17.(本小题满分10分)
在△ABC中,角A、B、C的对边分别为a、b、c,且满足(2a-c)cosB=bcosC.
(Ⅰ)求角B的大小;
|
(I)∵(2a-c)cosB=bcosC,∴(2sinA-sinC)cosB=sinBcosC
即2sinAcosB=sinBcosC+sinCcosB=sin(B+C)
∵A+B+C=π,∴2sinAcosB=sinA.∵0<A<π,∴sinA≠0. ∴cosB=
.
∵0<B<π,∴B=
.
(II)
=6sinA+cos2A.=-2sin2A+6sinA+1,A∈(0,
)设sinA=t,则t∈
.
则
=-2t2+6t+1=-2(t-
)2+
,t∈
.∴t=1时,
取最大值.5