maths!!!
回答 (2)
2007-12-22 11:57 pm
✔ 最佳答案
3cos2θ – 2cosθ – 1 = 3[cos2θ – 2(1 / 3)cosθ + (1 / 3)2] – (1 / 3) – 1
= 3(cosθ– 1 / 3)2 – 4 / 3
When cosθ = 1 / 3, 3cos2θ – 2cosθ – 1 is at minimum
So, the minimum value is – 4 / 3
參考: Myself~~~
2007-12-23 12:00 am
3cos²θ-2cosθ-1
=3(cos²θ-2/3cosθ)-1
=3(cos²θ-2/3cosθ+1/9)-1-3(1/9)
=3(cosθ-1/3)²-4/3
極小值=-4/3
=3(cos²θ-2/3cosθ)-1
=3(cos²θ-2/3cosθ+1/9)-1-3(1/9)
=3(cosθ-1/3)²-4/3
極小值=-4/3
收錄日期: 2021-04-23 20:32:32
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071222000051KK02653