簡化一條簡單三角比關係?

2016-02-19 10:30 am
sin(180°-θ)tan(90-θ)cos(-θ)


答案cos^2θ 求過程

回答 (2)

2016-02-19 10:49 am
✔ 最佳答案
Sol
Sin(180°-θ)=Sinθ
Tan(90°-θ)=Cotθ
Cos(-θ)=Cosθ
Sin(180°-θ)Tan(90°-θ)Cos(-θ)
=SinθCotθCosθ
=Sinθ*(Cosθ/Sinθ)*Cosθ
=Cos^2 θ
2016-03-21 4:39 am
sin(180°-θ)tan(90°-θ)cos(-θ)
=sin(θ)cos(θ)/tan(θ)
=sinθcos^2 θ/sinθ
=cos^2 θ


收錄日期: 2021-04-18 14:37:11
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160219023048AAfA7W9

檢視 Wayback Machine 備份