已知向量a=(5根号3cosx,cosx)b=(sinx,2cosx),函数f(x)=ab+b^2,

题目

已知向量a=(5根号3cosx,cosx)b=(sinx,2cosx),函数f(x)=ab+b^2,
当π/6

答案

ab=5根号3sinxcosx+2cos^2x
b^2=sin^2x+4cos^2x+4sinxcosx
f(x)=5根号3sinxcosx+2cos^2x+sin^2x+4cos^2x+4sinxcosx
=sin^2x+(5根号3+4)sinxcosx+6cos^2x
=1+(5根号3+4)sin2x/2+5cos^2x
=1+(5根号3+4)sin2x/2+5cos^2x-5/2+5/2
=7/2+(5根号3+4)sin2x/2+5cos2x/2
=7/2+根号(29+10根号3)sin(2x+a) sina=5/根号(29+10根号3)
π/6