三角函數求(cos80˚-cos20˚)/cos40˚的值
回答 (2)
2011-06-20 6:56 pm
✔ 最佳答案
和差化積公式cosA-cosB= -2*sin[(A+B)/2]*sin[(A-B)/2]
解題:
(cos80˚-cos20˚)/cos40˚
={-2sin[(80+20)/2]*sin[(80-20)/2]} /cos40
=[-2sin50*sin30]/cos40
=[-2cos40*(1/2)]/cos40
= -1
2011-06-20 5:48 pm
(Cos80˚－Cos20˚)/Cos40˚=?
Sol
Cos80˚－Cos20˚
=Cos80˚+Cos160˚
=Cos(120˚－40˚)+Cos(120˚+40˚)
=2Cos12˚0Cos40˚
=－Cos40˚
(Cos80˚－Cos20˚)/Cos40˚=－1
Sol
Cos80˚－Cos20˚
=Cos80˚+Cos160˚
=Cos(120˚－40˚)+Cos(120˚+40˚)
=2Cos12˚0Cos40˚
=－Cos40˚
(Cos80˚－Cos20˚)/Cos40˚=－1
收錄日期: 2021-04-30 16:01:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110620000010KK01725