Question

在△ABC中,角A、B、C所对的边是a,b,c,且a²+c²-b²=ac.

(1)求sin²+cos2B的值;

(2)若b=2,求△ABC面积的最大值.

Answer 1

解:(1)∵a²+c²-b²=AC,?

∴cosB==.                                                                                      ?

∴sin²+cos2B=［1-cos(A+C)］+(2cos²B-1)= (1+cosB)+(2cos²B-1). ?

=(1+)+(2×-1)=- .                                                                                   ?

(2)由cosB=得sinB=.                                                                                   ?

∵b=2,

∴a²+c²=AC+4≥2AC.(当且仅当a²=c²=时取“=”)∴AC≤.                                  ?

∴S_△ABC=AC·sinB≤××=.?

故△ABC面积的最大值为.