证：
因为：f(X)=cosx
所以：
4f(π/3-x)×f(x)×f(π/3+x)
=4cos(π/3-x)×cosx×cos(π/3+x)
=4[cos(π/3)cosx+sin(π/3)sinx]×cosx×[cos(π/3)cosx-sin(π/3)sinx]
=4[cos²(π/3)cos²x-sin²(π/3)sin²x]×cosx
=4[(1/4)cos²x-(3/4)sin²x]×cosx
=(cos²x-3sin²x)×cosx
=[cos²x-3(1-cos²x)]×cosx
=(4cos²x-3)×cosx
=4cos³x-3cosx
=cos(3x)
=f(3x)
即：4f(π/3-x)×f(x)×f(π/3+x)=f(3x)
证毕.