1.已知y+5sin²x+√3sinxcosx+6cos²x的最大值是1,求y.
2.求2sin(x/2)sin(a-x/2)的最大值.(a是常數)
數學不懂做13
2014-09-16 4:21 am
回答 (3)
2014-09-18 10:02 am
✔ 最佳答案
2.2sin(x/2)sin(a-x/2)
=-2/2 [ cos(x/2+a-x/2) - cos(x/2-a+x/2) ]
=cos(x-a) - cos a
-1≤cos(x-a)≤1
-1 - cos a ≤ cos(x-a) - cos a ≤ 1 - cos a
最大值=1 - cos a
2014-09-16 00:09:26 補充：
謝謝~
但而家我忙緊1D野 0.0
2014-09-18 02:02:07 補充：
Please Read：
圖片參考：https://s.yimg.com/rk/HA07344914/o/1236668282.png
2014-09-16 9:26 am
ASK YOUR TEACHER!!!!!!!!!!!!!!!
2014-09-16 7:44 am
HK~ 回答吧～
5sin²x+√3sinxcosx+6cos²x
= 1 + √3sinxcosx + cos²x
= 1 + √3/2 sin2x + (1 + cos2x)/2
= 3/2 + √3/2 sin2x + 1/2 cos2x
= 3/2 + sin(π/3)sin2x + cos(π/3)cos2x
= 3/2 + cos(2x - π/3)
5sin²x+√3sinxcosx+6cos²x
= 1 + √3sinxcosx + cos²x
= 1 + √3/2 sin2x + (1 + cos2x)/2
= 3/2 + √3/2 sin2x + 1/2 cos2x
= 3/2 + sin(π/3)sin2x + cos(π/3)cos2x
= 3/2 + cos(2x - π/3)
收錄日期: 2021-04-16 17:08:31
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140915000051KK00132