Question

△ABC中,a、b、c分别为角A、B、C的对边,S为△ABC的面积,且4sinBsin²(+)+cos2B=1+.

(1)求角B的大小;

(2)若a=4,S=8,求b的值.

Answer 1

解:(1)由4sinBsin²(+)+cos2B=1+,

可得4sinB×+cos2B=1+,即2sinB(1+sinB)+1-2sin²B=1+,

解得sinB=.                                                            

又0＜B＜π,

∴B=或.                                                            

(2)由S=acsinB=×4×c×=8,解得c=8.                            

当B=时,b===4;                    

当B=时,b===4.

∴b=4或b=4.