在△ABC中,若
|
由
=a cos A 2
=b cos B 2
结合正弦定理可得c cos C 2
=sinA cos A 2
=sinB cos B 2
,即sinC cos C 2
=2sin
cosA 2 A 2 cos A 2
=2sin
cosB 2 B 2 cos B 2
,2sin
cosB 2 B 2 cos C 2
化简得sin
=sinA 2
=sinB 2
,又C 2
,A 2
,B 2
∈(0,C 2
)此时正弦函数单调递增,π 2
故
=A 2
=B 2
,又A+B+C=π,故A=B=C=C 2
,即△ABC为等边三角形π 3
故选B.