把2sinθ-√12 cosθ表為 Rsin(θ+α)的形式,且θ和α以弧度為單位,
而 -π/2 < α < π/2 .
問數~關於三角函數~
2008-02-14 12:58 am
回答 (2)
2008-02-15 7:49 pm
✔ 最佳答案
Rsin(θ+α)=R[sinθcosα+cosθsinα]
比較得
Rcosα=2
Rsinα=-√12
R^2=16
R=4,α=-π/3
2sinθ-√12 cosθ=4sin(θ-π/3)
2008-02-14 1:14 am
=2(sinθ-√3cosθ)
=4(sinθsin(π/6)-cosθcos(π/6))
=-4(cosθcos(π/6)-sinθsin(π/6))
=-4cos(θ+(π/6))
=4(sinθsin(π/6)-cosθcos(π/6))
=-4(cosθcos(π/6)-sinθsin(π/6))
=-4cos(θ+(π/6))
參考: Me.......
收錄日期: 2021-04-14 00:06:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080213000051KK03017