Question

已知函数f(x)=2sinxcos(π/2-x)-√3sin(π+x)cosx+sin(π/2+x)cosx (1)求函数y=f(x)最小正周期和最值 (2)把

要过程啊

Answer 1

f(x)=2sinxcos(π/2-x)-√3sin(π+x)cosx+sin(π/2+x)cosx
=2sinxsinx+√3sinxcosx+cosxcosx
=2sin^2x+√3/2*sin2x+cos^2x
=sin^2x+√3/2*sin2x+1
=(1-cos2x)/2+√3/2*sin2x+1
=√3/2*sin2x-1/2cos2x+1/2+1
=√3/2*sin2x-1/2cos2x+3/2
=sin2xcosπ/6-cos2xsinπ/6+3/2
=sin(2x-π/6)+3/2
T=2π/2=π

-1<=sin(2x-π/6)<=1
1/2<=sin(2x-π/6)+3/2<=5/2

函数f(x)的最小值为：1/2
函数f(x)的最大值为：5/2

Answer 2

先化简吧！f(x)=2sinxsinx+根号3sinxcosx+cosxcosx＝1+sinx(sinx+根号3cosx)1+2sinxsin(x+pai/3)，你题不是写错了吧？不应该是这样啊！