求函数y=cos2x+2sinx-2值域.
问题描述:
求函数y=cos2x+2sinx-2值域.
答
∵y=cos2x+2sinx-2
=1-sin2x+2sinx-2
=-(sinx-1)2,
∵-1≤sin≤1,
∴-2≤sin-1≤0,
∴(sinx-1)2∈[0,4],-(sinx-1)2∈[-4,0].
∴函数y=cos2x+2sinx-2值域为[-4,0].