tan^2A-2tan^2B=1
Find the possible values of cosA/cosB
Trigonometry
2014-01-04 8:04 am
回答 (2)
2014-01-04 8:31 am
✔ 最佳答案
tan^2A-2tan^2B=1{[sec^2(A)]-1)}-2{[sec^2(B)]-1}=1
sec^2(A)-1-2sec^2(B)+2=1
sec^2(A)-2sec^2(B)=0
sec^2(A)=2sec^2(B)
[cos^2(B)]/[cos^2(A)]=2
(cosB/cosA)^2=2
cosB/cosA=√2
cosA/cosB=1/√2
2014-01-04 11:10:00 補充：
(cosB/cosA)^2=2
cosB/cosA=√2 or -√2
cosA/cosB=1/√2 or 1/-√2
2014-01-04 6:58 pm
係咪重有-1/√2？
X^2=9
X=+-3
X^2=9
X=+-3
收錄日期: 2021-04-26 21:53:36
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https://hk.answers.yahoo.com/question/index?qid=20140104000051KK00001