若sec²(π-θ)/[3cot²(π/2 +θ)-7]=1及π/2<θ<π
試求cosθ的值
一條三角函數問題
2009-01-12 5:26 am
回答 (1)
2009-01-12 5:40 am
✔ 最佳答案
sec(π-θ)/[3cot(π/2 +θ)-7]=1sec(θ)=[3cot(π/2 +θ)-7]
sec(θ) + 7 = 3cot(π/2 +θ)
tan(π/2 +θ)[sec(θ) + 7] = 3
cot(θ)[sec(θ) + 7] = 3
cos(θ)[sec(θ) + 7] = 3sin(θ)
1+7cos(θ)=3(1-cos(θ))
10cos(θ)=2
cos(θ)=1/5
cos(θ)=-1/SQRT(5)
收錄日期: 2021-04-26 13:06:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090111000051KK01898