根号6∕4sinx-根号2/4cosx,x属于（pie／6,2pie／3）,求值域
问题描述：
根号6∕4sinx-根号2/4cosx,x属于（pie／6,2pie／3）,求值域
答
∵y=(√6∕4)sinx-(√2/4)cosx
=(√2/4)[√3sinx-cosx]
=(√2/2)[(√3/2)sinx-(1/2)cosx]
=(√2/2)[sinπ/3sinx-(cosπ/3)cosx]
=(-√2/2)cos(x+π/3)
π/6≤x≤2π/3
∴π/2≤x+π/3≤π
∴-√2/2≤y≤0