(tan 240°+ sin 120°)(tan 240°-sin 120°)
求過程，感謝
求不用計算機三角比的值?
2016-02-19 12:49 pm
回答 (3)
2016-02-19 1:30 pm
✔ 最佳答案
(Tan240°+Sin120°)(Tan240°－Sin120°) =(Tan60°+Sin60°)(Tan60°－Sin60°)
=(Sin60°/Cos60°+Sin60°)/(Sin60°/Cos60°－Sin60°)
=(1/Cos60°+1)/(1/Cos60°－1)
=(1+Cos60°+1)/(1－Cos60°)
=(1+1/2)/(1－1/2)
=3
2016-03-21 4:34 am
(Tan240°+Sin120°)(Tan240°－Sin120°)
=(Tan60°+Sin60°)(Tan60°－Sin60°)
=(Sin60°/Cos60°+Sin60°)/(Sin60°/Cos60°－Sin60°)
=(1/Cos60°+1)/(1/Cos60°－1)
=(1+Cos60°+1)/(1－Cos60°)
=(1+1/2)/(1－1/2)
=3
=(Tan60°+Sin60°)(Tan60°－Sin60°)
=(Sin60°/Cos60°+Sin60°)/(Sin60°/Cos60°－Sin60°)
=(1/Cos60°+1)/(1/Cos60°－1)
=(1+Cos60°+1)/(1－Cos60°)
=(1+1/2)/(1－1/2)
=3
2016-02-19 2:45 pm
( tan240° + sin120° ) ( tan240°－sin120° )
= ( tan60° + sin60° ) ( tan60°－sin60° )
= ( sin60° / cos60° + sin60° ) / ( sin60° / cos60°－sin60° )
= ( 1 / cos60° + 1 ) / ( 1 / cos60°－1 )
= ( 1 + cos60° + 1 ) / ( 1－cos60° )
= ( 1 + 1/2 ) / ( 1－1/2 )
=3
= ( tan60° + sin60° ) ( tan60°－sin60° )
= ( sin60° / cos60° + sin60° ) / ( sin60° / cos60°－sin60° )
= ( 1 / cos60° + 1 ) / ( 1 / cos60°－1 )
= ( 1 + cos60° + 1 ) / ( 1－cos60° )
= ( 1 + 1/2 ) / ( 1－1/2 )
=3
收錄日期: 2021-04-18 14:29:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160219044909AA0UdhW