? 發問時間: 科學及數學數學 · 1 個月前

Please answer the question No.75?

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1 個解答

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  • 1 個月前
    最佳解答

    cos(π/9)[sin(π/9)-sin(2π/9)-sin(3π/9)+sin(4π/9)]

      = (sin(2π/9))/2-(sin(3π/9)+sin(π/9))/2

           -(sin(4π/9)+sin(2π/9))/2 + (sin(5π/9)+sin(3π/9))/2

      = -(sin(π/9))/2 - (sin(4π/9))/2 + (sin(5π/9))/2

      = -(sin(π/9))/2   (∵ sin(4π/9) = sin(π-5π/9) = sin(5π/9))

    ∴  sin(π/9)-sin(2π/9)-sin(3π/9)+sin(4π/9)

           = [-(sin(π/9))/2]/cos(π/9)

           = (-1/2)tan(π/9)

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