1、求值域要具体步骤：y=cos ①x∈[π/6,π/3]②x∈[-π/6,2π/3]③x∈[-2π/3,4π/3]

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1、求值域要具体步骤：y=cos ①x∈[π/6,π/3]②x∈[-π/6,2π/3]③x∈[-2π/3,4π/3]
2、y=-2cos（2x-π/4）+1,x∈[0,π/2]的值域

由y=cosx,
①当x∈[π/6,π/3]时,单调减函数,
f（π/6）=√3/2,
f（π/3）=1/2
∴1/2≤y≤√3/2.
②x∈[-π/6,2π/3]时,
x=2π/3时最小值y=-1/2,
x=0时,最大值y=1
∴-1/2≤y≤1
③x∈[-2π/3,4π/3]时,
x=0,最大值y=1
x=π,最小值y=-1.
∴-1≤y≤1.
（2）由y=-2cos（2x-π/4）+1
x∈[0,π/2],
其中：周期T=2π/2=π,
初相：Θ=π/4,
振幅：A=2（取正）
2x-π/4=0,即x=π/8时,
y取最小值y=-1
2x-π/4=π/2,即x=π/8时,
最大值y1=1.

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