帮我证明这题三角

kenny56

在三角形ABC中，证明

(1+cos(A-B)cosC)÷（1+cos(A-C)cosB)=(a^2+b^2)÷(a^2+c^2)

恳请各位大大帮小弟解答这题，

dunwan2tellu

提示 1 :

A + B + C = 180 ==> Cos C = Cos (A+B)

提示 2 :

Cos(A+B)Cos(A-B) = 1/2 * ( Cos 2A + Cos 2B )

提示 3:

Cos 2A = 1 - 2(sin A)^2

提示 4 :

sine rule , a/sin A = b/sin B = c/sin C



用上面的4个提示（照顺序1到4）来想想吧。 想不通的话，其他人再解答

kenny56

谢谢dunwan2tellu版主的提示

(1+cos(A-B)cosC)/(1+cos(A-C)cosB)
=(1-cos(A+B)cos(A-B))/(1-cos(A+C)cos(A-C))
=(1-1/2(cos2A+cos2B))/(1-1/2(cos2A-2C))
=(1-1/2(1-2sin^2A+1-2sin^2B)/(1-1/2(1-2sin^2A+1-2sin^2C)
=(sin^2A+sin^2B)/(sin^2A+sin^2C)
=(a^2sin^2C/c^2+b^2sin^2C/c^2)/(a^2sin^2B/b^2+c^2sin^2B/b^2)
=(sin^2C*b^2(a^2+b^2))/(sin^2B*c^2(a^2+c^2))
=(a^2+b^2)/(a^2+c^2)

[ 本帖最后由 kenny56 于 15-7-2007 04:28 PM 编辑 ]

mathlim

原帖由 dunwan2tellu 于 15-7-2007 01:28 AM 发表
提示 1 :

A + B + C = 180 ==> Cos C = Cos (A+B)

提示 2 :

Cos(A+B)Cos(A-B) = 1/2 * ( Cos 2A + Cos 2B )

提示 3:

Cos 2A = 1 - 2(sin A)^2

提示 4 :

sine rule , a/sin A = b/sin B =  ...

抱歉！无意间发现。

A + B + C = 180° ==> cosC = -cos(A+B)