Question

设函数f（X)=cos（2x+π/3)+sin方x。求函数f(x)的最大值和最小正周期

设函数f（X)=cos（2x+π/3)+sin方x。设A、B、C为△ABC的三个内角，若cosB=1/3，f（C/3)=-1/4，且C为锐角，求sinA

Answer 1

1）f(X)=cos(2x+π/3)+(sinx)^2

     =cos(2x)cos(π/3)-sin(2x)sin(π/3)+1-(cosx)^2

     =1/2(2(cosx)^2-1)-√3/2*sin(2x)+1-(cosx)^2

     =(cosx)^2-1/2-√3/2*sin(2x)+1-(cosx)^2

     =1/2-√3/2sin(2x)

∴当sin(2x)=-1时，f(x)取得最大值1/2+√3/2；最小正周期为T=2π/2=π

Answer 2

设函数f(X)=cos（2x+π/3)+sin²x。A、B、C为△ABC的三个内角，若cosB=1/3，f（C/3)=-1/4，且C为锐角，求sinA
解：f(x)=cos(2x+π/3)+sin²x=(1/2)cos2x-(√3/2)sin2x+(1-cos2x)/2=-(√3/2)sin2x+1/2
f(C/3)=-(√3/2)sin(2C/3)+1/2=-1/4
故sin(2C/3)=√3/2，∴2C/3=60°，∴C=90°，（题错！C不是锐角！）A=90°-B,
故sinA=sin(90°-B)=cosB=1/3.

Answer 3

1）f(X)=cos(2x+π/3)+(sinx)^2
=cos(2x)cos(π/3)-sin(2x)sin(π/3)+1-(cosx)^2
=1/2(2(cosx)^2-1)-√3/2*sin(2x)+1-(cosx)^2
=(cosx)^2-1/2-√3/2*sin(2x)+1-(cosx)^2
=1/2-√3/2sin(2x)
∴当sin(2x)=-1时，f(x)取得最大值1/2+√3/2；最小正周期为T=2π/2=π
2）∵C为锐角，∴0<C<π/2，0<2C/3<π/3
f(C/3)=1/2-√3/2sin(2C/3)=-1/4
=>sin(2C/3)=(1/2+1/4)/(√3/2)=√3/2
=>2C/3=π/3=>C=π/2 与已知矛盾，为什么？