若角a属于[-丌/12,丌/12],则y=cos(a+丌/4)十sin2a函数的最小值是?

1个回答

  • sin2a=2sinacosa=(sina+cosa)²-1=2(√2/2sina+√2/2cosa)²-1

    =2sin²(a+π/4)-1=2-2cos²(a+π/4)-1=1-2cos²(a+π/4)

    或sin2a=-cos(π/2+2a)=-[2cos²(π/4+a)-1]=1-2cos²(a+π/4)

    y=cos(a+π/4)十sin2a=cos(a+π/4)+1-2cos²(a+π/4)

    =-2[cos²(a+π/4)-2*1/4cos(a+π/4)+1/16]+9/8

    =-2[cos(a+π/4)-1/4]²+9/8

    a属于[-丌/12,丌/12]

    cos(a+π/4)>0

    故在a=-π/12时,

    cos(a+π/4)值最大

    cos(-π/12+π/4)=cosπ/6=√3/2

    原函数的最小值是 -2[√3/2-1/4]²+9/8=(√3-1)/2

    或 cos(--π/12+π/4)十sin(-π/6)=cosπ/6-sinπ/6=(√3-1)/2