1,求最小正周期和当x∈[π/8,3π/4]时f(x)的值域.

提问时间：2021-04-19

答案

f(x)=2sinx(sinx-cosx)+1,
=2sin²x-2sinxcosx+1
=1-cos2x-sin2x+1
=-(sin2x+cos2x)+2
= -√2sin(2x+π/4)+2
T=2π/2=π
x∈[π/8,3π/4]
2x+π/4∈[π/2,7π/4]
sin(2x+π/4)∈[-1,1]
所以 y∈【2-√2,2+√2】