1.m·n=√3sin(x/4)cos(x/4)+cos²(x/4)
=(√3/2)sin(x/2)+(1/2)cos(x/2)+1/2
=cos(x/2-π/3)+1/2=1
cos(x/2-π/3)=-1/2.x/2-π/3=±2π/3+2kπ,x/2= ±2π/3+2kπ+ π/3
x=±4π/3+4kπ+ 2π/3, x+π/3=±4π/3+4kπ+ π
cos(∏/3+x)=1/2
2,f(A)=cos(∠A/2-π/3)+1/2
(2a-c)cosB=bcosC,从正弦定理,(2sinA+sinC)cosB=sinBcosC
可得sinA(2cosB-1)=0 sinA≠0,2cosB-1=0.∠B=π/3
0<∠A<2π/3.-π/3<[∠A/2-π/3]<0.-1/2<cos[∠A/2-π/3]<0
0<f(A)<1/2.
m*n=1
化简:
(√3sin(x/4),1)*(cos(x/4),(cos(x/4))^2)=1
√3sin(x/4)*cos(x/4)+ [cos(x/4)]^2 =1
(√3/2)sin(x/2) +(1/2) cos(x/2) +1 /2=1
(√3/2)sin(x/2)+(1/2)cos(x/2)= 1/2
sin(x/2+π/6)=1/2
[sin(x/2+π/6)]^2=1/4
利用公式(1-2sin^2a=cos2a):
∴[sin(x/2+π/6)]^2=1/2-1/2cos(x+π/3)=1/4
∴cos(x+π/3)=(1/4-1/2)*(-2)=1/2