关於正弦函数的值域

π/12<=x<=π/2
π/12+π/6<=x+π/6<=π/2+π/6
即
π/4<=x+π/6<=2π/3
π/4<=x+π/6<π/2时是增函数
π/2<=x+π/6<2π/3时是减函数
所以x+π/6=π/2,sin(x+π/6)最大=1
sinπ/4<sin2π/3
所以sin(x+π/6)最小=√2/2

再乘2
所以值域[√2,2]

x∈[π/12,π/2]
则
π/12+π/6<=x+π/6<=π/2+π/6

则对2sin(x+π/6)
x=π/3时，取得最大值，2sin(π/3+π/6)=2sinπ/2=2
当x=π/12时，取得最小值，2sin(π/12+π/6)=2sinπ/4=√2

【根号2，2】

x∈[π/12,π/2]
x+π∈[π/4,2π/3]
根据正弦图像，sin(x+π/6),此区间内有最大值1，最小值是当x=π/4时=根号2/2
f(x) ∈[根号2,2]
ps：sorry,不会打根号