sin(θ-π/4)+√3sin((θ+π/4)=2
更新1:
sin(θ-π/4)+√3sin(θ+π/4)=2
a-maths!!!!!
回答 (2)
2008-04-30 1:55 am
✔ 最佳答案
如下:圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Apr08/Crazytrigo6.jpg
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參考: My Maths knowledge
2008-04-20 5:35 am
sin(θ-π/4)+√3sin(θ+π/4)=2
- sin(π/4 - θ)+√3sin(θ+π/4)=2
- sin(π/2 - (π/4 + θ)) ) +√3sin(θ+π/4)=2
-cos(π/4 + θ) +√3sin(θ+π/4) = 2
(1/2)cos(π/4 + θ) - [(√3)/2]sin(π/4 + θ) = -1
cos(π/3)cos(π/4 + θ) - sin(π/3)sin(π/4 + θ) = -1
cos[(π/3) + (π/4 + θ)] = -1
cos(7π/12 + θ) = -1
7π/12 + θ = 2nπ ± π
7π/12 + θ = (2n+1)π
θ = (2n+1)π- 7π/12
- sin(π/4 - θ)+√3sin(θ+π/4)=2
- sin(π/2 - (π/4 + θ)) ) +√3sin(θ+π/4)=2
-cos(π/4 + θ) +√3sin(θ+π/4) = 2
(1/2)cos(π/4 + θ) - [(√3)/2]sin(π/4 + θ) = -1
cos(π/3)cos(π/4 + θ) - sin(π/3)sin(π/4 + θ) = -1
cos[(π/3) + (π/4 + θ)] = -1
cos(7π/12 + θ) = -1
7π/12 + θ = 2nπ ± π
7π/12 + θ = (2n+1)π
θ = (2n+1)π- 7π/12
收錄日期: 2021-04-23 20:32:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080419000051KK03266