问题补充：
已知三角形ABC中的内角A,B,C的对边分别为a,b,c,定义向量m=(2sinB,-根号3),定量n=(cos2B,2cos平方B/2-1)且定量m//定量n （1）求角B的大小; (2)如果b=2,求三角形ABC的面积的最大值是?
答案：
(1)m//n=>2sinB/(-√3 ) = cos2B/(2[cos(B/2)]^2-1)2sinB(2[cos(B/2)]^2-1) =-√3cos2Bsin2B = -√3cos2Btan2B = -√3B = π/3(2)b=2by sine-rulea/sinA = b/sinBa= (b/sinB)sinA = (4√3/3)sinAc= (b/sinB)sinC = (...
======以下答案可供参考======
供参考答案1:
楼下的错了 2sinB/(2[cos(B/2)]^2-1)= (-√3 ) cos2B
2BcosB +√3cos2B=0
2sin（2B+π/3）=0
因为在三角形中 所以 2B+π/3=0 B=Kπ/2-π/6B=π/3或π5/6
