在△ABC中,猜想T=sinA+sinB+sinC的最大值,并证明.

4个回答

  • ∵sinA+sinB+sinC=2sin[(A+B)/2]cos[(A-B)/2]+2sin(C/2)cos(C/2)

    =2sin[(180°-C)/2]cos[(A-B)/2]+2sin(C/2)cos(C/2)

    =2sin(90°-C/2)cos[(A-B)/2]+2sin[(180°-A-B)/2]cos(C/2)

    =2cos(C/2)cos[(A-B)/2]+2sin[90°-(A+B)/2]cos(C/2)

    =2cos(C/2){cos[(A-B)/2]+cos[(A+B)/2]}

    =4cos(C/2)cos{[(A-B)/2+(A+B)/2]/2}cos{[(A-B)/2+(A+B)/2]/2}

    =4cos(C/2)cos(A/2)cos(B/2)

    显然,cos(C/2)、cos(A/2)、cos(B/2)都是正数,

    ∴cos(C/2)cos(A/2)cos(B/2)

    ≤{[cos(C/2)]^3+[cos(A/2)]^3+[cos(C/2)]^3}/3

    当cos(C/2)=cos(A/2)=cos(B/2)时,cos(C/2)cos(A/2)cos(B/2)有最大值.

    由cos(C/2)=cos(A/2)=cos(B/2)得:A=B=C=60°,

    ∴此时{[cos(C/2)]^3+[cos(A/2)]^3+[cos(C/2)]^3}/3

    =(cos30°)^3=(√3/2)^2=3√3/8.

    ∴cos(C/2)cos(A/2)cos(B/2)的最大值是3√3/8.

    得:4cos(C/2)cos(A/2)cos(B/2)的最大值是3√3/2.

    即:sinA+sinB+sinC的最大值是3√3/2.