Question

已知函数f(x)=2cosxsin(x+π/3)-√3sin^2x+sinxcosx求函数f(x)的最小正周期及最值

Answer 1

f(x)=2cosxsin(x+π/3）-√3sin^2x+sinxcos
=sin(x+π/3+x）+sin(x+π/3-x）-√3(1-cos^2x)+sinxcos
=sin(x+π/3+x）+sin(π/3）-√3+√3cos^2x+0.5sin2x
=sin(2x+π/3）+√3/2-√3+√3cos^2x+0.5sin2x
=sin(2x+π/3）-√3/2+√3*(1+cos2x)/2+0.5sin2x
=sin2x*cos(π/3）+cos2x*sin(π/3）+(√3/2)cos2x+0.5sin2x
=0.5sin2x+(√3/2)cos2x+(√3/2)cos2x+0.5sin2x
=sin2x+(√3)cos2x
=(2/2)*[sin2x+(√3)cos2x]
=2*[(1/2)*sin2x+(√3/2)cos2x]
=2*[sin(π/6)*sin2x+cos(π/6)cos2x]
=2cos(2x-π/6)

到这里最值周期自己求吧

Answer 2

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