求函数ƒ(x)=cosx[(cosπ/5)2-1/2]-（1/2）sinxsin(2π/5)的最小值．

求函数ƒ(x)=cosx[(cosπ/5)²-1/2]-（1/2）sinxsin(2π/5)的最小值．
【正确答案】：ƒ(x)=(1/2)cosx(2cos²π/5-1)-(1/2)sinxsin(2π/5) =1/2(cosxcos2π/5-sinxsin2π/5) =1/2cos(x+2π/5) 当cos(x+2π/5)=-1时, ƒ (x)取最小值-(1/2)．