Question

已知向量a=（2根号3sinx,cosx+sinx)，b=(cosx，cosx-sinx），函数f(x)=a·b，求f(

（2）求f(x)的单调增区间，（3）若x属于【0,pai/2],求函数f(x)的值域

Answer 1

解
a=(2√3sinx,cosx+sinx),b=(cosx,cosx-sinx）
f(x)=a·b=2√3sinxcosx+cos^2x-sin^2x
=√3sin2x+cos2x
=2sin(2x+π/6)
令2kπ-π/行察2<=2x+π/6<=2kπ+π/2
解得kπ-π/3<=x<=kπ+π/6
∴f(x)的单调增区间唤中为[kπ-π/3,kπ+π/和带山6]
x∈[0,π/2],2x+π/6∈[π/6,7π/6]
∴-1/2<=f(x)<=1
∴函数f(x)的值域是[-1/2,1]

追问

求f(x)的解析式及周期

追答

f(x)=a·b
=2√3sinxcosx+cos^2x-sin^2x
=√3sin2x+cos2x
=2sin(2x+π/6)

周期T=2π/2=π