求y=cos2x+2sinx-3的值域,
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/09 00:30:50
![求y=cos2x+2sinx-3的值域,](/uploads/image/z/13297700-20-0.jpg?t=%E6%B1%82y%3Dcos2x%2B2sinx-3%E7%9A%84%E5%80%BC%E5%9F%9F%2C)
x){69بBۨ83B
{ΟcT_~
fTe`*ⶺFa]aogk
dhu@tOvzgHɣh]CX$:^
Tˁ@?7hHX<; Ày
求y=cos2x+2sinx-3的值域,
求y=cos2x+2sinx-3的值域,
求y=cos2x+2sinx-3的值域,
y=1-2sinx^2+2sinx-3
=-2[sinx^2-sinx+(1/2)^2]-2+1/2
=-2(sinx-1/2)^2-3/2
因为:sinx∈[-1,1]
所以:y=cos2x+2sinx-3的值域求得为[-6,-3/2]