Question

已知向量m=(2√3sinx/4，2)，向量n=(cosx/4，cos^2x/4)

(1)若向量m*n=2,求cos(x+π/3)的值
(2)记f(x)=向量m*n，在△ABC中，角A,B,C的对边分别是a,b,c,且满足
(2a-c)cosB=bcosC,求f(x)的取值范围

Answer 1

(1)
m*n=2√3sinx/4cosx/4+2cos^2x/4
=√3sinx/2+2cos^2x/4-1+1
=√3sinx/2+cosx/2+1
=2sin(x/2+π/6)+1

2sin(x/2+π/6)+1=2
sin(x/2+π/6)=1/2

cos(x+π/3)=cos2(x/2+π/6)=1-2sin^2(x/2+π/6)=1/2

(2)
f(x)的取值范围[-1,3]