f.4 Amaths一問
2007-03-14 2:15 am
It is given that tanθ+cotθ=7/2.Without solving for θ,find the value of sin2θ.
回答 (2)
2007-03-14 2:26 am
✔ 最佳答案
tanθ+cotθ=7/2tanθ+1/tanθ=7/2
sinθ/cosθ+cosθ/sinθ=7/2
sinθ^2+cosθ^2/sinθcosθ=7/2
1/sinθcosθ=7/2
2/7=sinθcosθ
2007-03-13 18:40:00 補充：
sinθcosθ=sin2θ/2e.gsin30*cos30=sin60 /2我冇公式要一個個ｔｅｓｔ４／７＝２ｓｉｎθｃｏｓθ４／７＝ｓ ｉｎ２θ
2007-03-14 2:28 am
sin2θ = 2 sinθcosθ
tan θ + cot θ = 7/2
(sin^2 θ + cos^2 θ) / (sin θ cos θ) = 7/2
1/(sin θ cos θ) = 7/2
7 sin θ cos θ = 2
sin θ cos θ = 2/7
2 sin θ cos θ = 4/7
Therefore sin 2θ = 4/7
tan θ + cot θ = 7/2
(sin^2 θ + cos^2 θ) / (sin θ cos θ) = 7/2
1/(sin θ cos θ) = 7/2
7 sin θ cos θ = 2
sin θ cos θ = 2/7
2 sin θ cos θ = 4/7
Therefore sin 2θ = 4/7
參考: 自己
收錄日期: 2021-04-18 21:02:43
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070313000051KK02722