在三角形ABC中,内角A.B.C的对边分别为a,b,c,已知B=C,2b=(根号3)a.----求cos(2A+45

B=C ；2b=√3a；∴ b = √3/2 a
sin(A/2) = (a/2) / b = √3/3
cos(A/2)= √（1-1/3）= √6/3
sinA = 2sin(A/2)cos(A/2) = 2√2/3
cosA = √（1-8/9）=1/3
sin2A = 2sinAcosA = 4√2/9
cos2A = -√（1-32/81）= - 7/9
cos（2A+45°）= cos2Acos45°-sin2Asin45° = √2/2（-7/9-4√2/9）= -(8-7√2)/18