已知函数f(x)=cos^2(x+π/12)+sinxcosx.求f(x)的最值.

网友回答

f(x)=[1+cos(2x+π/6)]/2+(sin2x)/2
=1/2+1/2[sin(π/3-2x)+sin2x]
=1/2+sin(π/6)cos(π/6-2x)
=1/2+1/2* cos(2x-π/6)
cos(2x-π/6)=1,fmax=1/2+1/2=1
cos(2x-π/6)=-1,fmin=1/2-1/2=0