(1) f(x)= 1 2 - 3 2 sin2ωx-co s 2 ωx= 1 2 - 3 2 sin2ωx- 1+cos2ωx 2 = -( 3 2 sin2ωx+ 1 2 cos2ωx)=-sin(2ωx+ π 6 ) . (3分) ∵ T= 2π 2ω =4π ,∴ ω= 1 4 .(5分) (2)∵ 2a-c b = cosC cosB ,∴ (2a-c)cosB=bcosC , (2sinA-sinC)cosB=sinBcosC , ∴ 2sinAcosB=sinBcosC+cosBsinC=sin(B+C)=sinA .(7分) ∵sinA≠0,∴ cosB= 1 2 ,∴ B= π 3 .(10分) ∴ f(A)=-sin( 1 2 A+ π 6 ),0<A< 2π 3 ,∴ π 6 < A 2 + π 6 < π 2 , ∴ 1 2 <sin( A 2 + π 6 )<1 ,∴ f(A)∈(-1,- 1 2 ) .(12分)
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