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y=cos^2x-1/2sinx+1 求值域
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y=cos^2x-1/2sinx+1 求值域
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答案和解析
假定为y = cos²x - (1/2)sinx + 1
y = 1 - sin²x - (1/2)sinx + 1
= - sin²x - (1/2)sinx + 2
= -[sin²x + (1/2)sinx + 1/16 - 1/16] + 2
= -(sinx + 1/4)² + 33/16
sinx = -1/4时,y取最大值33/16
sinx = 1时,y取最小值1/2
值域：[1/2,33/16]
y = 1 - sin²x - (1/2)sinx + 1
= - sin²x - (1/2)sinx + 2
= -[sin²x + (1/2)sinx + 1/16 - 1/16] + 2
= -(sinx + 1/4)² + 33/16
sinx = -1/4时,y取最大值33/16
sinx = 1时,y取最小值1/2
值域：[1/2,33/16]
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