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[超急]三角形恆等式數學兩題(15分)
2008-05-30 2:43 am
回答 (2)
2008-05-30 2:55 am
✔ 最佳答案
18)sinθcosθtanθ﹣1= sinθcosθ(sinθ/ cosθ)﹣1
= (sin2θcosθ/ cosθ)﹣1
= sin2θ﹣1
(或 = 1﹣cos2θ﹣1 = -cos2θ)
20)sinθ= 8 / 17
你可以畫一三角形ABC，其中∠ABC為銳角，AB為斜邊，及∠ACB是直角
AB2 = BC2 + AC2
172 = BC2 + 82
BC2 = 172﹣82
BC2 = 225
BC = 15
∴cosθ= 15 / 17
及tanθ= 8 / 15
2008-05-29 18:58:21 補充：
第2題圖片：
http://s248.photobucket.com/albums/gg191/ncy_0916/?action=view¤t=trigo10-2.jpg
2008-06-03 3:26 am
1. sinθcosθtanθ - 1
= sinθcosθ x (sinθ/ cosθ) -1
= sin^2θ - 1
= -(1 - sin^2θ)
= -cos^2θ
2. sinθ= (8/17)
The length of the unknown side is:
L^2 = 17^2 - 8^2 (Pyth. theorem)
L = 15
cosθ = (15/17)
tanθ = (8/15)
= sinθcosθ x (sinθ/ cosθ) -1
= sin^2θ - 1
= -(1 - sin^2θ)
= -cos^2θ
2. sinθ= (8/17)
The length of the unknown side is:
L^2 = 17^2 - 8^2 (Pyth. theorem)
L = 15
cosθ = (15/17)
tanθ = (8/15)
參考: I hope it can help for your homework!!! Me!I am learning this page too, but in English version.
收錄日期: 2021-04-23 20:21:40
原文連結 [永久失效]:
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