已知函数f(x)=cos(x+π/4)cos(x-π/4)+2tsinxcosx-sin(x+π/4)sin(x-π/4)
问题描述：
已知函数f(x)=cos(x+π/4)cos(x-π/4)+2tsinxcosx-sin(x+π/4)sin(x-π/4)
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
答
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