设△ABC的内角A,B,C的对应边分别是a,b,c,已知cos2B-cos2A=2sin(60°+B)sin(60°-B)求角A的大小

问题描述:

设△ABC的内角A,B,C的对应边分别是a,b,c,已知cos2B-cos2A=2sin(60°+B)sin(60°-B)
求角A的大小

cos2B-cos2A=2sin(60°+B)sin(60°-B)
cos2B-cos2A=2sin((2×60°+2B)/2)sin((2×60°-2B)/2)
cos2B-cos2A=cos2B-cos120°
cos2A=cos120°
2A=120°
A=60°