若
sin(A+B)=1/2，sin(A-B)=1/3，
試求tanA:tanB
A-maths 複角
2009-04-14 1:58 am
回答 (2)
2009-04-17 1:48 am
若
sin(A+B)=1/2，sin(A-B)=1/3，
試求tanA:tanB
解:
圖片參考：http://f.imagehost.org/0148/ScreenHunter_07_Apr_16_17_36.gif
sin(A+B)=1/2，sin(A-B)=1/3，
試求tanA:tanB
解:
圖片參考：http://f.imagehost.org/0148/ScreenHunter_07_Apr_16_17_36.gif
2009-04-14 2:24 am
For sin(A+B)=1/2
sinAcosB+cosAsinB=1/2 ------(1)
For sin(A-B)=1/3
sinAcosB-cosAsinB=1/3 ------(2)
(1)+(2)
2sinAcosB=5/6
sinAcosB=5/12 ------(3)
(1)-(2)
2cosAsinB=1/6
cosAsinB=1/12 ------(4)
(3)/(4)
(sinAcosB)/(cosAsinB)=5
tanAcotB=5
tanA/tanB=5 (5=5/1)
As the result, tanA:tanB = 5:1
唔知唔呢..
希望幫到你=]
sinAcosB+cosAsinB=1/2 ------(1)
For sin(A-B)=1/3
sinAcosB-cosAsinB=1/3 ------(2)
(1)+(2)
2sinAcosB=5/6
sinAcosB=5/12 ------(3)
(1)-(2)
2cosAsinB=1/6
cosAsinB=1/12 ------(4)
(3)/(4)
(sinAcosB)/(cosAsinB)=5
tanAcotB=5
tanA/tanB=5 (5=5/1)
As the result, tanA:tanB = 5:1
唔知唔呢..
希望幫到你=]
收錄日期: 2021-04-23 20:37:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090413000051KK01534