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求函数y=2sin2x+sinx-cosx,当x属于中括号-π/2到π/2中括号的最大值和最小值
人气:180 ℃ 时间:2020-04-05 07:03:22
解答
y=2sin2x+sinx-cosx=2-2(1-sin2x)+sinx-cosx
=2-2(sin²x+cos²x-2sinxcosx)+(sinx-cosx)
=﹣2(sinx-cosx)²+(sinx-cosx)+2
=﹣2sin²(x-π/4)+sin(x-π/4)+2
=﹣2[sin(x-π/4)-1/4]²+17/8
∵x∈[﹣π/2,π/2] ∴x-π/4∈[﹣3π/4,π/4] ∴sin(x-π/4)∈[﹣√2/2,√2/2]
∴ 当sin(x-π/4)=1/4时,y取得最大值,为17/8
当sin(x-π/4)=﹣√2/2时,y取得最小值,为﹣1
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