Question

已知向量a=(sinx/3,cosx/3),向量b=(cosx/3,根号3 cosx/30 ,函数f(x)=向量a*b

（1）将f（x）写成Asin（wx+）的形式，并求其图象对称中心的坐标；
（2）如果△ABC 的三边a，b，c 满足b^2=ac ，且边b 所对的角为x ，试求x 的范围及此时函数f（x) 的值域.

Answer 1

1) f(x)=a*b=sin(x/3)*cos(x/3)+cos(x/3)*√3cos(x/3)
=2cos(x/3)[1/2sin(x/3)+√3/2cos(x/3)]
=2cos(x/3)sin(x/3+π/3)
=sin(x/3+x/3+π/3)-sin(x/胡郑庆3-x/3-π/3)
=sin(2x/3+π/3)+sin(π/3)
=sin(2x/3+π/裤握3)+√3/2
图像对称中心为 2x/3+π/3=kπ+π/2
解得x=3kπ/2+π/4
2) △ABC 的三边a，b，c，b所对应角为x，则0<x<π
由b^2=ac及余弦定理得
b^2=a^2+c^2-2ac*cosB
ac=a^2+c^2-2ac*cosx
cosx=(a^2+c^2-ac)/2ac
≥(2ac-ac)/2ac
=1/2
即cosx≥1/2，∴0<x≤π/3
f(x)=sin(2x/3+π/3)+√3/2
π/3<2x/3+π/3≤5π/9
∵5π/9>π/2, ∴sin(2x/3+π/3)在定义域内的最大值为sin(π/2)=1
又sin(5π/9)=cos(π/2-5π/9)=cos(π/18)>cos(π/6)=cos(π/2-π/3)=sin(π/3)
∴sin(2x/3+π/3)在定义域内的最小值为sin(π/3)=√3/2
∴f(x)=sin(2x/丛握3+π/3)+√3/2的值域为[√3,1+√3/2]

追问

=2cos(x/3)sin(x/3+π/3)
=sin(x/3+x/3+π/3)-sin(x/3-x/3-π/3)
怎么算

追答

三角公式的积化和差公式：2cosαsinβ=sin(α+β)-sin(α-β)

Answer 2

1)
f(x)=a*b=sin(x/3)*cos(x/3)+cos(x/3)*√3cos(x/3)
=2cos(x/3)[1/2sin(x/3)+√3/2cos(x/3)]
=2cos(x/3)sin(x/3+π/3)
=sin(x/3+x/3+π/3)-sin(x/3-x/3-π/3)
=sin(2x/3+π/3)+sin(π/3)
=sin(2x/3+π/3)+√3/2
图像对称中心为
2x/3+π/3=kπ+π/2
解得x=3kπ/2+π/4
2)
△ABC
的三边a，b，c，b所锋数对应角拦基烂为x，则0<x<π
由b^2=ac及余弦简漏定理得
b^2=a^2+c^2-2ac*cosB
ac=a^2+c^2-2ac*cosx
cosx=(a^2+c^2-ac)/2ac
≥(2ac-ac)/2ac
=1/2
即cosx≥1/2，∴0<x≤π/3
f(x)=sin(2x/3+π/3)+√3/2
π/3<2x/3+π/3≤5π/9
∵5π/9>π/2,
∴sin(2x/3+π/3)在定义域内的最大值为sin(π/2)=1
又sin(5π/9)=cos(π/2-5π/9)=cos(π/18)>cos(π/6)=cos(π/2-π/3)=sin(π/3)
∴sin(2x/3+π/3)在定义域内的最小值为sin(π/3)=√3/2
∴f(x)=sin(2x/3+π/3)+√3/2的值域为[√3,1+√3/2]