函数f(x)=sin(x+π/3)cos(x-π/3)的最大值。速答，谢谢 ！
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f(x)=0.5*sin(x+π/3+x-π/3)+0.5*sin(x+π/3-x+π/3)
=0.5sin2x+0.5sin(2π/3)
=0.5sin2x+√3/4
当2x=π/2+2kπ 即x=π/4+kπ时 k属于整数
f(x)max=0.5*1+√3/4=0.5+√3/4
=0.5sin2x+0.5sin(2π/3)
=0.5sin2x+√3/4
当2x=π/2+2kπ 即x=π/4+kπ时 k属于整数
f(x)max=0.5*1+√3/4=0.5+√3/4
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f(x)=sin(x+π/3)cos(x-π/3)
=sin(x+π/3)cos(x+π/3-2π/3)
=sin(x+π/3)[cos(x+π/3)(cos2π/3)+sin(x+π/3)sin(2π/3)]
=sin(x+π/3)[(√3/2)sin(x+π/3)-(1/2)cos(x+π/3)]
=[√3-√3cos(2x+2π/3)-sin(2x+2π/3)]/4
=(√3+2sin2x)/4
≤(√3+2)/4
=sin(x+π/3)cos(x+π/3-2π/3)
=sin(x+π/3)[cos(x+π/3)(cos2π/3)+sin(x+π/3)sin(2π/3)]
=sin(x+π/3)[(√3/2)sin(x+π/3)-(1/2)cos(x+π/3)]
=[√3-√3cos(2x+2π/3)-sin(2x+2π/3)]/4
=(√3+2sin2x)/4
≤(√3+2)/4
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