Question

高一数学三角函数问题

已知函数f(x)=√3sin2x+2sin(x-π/4)sin(x+π/4),将f(x)化为y=sinx.
若不等式｜f(x)-m｜<2在 [-π/12,π/2]上恒成立,求实数m的取值范围.
(注:｜是绝对值符号)

Answer 1

（1）化：
f(x)=√3sin2x + cos(x-π/4-x-π/4) - cos(x-π/4+x+π/4)
=√3sin2x + 0 - cos2x
=2[cos(π/6)sin2x - sin(π/6)cos2x]
=2sin(2x-π/6)
(2)不等式可化为
m-2 < 2sin(2x-π/6) < m+2
由于 X取值在 [-π/12,π/2]
则 (2x-π/6)的取值在 [-π/3，5π/6]
2sin(2x-π/6)取值范围为 [-√3,2]
因此
m-2 < -√3 且 m+2>2
即 0<m<2-√3

Answer 2

f(x)=√3sin2x+2sin(x-π/4)sin(x+π/4)
=√3sin2x+2(sinxcosπ/4-cosxsinπ/4)(sinxcosπ/4+cosxsinπ/4)
=√3sin2x-cos2x
=2sin(2x-π/6)
当x∈[-π/12,π/2],(2x-π/6)∈[-π/3,5π/6],则f(x)∈[-√3,2]
于是-√3-m≤f(x)-m≤2-m
所以-√3-m>-2且2-m<2
于是0<m<2-√3

Answer 3

f(x)=√3sin2x+(√2sinx-√2cosx)*(√2/2sinx+√2/2cosx)
=√3sin2x-cos2x
=2sin(2x-π/6)
f(x)-m的绝对值小于2
即-2<f(x)-m<2
即m-2<f(x)<2+m
求出f(X)在 [-π/12,π/2]的范围即可求解