已知向量a=(cosx,sinx),b=(sin2x,1-cos2x),c=(0,1),x∈(0,π).求函数f(x)=

∵|b|²=sin²2x+(1-cos2x)²
=sin²2x+1+cos²2x-2co2x
=2-2cos2x
=2-2(1-2sin²x)
=4sin²x
又 x∈(0,π)
∴|b|=2sinx
∵(a+b)*c=(cosx+sin2x)×0+(sinx+1-co2x)×1
=sinx+1-cos2x
=sinx+1-(1-2sin²x)
=2sin²x+sinx
∴函数f(x)=|b|-(a+b)*c
=2sinx-(2sin²x+sinx)
=-2sin²x+sinx
=-2(sinx-1/4)²+1/8
函数f(x)可看成是自变量为sinx的二次函数,定义域为(0,1]
且在区间(0,1/4]为增函数,在[1/4,1]内为减函数
所以,当sinx=1,即x=π/2时,函数f(x)取得最小值,最小值f(π/2)=-1