Prove
tanA+tanB over tanA-tanB = sin(A+B) over sin(A-B)
A. Maths Compound Angle
2008-04-01 2:28 am
回答 (3)
2008-04-01 2:40 am
✔ 最佳答案
By multiplyingcosAcosB/cosAcosB
(tanA+tanB)(cosAcosB)/(tanA-tanB)(cosAcosB)
=(sinAcosB+sinAcosB)/(sinAcosB-sinBcosA)
=sin(A+B)/sin(A-B)
2008-03-31 18:42:02 補充：
這個方法簡單很多,是從L.H.S.做起
只不過慢了一分鐘
2008-03-31 18:52:44 補充：
這條最好用擴分的方法來做
而且做證明應該由較繁複那邊做起(多數是L.H.S.),否則大有可能要砌數
參考: , 自己,堅決不抄人
2008-04-01 2:42 am
L.H.S=(tanA tanB) / (tanA-tanB)
=[(sinA/cosA) (sinB/cosB)] / [(sinA/cosA)-(sinB/cosB)]
=(sinAcosB cosAsinB) / (sinAcosB-cosAsinB)
= sin(A B) / sin(A-B)
=R.H.S
=[(sinA/cosA) (sinB/cosB)] / [(sinA/cosA)-(sinB/cosB)]
=(sinAcosB cosAsinB) / (sinAcosB-cosAsinB)
= sin(A B) / sin(A-B)
=R.H.S
2008-04-01 2:39 am
RHS
=sin(A+B) sin(A-B)
=(sin A cos B + cos A sin B) (sin A cos B - cos A sin B)
=[(sin A cos B + cos A sin B)cos A cos B] [(sin A cos B - cos A sin B)cos A cos B]
=(sin Acos A + sin Bcos B)(sin Acos A - sin Bcos B)
=(tan A + tan B)(tan A - tan B)
=RHS
2008-03-31 18:40:16 補充：
最後那行應為=LHS
=sin(A+B) sin(A-B)
=(sin A cos B + cos A sin B) (sin A cos B - cos A sin B)
=[(sin A cos B + cos A sin B)cos A cos B] [(sin A cos B - cos A sin B)cos A cos B]
=(sin Acos A + sin Bcos B)(sin Acos A - sin Bcos B)
=(tan A + tan B)(tan A - tan B)
=RHS
2008-03-31 18:40:16 補充：
最後那行應為=LHS
收錄日期: 2021-04-23 20:33:41
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