Question

求解下题

Answer 1

向量m•n=√3sin(x/4)cos(x/4)
+ 1•cos²(x/4)
=(√3/2)sin(x/2) + [1+cos(x/2)]/2
=(√3/2)sin(x/2) + (1/2)cos(x/2) + 1/2
=sin(x/2 + π/3) + 1/2=1
∴sin(x/2 + π/3)=1/2
则cos(x/2 + π/3)=±√3/2
cos(x + 2π/3)=cos[2(x/2 + π/3)
=1 - 2sin²(x/2 + π/3)=1/2
sin(x + 2π/3)=±√3/2
cos[-(x - 2π/3)]=cos(x - 2π/3)
=cos[(x + 2π/3) - 4π/3]
=-1或1/2

追答

由已知：f(A)=sin(A/2 + π/3) + 1/2
根据正弦定理：
(2sinA - sinC)cosB=sinBcosC
2sinAcosB - sinCcosB=sinBcosC
2sinAcosB=sinBcosC + sinCcosB
2sinAcosB=sin(B+C)
2sinAcosB=sin(π-A)
2sinAcosB=sinA
sinA(2cosB - 1)=0
∴sinA=0或cosB=1/2
∵A和B是△ABC的内角
∴B=π/3
∴0<A<2π/3
则0<A/2<π/3
π/3<A/2 + π/3<2π/3
∴f(A)∈((1+√3)/2，3/2]