解
a=(sinωx,-cosωx),b=(sinωx,-3cosωx),c=(-cosωx,sinωx)
f(x)=a·(b+c)=(sinωx,-cosωx)(sinωx-cosωx,-3cosωx+sinωx)
=sinωx*sinωx-sinωx*cosωx+3cosωxcosωx-cosωx*sinωx
=1+2cosωxcosωx-2sinωx*cosωx
=cos2ωx-sin2ωx+2
=√2cos(2ωx+π/4)+2
當(dāng)cos(2ωx+π/4)=1
f(x)取得最大值√2+2
∵P,Q兩點(diǎn)間距離|PQ|的最大值是π,即周期為π,T=2π/2ω=π,解得ω=1
∴f(x)=√2cos(2x+π/4)+2
f(π/6)=√2cos(2π/6+π/4)+2=√2[(1/2)*(√2/2)-(√3/2)*(√2/2)]+2=(5-√3)/2