(a)若cosθ+cosβ=a及sinθ+sinβ=b(b≠0),証明
cos(θ-β)=1/2 (a²+b²-2)及tan(θ+β)/2=b/a。
(b)解方程組
cosθ+cosβ=√2
sinθ+sinβ=√2
其中0≤θ≤2π及0≤β≤2π。
a-maths~~~~
2007-11-10 2:50 am
回答 (2)
2007-11-13 9:00 am
✔ 最佳答案
(a)圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Nov07/Crazytrigo3.jpg
(b)
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Nov07/Crazytrigo4.jpg
參考: My Maths knowledge
2007-11-10 3:16 am
(a)cosθ+cosβ
cos(θ-β) = cosθcosβ +sinθsinβ
=[(cosθ+cosβ)^2 + (sinθ+sinβ)^2)-2]/2
=(a^2 + b^2 -2)/2
b/a
=(sinθ+sinβ)/(cosθ+cosβ)
= 2{[sin(θ+β)/2] [cos(θ-β)/2] }/2{[cos(θ+β)/2] [cos(θ-β)/2] }
=tan[(θ+β)/2]
(b)
cos(θ-β) = 1
tan[(θ+β)/2] = 1
so
θ - β= 0
θ+β = π/2 or 5π/2
then
θ = β = π/4
or
θ = β= 5π/4
cos(θ-β) = cosθcosβ +sinθsinβ
=[(cosθ+cosβ)^2 + (sinθ+sinβ)^2)-2]/2
=(a^2 + b^2 -2)/2
b/a
=(sinθ+sinβ)/(cosθ+cosβ)
= 2{[sin(θ+β)/2] [cos(θ-β)/2] }/2{[cos(θ+β)/2] [cos(θ-β)/2] }
=tan[(θ+β)/2]
(b)
cos(θ-β) = 1
tan[(θ+β)/2] = 1
so
θ - β= 0
θ+β = π/2 or 5π/2
then
θ = β = π/4
or
θ = β= 5π/4
參考: me
收錄日期: 2021-04-23 20:37:24
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071109000051KK02586