A.maths 高手請答~~~~~~~~~~~~~~~~~!!!!

∵tanA=sinA/cosA,tanB=sinB/cosB
a/sinA=b/sinB==>sinA/sinB=a/b,
cosA=(b^2+c^2-a^2)/2bc,cosB=(a^2+c^2-b^2)/2ac
∴L.H.S.
=sinAcosB/sinBcosA
=(a/b){[(a^2+c^2-b^2)/2ac]/[(b^2+c^2-a^2)/2bc]}
=(a/b)[2bc(a^2+c^2-b^2)/2ac(b^2+c^2-a^2)]
=(a^2+c^2-b^2)/(b^2+c^2-a^2)
=R.H.S.

(1-c)^4 +(1+c)^4 =17 c^2
(1-c)^4 -2(1-c)^2(1+c)^2+(1+c)^4 =17 c^2-2(1-c)^2(1+c)^2
[(1-c)^2-(1+c)^2]^2=17 c^2-2(1-c^2)^2
[2(-2c)]^2=17 c^2-2+4c^2-2c^4
16c^2-17 c^2+2-4c^2+2c^4=0
2c^4-5c^2+2=0
(2c^2-1)(c^2-2)=0
c=±√2/2 or
c=±√2

1. Hint.

tanA/tanB
= sinA/cosA x cosB/sinB
= cosB/cosA x sinA/sinB
之後用cos formula and sin formula就計到.

不明白我可以再解.

2007-03-28 01:25:19 補充：
2.本人認為, 用pascal coefficient已經好快. 無需另求他法.即 coefficient 1 4 6 4 1.如果你要求另類計法, 可考慮將17c^2拆成16^2 + c^2:(1-c)^4 - c^2 = 16c^2 - (1+c)^4再用: a^2 - b^2 = (a+b)(a-b)來計. 但不見得快和簡單.