Is there a simple way to solve sin^8θ+cos^8θ or sin^8θ-cos^8θ?
this is all i have to solve it
sinθ/cosθ = tanθ
sin^2θ+cos^2θ = 1
sinθ = cos(90-θ)
cosθ = sin(90-θ)
tanθ = 1/tan(90-θ)
and identities of polynomials
a^2 - b^2= (a+b)(a-b)
(a+b)^2 = a^2 + 2ab + b^2
(a-b)^2 = a^2- 2ab + b^2
sin^8θ+cos^8θ / sin^8θ-cos^8θ
2009-05-21 4:08 am
回答 (4)
2009-05-21 7:40 pm
✔ 最佳答案
圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/May09/Crazytrigo1.jpg
希望我無理解錯樓主想問的.
參考: Myself
2009-05-21 7:07 am
You mean "simplify sin^8θ+cos^8θ or sin^8θ-cos^8θ in terms of sinθ and cosθ "??
2009-05-21 5:05 am
sin^8θ-cos^8θ
=(sin^4θ+cos^4θ)(sin^4θ-cos^4θ)
=(sin^4θ+cos^4θ)(sin^2θ-cos^2θ)(sin^2θ+cos^2θ)
=(sin^4θ+cos^4θ)(sinθ-cosθ)(sinθ+cosθ)//
=(sin^4θ+cos^4θ)(sin^4θ-cos^4θ)
=(sin^4θ+cos^4θ)(sin^2θ-cos^2θ)(sin^2θ+cos^2θ)
=(sin^4θ+cos^4θ)(sinθ-cosθ)(sinθ+cosθ)//
2009-05-21 4:22 am
Both sin^8θ+cos^8θ and sin^8θ-cos^8θare NOT equations.
Therefore, there is NO solution.
Therefore, there is NO solution.
收錄日期: 2021-04-22 00:39:43
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090520000051KK01477