1.解以下方程,其中0度<=x<=360度(360度大過等於x,x大過等於0度)答案準確至四位有數 字
3-4(cos2x)+cos4x=1/2
2.證明以下恆等式:
2(tanA)/1+tan^2A=sin2A
不要跳步~~~!!!
最好在算式旁邊寫明用過那一條公式
F.4 a.maths2條(複角)
2007-07-24 1:02 am
回答 (2)
2007-07-24 1:16 am
✔ 最佳答案
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參考: My Maths knowledge
2007-07-24 1:51 am
1.
Let y = cos 2x
cos 4x = 2 (cos 2x)^2 - 1 = 2 y^2 - 1
=> 3 - 4y + (2y^2 - 1) = 2 - 4y + 2y^2
=> 3 - 8y + 4y^2 = 0
=> y = 0.5 or y = 1.5 , but, -1 <= cos k <= 1 ==> y = 0.5
cos 2x = 0.5 , 2x = 60 , x = 30
2.
(tan A)^2 = (1 - cos 2A) / (1 + cos 2A)
=> 1 + (tan A)^2 = 2 / (1 + cos 2A)
1 + cos 2A = 2 (cos A)^2
=> 1 + (tan A)^2 = 1 / (cos A)^2
then,
2 (tan A) / [1 + (tan A)^2] = 2 (tan A) (cos A) ^ 2 = 2 (sin A)(cos A) = sin 2A
Let y = cos 2x
cos 4x = 2 (cos 2x)^2 - 1 = 2 y^2 - 1
=> 3 - 4y + (2y^2 - 1) = 2 - 4y + 2y^2
=> 3 - 8y + 4y^2 = 0
=> y = 0.5 or y = 1.5 , but, -1 <= cos k <= 1 ==> y = 0.5
cos 2x = 0.5 , 2x = 60 , x = 30
2.
(tan A)^2 = (1 - cos 2A) / (1 + cos 2A)
=> 1 + (tan A)^2 = 2 / (1 + cos 2A)
1 + cos 2A = 2 (cos A)^2
=> 1 + (tan A)^2 = 1 / (cos A)^2
then,
2 (tan A) / [1 + (tan A)^2] = 2 (tan A) (cos A) ^ 2 = 2 (sin A)(cos A) = sin 2A
收錄日期: 2021-04-13 13:58:36
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