F.4 trigo amath solve cos4Θ+cos2Θ=cosΘ

2007-02-19 1:43 am

Find the general solution of cos4Θ+cos2Θ=cosΘ.
Those number are NOT indice.

更新1:

To: 納粹戰爭主義革命軍元首你爭少少...cosΘ唔可以就咁樣約左佢你咁樣會少左個solution. 應該抽common factor再做一次.... 很對不起呀！一半ge答案係錯ge答案...

更新2:

To: gladysbwlo1116 你好好..識抽common factor. 但只差最後一步。試想一下：cos x = 0, x 真的等於360°n嗎？那個是sine function而已... 所以你的答案也是一個錯的答案，但較上者好一點。我認為cosine function 用±號有時會好D. 不過sine就真係用左都多餘...

更新3:

To:納粹戰爭主義革命軍元首很高興你會返黎睇一次.. 你最後做o個次就岩啦...

回答 (2)

用戶19809

2007-02-19 1:51 am

✔ 最佳答案

cos4Θ+cos2Θ=cosΘ
2 cos (4Θ+2Θ)/2 cos (4Θ-2Θ)/2=cos Θ
2 cos 3Θ cos Θ=cos Θ
2 cos 3Θ=1
cos 3Θ=1/2
cos 3Θ=cos 60 (the numbers denotes degree)
3Θ=360n士60 where n is any integers.

2007-02-18 18:08:48 補充：
cos4Θ cos2Θ=cosΘ2 cos (4Θ 2Θ)/2 cos (4Θ-2Θ)/2=cos Θ2 cos 3Θ cos Θ=cos Θ2 cos 3Θ cos Θ-cos Θ=0cos Θ (2 cos 3Θ-1)=0cos Θ=0 or 2 cos 3Θ-1=0Θ=360n士90 or cos 3Θ=1/2Θ=360n士90 or Θ=120n士20

2007-02-18 18:10:05 補充：
cos4Θ＋cos2Θ=cosΘ2 cos (4Θ＋2Θ)/2 cos (4Θ-2Θ)/2=cos Θ2 cos 3Θ cos Θ=cos Θ2 cos 3Θ cos Θ-cos Θ=0cos Θ (2 cos 3Θ-1)=0cos Θ=0 or 2 cos 3Θ-1=0Θ=360n士90 or cos 3Θ=1/2Θ=360n士90 or Θ=120n士20

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[匿名]

2007-02-19 2:01 am

cos4x+cos2x=cosx
2[cos(4+2)x/2][cos(4-2)x/2]=cosx
2(cos3x)(cosx)=cosx
cos3x=1/2 or cosx=0
3x=360n+60 or 3x=360n-60 or x=360n
x=120n+20 or x=120n-20 or x=360n
where n is an integer

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