Question

向量a=(sinx/3,cosx/3)向量b= (cosx/3.根号3cosx/3)函数f(x)=向 量a*向量b

Answer 1

f(x)=sin((2x+pi)/3)+sqrt(3)/2
对应的减区间即可求得
b^2=a^2+c^2-2accos(x)
a^2+c^2-ac=2accos(x)
cosx>0
a^2+c^2-ac>=ac
2accos(x)>=ac
cosx>=1/2
x>=π/3
x<=π/2

Answer 2

向量a=(sin(x/3)，cos(x/3))向量b= (cos(x/3)，(√3)cos(x/3))，求函数f(x)=a•b
解：f(x)=a•b=sin(x/3)cos(x/3)+(√3)cos²(x/3)=(1/2)sin(2x/3)+(√3/2)[1+cos(2x/3)]
=sin(2x/3)cos(π/3)+cos(2x/3)sin(π/3)+√3/2=sin[(2x/3)+π/3] +√3/2

Answer 3

f(x)
=ab
=sinx/3cosx/3-√3cos²x/3
=1/2sin2x/3-√3/2(1+cos2x/3)
=1/2sin2x/3-√3/2cos2x/3-√3/2
=sin(2x/3-π/3)-√3/2