y=sin^2 x+3sinx-1 2.y=-cos^2 x+sinx+3,求下列函数值域
来源:百度知道 编辑:UC知道 时间:2024/06/12 20:51:51
(1)y=sin^2 x+3sinx+9/4-9/4-1
=(sinx+3/2)^2-13/4
因为sinx在[-1,1]之间,由二次函数图像得
[-3,3]
(2)y=-(1-sin^2 x)+sinx+3
=sin^2 x+sinx+2
=sin^2 x+sinx+1/4-1/4+2
=(sinx+1/2)^2+7/4
因为sinx在[-1,1],由图像得[7/4,4]
1.[-3,3]
2.[7/4,4]
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