要計算過程
a.1 over sinθ - sin（90° - θ）over tanθ
b.(1-cosθ)(1+cosθ) over tan^2θ
c.己知cos θ=3 over 5,其中θ是一個銳角,求sinθ及tanθ的值
plz...三角學練習θ
2012-08-26 10:19 am
回答 (1)
2012-08-27 4:13 am
✔ 最佳答案
a.1/sinθ - sin(90°-θ)/tanθ=1/sinθ - cosθ/tanθ
=1/sinθ-cos^2θ/sinθ
=sin^2θ/sinθ
=sinθ
b.(1-cosθ)(1+cosθ)/tan^2θ
=(1-cos^2θ)/tan^2θ
=sin^2θ/tan^2θ
=cos^2θ
c.己知cos θ=3/5,其中θ是一個銳角,求sinθ及tanθ的值
sinθ
=√(1-cos^2θ)
=√(1-(3/5)^2)
=4/5
tanθ
=sinθ/cosθ
=(4/5) / (3/5)
=4/3
參考: me
收錄日期: 2021-04-20 13:14:30
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120826000051KK00071