cosA+2sin2AsinA/2cosA
回答 (2)
2011-01-20 11:18 pm
✔ 最佳答案
As follows:圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/ScreenHunter_01Jan201517.gif
2011-01-22 12:24 am
Do you want to ask how to calculate cosA+2sin2AsinA/2cosA or (cosA+2sin2AsinA)/2cosA?
If you want to calculate cosA+2sin2AsinA/2cosA,
then cosA+2sin2AsinA/2cosA
= cosA + 2(2sinAcosA)sinA/2cosA
= cosA + 2 sin^2 A
If you want to calculate (cosA+2sin2AsinA)/2cosA,
then (cosA+2sin2AsinA)/2cosA
= [cosA+2(2sinAcosA)sinA]/2cosA
= (cosA + 4 sin^2 A cosA) / 2cosA
= (1 + 4 sin^2 A)/2cosA
If you want to calculate cosA+2sin2AsinA/2cosA,
then cosA+2sin2AsinA/2cosA
= cosA + 2(2sinAcosA)sinA/2cosA
= cosA + 2 sin^2 A
If you want to calculate (cosA+2sin2AsinA)/2cosA,
then (cosA+2sin2AsinA)/2cosA
= [cosA+2(2sinAcosA)sinA]/2cosA
= (cosA + 4 sin^2 A cosA) / 2cosA
= (1 + 4 sin^2 A)/2cosA
收錄日期: 2021-04-22 00:52:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110120000051KK01936