1)Find the values of the following expressions!
(sin19tan71-cos19)^2
Maths (F.2)
2010-04-17 10:26 pm
回答 (3)
2010-04-17 10:39 pm
2010-04-17 11:04 pm
(sin19tan71-cos19)^2
= {(sin19)×[1/tan(90-71)]-cos19}^2
= [sin19×(1/tan19)-cos19]^2
= (sin19×cos19÷sin19-cos19)^2
= (cos19-cos19)^2
= 0^2
= 0
= {(sin19)×[1/tan(90-71)]-cos19}^2
= [sin19×(1/tan19)-cos19]^2
= (sin19×cos19÷sin19-cos19)^2
= (cos19-cos19)^2
= 0^2
= 0
收錄日期: 2021-04-23 20:40:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100417000051KK00725