Two questions about Simplication of Trigonometric Expression:
1. sin^4θ - cos^4θ (steps please)
2. 1 over 1-cosθ + 1 over1+cosθ (stepts please)
HELP ME PLEASE!!
2 question want to ask for help (about Simplication Trig)
2008-01-09 3:25 am
回答 (2)
2008-01-09 3:36 am
✔ 最佳答案
sin^4θ - cos^4θ=(sin^2θ-cos^2θ)(sin^2θ+cos^2θ)
=cos2θ x 1
=cos2θ//
1/(1-cosθ)+1/(1+cosθ)
=(1+cosθ)/(1-cosθ)(1+cosθ)+(1-cosθ)/(1+cosθ)(1-cosθ)
=(1+cosθ+1-cosθ)/(1-cos^2θ)
=2/sin^2 θ
=2csc^2θ//
2008-01-08 20:50:20 補充：
sin^4θ - cos^4θ=(sin^2θ-cos^2θ)(sin^2θ+cos^2θ)=-cos2θ x 1=-cos2θ//
2008-01-09 3:42 am
(sinθ)^4 - (cosθ)^4
=[(sinθ)^2 - (cosθ)^2][(sinθ)^2 + (cosθ)^2]
= -cos(2θ)
1 / (1-cosθ) + 1 / (1+cosθ)
= [(1+cosθ) + (1-cosθ)] / [(1-cosθ)(1+cosθ)]
= 2 / [1-(cosθ)^2]
= 2 / (sinθ)^2
= 4 / [1-cos(2θ)]
=[(sinθ)^2 - (cosθ)^2][(sinθ)^2 + (cosθ)^2]
= -cos(2θ)
1 / (1-cosθ) + 1 / (1+cosθ)
= [(1+cosθ) + (1-cosθ)] / [(1-cosθ)(1+cosθ)]
= 2 / [1-(cosθ)^2]
= 2 / (sinθ)^2
= 4 / [1-cos(2θ)]
收錄日期: 2021-04-16 22:59:38
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080108000051KK03104