已知函数f(x)=cos(x+π/4)cos(x-π/4)+2tsinxcosx-sin(x+π/4)sin(x-π/4)

来源：百度知道 编辑：UC知道 时间：2024/06/10 19:54:24
1.当t=2时，求f(x)在[0,π/2]上的最大值和最小值
2.若函数f(x)在区间（π/12,π/6)上时增函数，求实数t的取值范围

感谢帮助

f(x)=cos(x+π/4+x-π/4)+tsin2x=cos2x+tsin2x

1.t=2时，f(x)=cos2x+2sin2x=√5sin(2x+arctan0.5)
在x∈[0,π/2]上，f(x)最大值为√5
f(x)最小值为√5sin(2*π/2+arctan0.5)=-1

2.f(x)=√(t^2+1)sin(2x+arctan(1/t))
f(x)在x∈[π/12,π/6]上为增函数
-π/2<=2x+arctan(1/t)<=π/2
-2π/3<=arctan(1/t)<π/6
-√3<=1/t<=1/√3
t>=√3或t<=-1/√3