Question

已知向量a=(sinx,根号3cosx),向量b=(cosx,cosx),f(x)=向量a*向量b,求f(x)的解析式和

已知向量a=(sinx,根号3cosx),向量b=(cosx,cosx),f(x)=向量a*向量b,求f(x)的解析式和递增区间

Answer 1

向量a=(sinx,根号3cosx),向量b=(cosx,cosx),
f(x)=向量a*向量b
=sinxcosx+√3cos²x
=1/2sin2x+√3/2(1+cos2x)
=1/2sin2x+√3/2cos2x+√3/2
=sin(2x+π/3)+√3/2
由2kπ-π/2≤2x+π/3≤2kπ+π/2,k∈Z
2kπ-5π/6≤2x≤2kπ+π/6,k∈Z
∴kπ-5π/12≤x≤kπ+π/12,k∈Z
∴f(x)递增区间为[kπ-5π/12,kπ+π/12],k∈Z

希望帮到你，不懂请追问

Answer 2

f(x)=sinxcosx+√3cos²x
=(1/2)sin2x+(√3/2)(cos2x+1)
=(1/2)sin2x+(√3/2)cos2x+√3/2
=sin(2x+π/3)+√3/2
令 -π/2+2kπ≤2x+π/3≤π/2+2kπ，
解得 -5π/12+kπ≤x≤π/12+kπ，
即增区间为[-5π/12+kπ，π/12+kπ]，k∈Z
同理，得减区间为[π/12+kπ，7π/12+kπ]，k∈Z

Answer 3

上面的回答好精彩，学到了。