字體：小 中 大 |
|
|
|
| 2017/08/07 00:06:47瀏覽84｜回應0｜推薦0 | |
| 標題: A.Maths 發問: In triangle ABC , A , B and C are the interior angles We know that: 1. cot x = tan (90-x) 2. A + B + C = 180 3. cot (A/2) = tan [(B/2)+(C/2)] 4. cot (A/2) = [(tan(B/2) + tan(C/2)] / [1- tan(B/2)tan(C/2)] Hence show that : cot(A/2)+cot(B/2)+cot(C/2) = cot(A/2)cot(B/2)cot(C/2) 最佳解答: PART 1 cot(A/2) =tan(B/2+C/2 )...(3) =[sin(B/2+C/2)]/[cos(B/2+C/2)] =[sin B/2 cos C/2 +cos B/2 sin C/2]/[ cos B/2 cos C/2 - sin B/2 sin C/2] =[cot C/2 + cot B/2]/[cot B/2 cot C/2 -1] PART 2 cot(A/2)+cot(B/2)+co t(C/2) =[cot C/2 + cot B/2]/[cot B/2 cot C/2 -1] + cot B/2 + cot C/2 [PART 1 result] =[cot B/2 + cot C/2][1/(cot B/2 cot C/2 -1) + 1] =[cot B/2 + cot C/2][1+(cot B/2 cot C/2 - 1)]/(cot B/2 cot C/2 -1)] =[cot B/2 + cot C/2][(cot B/2 cot C/2)/(cot B/2 cot C/2 -1) ] =cot B/2 cot C/2 [(cot B/2 + cot C/2)/(cot B/2 cot C/2 -1)] =cot A/2 cot B/2 cot C/2 [PART 1 result again]
此文章來自奇摩知識+如有不便請留言告知 其他解答: |
|
| ( 創作｜繪圖 ) |
