請大家教下我點樣做呢3題數。多謝各位!
1,cos2次x - (tan2次x)(sin2次x)
2 , 1 over 3^1-sinx
3, 2^cosx ( 求 0度<360度 (包括0同360..唔識打大於或等於的符號)
三角函數的問題.....請大家解難題
2007-07-10 3:21 am
回答 (1)
2007-07-10 11:38 pm
✔ 最佳答案
1. cos^2 x - tan^2 x * sin^2 x= cos^2 x - sin^4 x / cos^2 x
= (cos^4 x - sin^4 x) / cos^2 x
= (cos^2 x + sin^2 x)(cos^2 x - sin^2 x) / cos^2 x
=(1)(cos^2 x - sin^2 x) / cos^2 x
=1 - tan^2 x
2. Can it be solved?
3.2^cos x = ??? What does 2^cos x equal to? If this information is not known, the question is not an equation. It cannot be solved.
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070709000051KK03893