由和差化积公式,cos2C-cos2A=-2sin(C+A)·sin(C-A)=-2sin(π-B)·sin(C-A)=-2sinB·sin(C-A).=-2sinB·(sinC·cosA-cosC·sinA)
则由题意:
-10RsinB·(sinC·cosA-cosC·sinA)=(6a-5b)sinB
1
10R·(sinC·cosA-cosC·sinA)=5b-6a
sinC·cosA-cosC·sinA=(1/2)(b/R)-(3/5)(a/R)
=sinB - (6/5)sinA=sin(π-B) - (6/5)sinA=sin(A+C) - (6/5)sinA=sinA·cosC +cosA·sinC - (6/5)sinA
则 2sinA·cosC = (6/5)sinA
所以 cosC = 3/5
2
cosC = 3/5→cos2C=2cos^2 C -1 = -7/25;
sinC = 4/5.
c=2R·sinC = 8/5;
√[R^2 -(c/2)^2]=3/5;
3/5+R=8/5;
则
S≤(8/5)·C /2=32/25.