求函數(shù)y=2sin2x+sinx-cosx,當x屬于中括號-π/2到π/2中括號的最大值和最小值

數(shù)學人氣：536 ℃時間：2020-04-05 07:03:22

優(yōu)質(zhì)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]
∴ 當sin(x－π/4)＝1/4時,y取得最大值,為17/8
當sin(x－π/4)＝﹣√2/2時,y取得最小值,為﹣1