? 發問於 科學及數學數學 · 2 月前

How to prove cos(3pi/5)=((1-sqrt5)/4)?

Already found 

(a) is (x+1)(4x^2 - 2x -1)

(b) 4cos^3 (theta) - 3cos(theta)

(c) if let x=cos(pi/5), then cos(3theta) = 4cos^3(pi/5) - 3cos(pi/5)=4x^3 - 3x

But then how to proceed? where does the factor x+1 come in?

更新:

Hi, Gong, yes of course.

Attachment image

2 個解答

評分
  • ?
    Lv 7
    2 月前
    最愛解答

    昨晚的回答不知何故又被隱了,

    簡單回答吧.

    令 x = cos(3π/5) 則

    cos(9π/5) = 4x^3 - 3x

    cos(6π/5) = 2x^2 - 1

    cos(9π/5) + cos(6π/5) = 0

    即  

       4x^3 + 2x^2 - 3x - 1

          =  (x+1)(4x^2 - 2x - 1 ) = 0.

    因 π/2 < 3π/5 < π, 故

        0 > cos(3π/5) > -1

    故 cos(3π/5) = (1-√5)/4.

  • GONG
    Lv 6
    2 月前

    用中文回答可以嗎?..........

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