• 设f1(x)=sinx,定义fn+1(x)=为fn(x)的导数,即fn+1(x)=f′n(x),n∈N*若△ABC的内角A满足f1(A)+f2(A)+…+f2014(A)=√22,则tanA的值是 .试题及答案-填空题-云返教育

    • 试题详情

      设f1(x)=sinx,定义fn+1(x)=为fn(x)的导数,即fn+1(x)=f′n(x),n∈N*若△ABC的内角A满足f1(A)+f2(A)+…+f2014(A)=
      2
      2
      ,则tanA的值是         

      试题解答


      -(2+
      3
      )
      解:∵f1(x)=sinx,fn+1(x)=f′n(x),
      ∴f
      2(x)=f′1(x)=cosx,
      f
      3(x)=f′2(x)=-sinx,
      f
      4(x)=f'3(x)=-cosx,
      f
      5(x)=f′4(x)=sinx,
      f
      6(x)=f′5(x)=cosx,
      ∴f
      n+1(x)=f′n(x),具备周期性,周期性为4.
      且f
      1(x)+f2(x)+f3(x)+f4(x)=cosx-sinx+sinx-cosx=0,
      f1(A)+f2(A)+…+f2014(A)=
      2
      2

      ∴f
      1(A)+f2(A)=
      2
      2

      即sinA+cosA=
      2
      2

      2
      sin?(A+
      π
      4
      )=
      2
      2

      即sin(A+
      π
      4
      )=
      1
      2

      ∵A是△ABC的内角,
      ∴A+
      π
      4
      =
      6

      解得A=
      6
      -
      π
      4
      =
      7
      12
      π.
      ∴tanA=tan?(
      6
      -
      π
      4
      )=
      tan?
      6
      -tan?
      π
      4
      1+tan?
      6
      tan?
      π
      4
      =
      -
      3
      3
      -1
      1-
      3
      3
      =-(2+
      3
      ).
      故答案为:-(2+
      3
      ).
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