y=sinx*sin(x+三分之二π)且x属于【0.π】的值域
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发布时间：2024-10-23 22:35
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热心网友
时间：2024-11-07 07:33
解：∵y=sinx*sin(x+2π/3)
=[cos(2π/3)-cos(2x+2π/3)]/2
(
积化和差公式
)
=[-1/2-cos(2x+2π/3)]/2
∴2y=-1/2-cos(2x+2π/3)
==>2y+1/2=-cos(2x+2π/3)
∵x属于[0，π]
==>2x+2π/3属于[2π/3,8π/3]
∴│-cos(2x+2π/3)│≤1
==>│2y+1/2│≤1
==>-1≤2y+1/2≤1
==>-3/2≤2y≤1/2
==>-3/4≤y≤1/4
故函数y=sinx*sin(x+2π/3)的
值域
是[-3/4，1/4]。
