sinθcosθ=?
回答 (2)
2007-04-29 5:26 am
✔ 最佳答案
sinθcosθ=(sin2θ)/2
證明以下恆等式:
(3cosθ+4sinθ)^2+(4cosθ-3sinθ)^2=25
L.H.S.
=(3cosθ+4sinθ)^2+(4cosθ-3sinθ)^2
=9cos^2θ+24sinθcosθ+16sin^2θ+16cos^2θ-24sinθcosθ+9sin^2θ
=9sin^2θ+9cos^2θ+16sin^2θ+16cos^2θ
=(9+16)(sin^2θ+cos^2θ)
=25
=R.H.S.
參考: me
2007-04-29 5:20 am
sin2θ=2sinθcosθ
F4 A maths
問題問咩?
2007-04-28 21:33:26 補充：
LHS=(3cosθ 4sinθ)^2 (4cosθ 3sinθ)^2=[(3cosθ)^2 2(3cosθ)(4sinθ) (4sinθ)^2] [(4cosθ)^2 2(4cosθ)(3sinθ) (3sinθ)^2]=9(cosθ)^2 24sinθcosθ 16(sinθ)^2 16(cosθ)^2-24sinθcosθ 9(sinθ)^2=25(sinθ)^2 25(cosθ)^2=25=RHS
F4 A maths
問題問咩?
2007-04-28 21:33:26 補充：
LHS=(3cosθ 4sinθ)^2 (4cosθ 3sinθ)^2=[(3cosθ)^2 2(3cosθ)(4sinθ) (4sinθ)^2] [(4cosθ)^2 2(4cosθ)(3sinθ) (3sinθ)^2]=9(cosθ)^2 24sinθcosθ 16(sinθ)^2 16(cosθ)^2-24sinθcosθ 9(sinθ)^2=25(sinθ)^2 25(cosθ)^2=25=RHS
參考: me
收錄日期: 2021-04-13 17:10:30
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070428000051KK04865