Let ∆ABC be an acute-angles triangle. Prove that
cos²A + cos²B + cos²C = 1 － 2 cos A cos B cos C
Hence deduce that
cos²A cos²B cos²C < (1/27)
inequality
2009-07-18 5:53 am
回答 (1)
2009-07-18 7:20 am
✔ 最佳答案
cos2A + cos2B + cos2C= cos2(180 – B – C) + cos2B + cos2C
= cos2[(90 – B) + (90 – C)] + cos2B + cos2C
= {cos[(90 – B) + (90 – C)]}2 + cos2B + cos2C
= [cos(90 – B)cos(90 – C) – cos(90 – B)cos(90 – C)]2 + cos2B + cos2C
= (sinBsinC – cosBcosC)2 + cos2B + cos2C
= sin2Bsin2C – 2sinBsinCcosBcosC + cos2Bcos2C + cos2B + cos2C
= (1 – cos2B)(1 – cos2C) – 2sinBsinCcosBcosC + cos2Bcos2C + cos2B + cos2C
= 1 – cos2B – cos2C + cos2Bcos2C – 2sinBsinCcosBcosC + cos2Bcos2C + cos2B + cos2C
= 1 + 2cos2Bcos2C – 2sinBsinCcosBcosC
= 1 + 2cosBcosC(cosBcosC – sinBsinC)
= 1 + 2cosBcosCcos(B + C)
= 1 – 2cosBcosCcos(180 – B – C)
= 1 – 2cosAcosBcosC
AM >= GM
(cos2A + cos2B + cos2C)/3 >= (cos2Acos2Bcos2C)1/3
(cos2A + cos2B + cos2C)3/27 >= cos2Acos2Bcos2C
(1 – 2cosAcosBcosC)3/27 >= cos2Acos2Bcos2C
Since A,B and C are acute, cosA,cosB,cosC>0
cosAcosBcosC>0
-2cosAcosBcosC<0
0<=1-2cosAcosBcosC<1
(1-2cosAcosBcosC)3<1
Therefore, 1/27 > (1 – 2cosAcosBcosC)3/27 >= cos2Acos2Bcos2C
收錄日期: 2021-04-23 23:18:29
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https://hk.answers.yahoo.com/question/index?qid=20090717000051KK02001