y=Sin(π/4-wx)且(W>0) Y最小正周期为π,求单调递增区间?若X∈【0,π/2】求函数值域?
问题描述:
y=Sin(π/4-wx)且(W>0) Y最小正周期为π,求单调递增区间?若X∈【0,π/2】求函数值域?
答
①最小正周期T=2π/w=π,所以w=2,即y=sin(π/4-2x)②y=sin(π/4-2x)=-sin(2x-π/4)令t=2x-π/4,则2kπ+π/2≤t≤2kπ+3π/2为增区间,从而求出kπ+3π/8≤x≤kπ+7π/8③因为x∈[0,π/2],所以(π/4-2x)∈[-3π/4,π/4...