1.1/sinθ- sinθ/tan^2θ
ans:sinθ
2.(sinθ+cosθ)^2-1
ans: 2sinθcosθ
我要步驟謝謝
問2題數學題,快d黎啦!!!!!!!!!!!!!!!!!!!!!!!!
2008-04-15 4:08 am
回答 (2)
2008-04-15 4:17 am
✔ 最佳答案
Ans: 1. 1/sinθ- sinθ/tan^2θ
=1/sinθ- sinθ/(sin^2θ/cos^2θ)
=1/sinθ- sinθ(cos^2θ/sin^2θ)
=1/sinθ- (cos^2θ/sinθ)
=(1-cos^2θ)/sinθ
=sin^2θ/sinθ
=sinθ
2. (sinθ cosθ)^2-1
=(sinθ)^2 2sinθcosθ (cosθ)^2 -1
= 2sinθcosθ 1-1
= 2sinθcosθ
參考: 自己
2008-04-15 5:08 am
答案:
1. 1/sinθ- sinθ/(tanθ)^2
=1/sinθ- sinθ/((sinθ)^2/(cosθ)^2)
=1/sinθ- sinθ((cosθ)^2/(sinθ)^2)
=1/sinθ- ((cosθ)^2/sinθ)
=(1-(cosθ)^2)/sinθ
=((sinθ)^2 + (cosθ)^2 -cos^2θ)/sinθ(因為(sinθ)^2 + (cosθ)^2=1)
=(sinθ)^2/sinθ
=sinθ
2. (sinθ+cosθ)^2-1
=(sinθ)^2+2sinθcosθ+(cosθ)^2 -1
= 2sinθcosθ+1-1(因為(sinθ)^2 + (cosθ)^2 = 1 , 要背嫁)
= 2sinθcosθ
1. 1/sinθ- sinθ/(tanθ)^2
=1/sinθ- sinθ/((sinθ)^2/(cosθ)^2)
=1/sinθ- sinθ((cosθ)^2/(sinθ)^2)
=1/sinθ- ((cosθ)^2/sinθ)
=(1-(cosθ)^2)/sinθ
=((sinθ)^2 + (cosθ)^2 -cos^2θ)/sinθ(因為(sinθ)^2 + (cosθ)^2=1)
=(sinθ)^2/sinθ
=sinθ
2. (sinθ+cosθ)^2-1
=(sinθ)^2+2sinθcosθ+(cosθ)^2 -1
= 2sinθcosθ+1-1(因為(sinθ)^2 + (cosθ)^2 = 1 , 要背嫁)
= 2sinθcosθ
收錄日期: 2021-04-24 10:54:35
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080414000051KK02273