Given that in a triangle ABC,
(b+c)cos A + (c+a) cos B +(a+b) cos C=a+b+c
and 4 sin (A/2) sin (B/2) sin (C/2) = cos A + cos B + cos C -1
Prove that 4(a+b+c)sin (A/2) sin (B/2) sin (C/2) =a cos A +b cos B +c cos C
TRIGONOMETRY
2011-06-07 12:24 am
回答 (1)
2011-06-07 1:15 am
✔ 最佳答案
4 (a + b + c) sin (A/2) sin (B/2) sin (C/2) = (a + b + c) (cos A + cos B + cos C - 1)= (a cos A + b cos B + c cos C) + (b+c)cosA + (a+b)cosB + (a+c)cosC- (a + b + c)= (a cos A + b cos B + c cos C) + (a + b + c) - (a + b + c)= a cos A + b cos B + c cos C
2011-06-06 17:18:01 補充：
line 3 :
(a+b)cosB should be (a+c)cosB
Sorry!
2011-06-06 17:20:28 補充：
(a+c)cosC should be (a+b)cosC
收錄日期: 2021-04-21 22:20:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110606000051KK00700