Question

已知函数f(x)=cosx(根号3sinx+cosx)-1/2(x∈R).

Answer 1

1、

f(x)=(√3/2)sin2x+(1/2)cos2x

=sin(2x+π/6)

周期T=2π/2=π

当x∈【0,π/2】时,

2x+π/6∈【π/6,7π/6】

则sin(2x+π/6)∈【-1/2,1】

所以,f(x)在区间【0,π/2】上的最大值为1,最小值为-1/2；

2、

即：sin(2x0+π/6)=5/13

x0∈【π/4,π/2】,则2x0+π/6∈【2π/3,7π/6】

所以,cos(2x0+π/6)=-12/13

cos2x0=cos[(2x0+π/6)-π/6]

=cos(2x0+π/6)cos(π/6)+sin(2x0+π/6)sin(π/6)

=(-12/13)(√3/2)+(5/13)(1/2)

=(5-12√3)/26