M 2 maths 快40分!
2012-01-29 9:02 am
圖片參考：http://imgcld.yimg.com/8/n/HA00724002/o/701201290004613873432310.jpg
Please help me! Happy New Year & dream come true!
回答 (1)
2012-01-30 1:50 am
✔ 最佳答案
∫(0->π/2) (acos^2 θ + bsin^2 θ - asin^2 θ - bcos^2 θ)dθ= ∫(0->π/2) (a-b)(cos^2 θ-sin^2 θ)dθ
= ∫(0->π/2) (a-b)(cos2θ)dθ
= (1/2) ∫(0->π/2) (a-b) cos2θ d2θ
= (1/2)(a-b) [-sin2θ](0->π/2)
= 0
Therefore,
∫(0->π/2) (acos^2 θ + bsin^2 θ - asin^2 θ - bcos^2 θ)dθ = 0
∫(0->π/2) (acos^2 θ + bsin^2 θ)dθ = ∫(0->π/2) (asin^2 θ + bcos^2 θ)dθ
參考: Hope the solution can help you^^”
收錄日期: 2021-04-20 12:12:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120129000051KK00046