A.Maths - Ans.fyi 參考答案

A.Maths

2007-12-26 11:26 pm

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)

回答 (1)

myisland8132

2007-12-26 11:44 pm

✔ 最佳答案

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]

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