(a²+b²)sin(A-B)=(a²-b²)sinC 求三角形abc的形状

4个回答

  • (a^2+b^2)sin(A-B)=(a^2-b^2)sinC

    ∵sinC = sin{180°-(A+B)} = sin(A+B)

    ∴(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B)

    a^2sin(A-B) + b^2sin(A-B) = a^2sin(A+B) - b^2sin(A+B)

    a^2 { sin(A+B)-sin(A-B) } = b^2 { sin(A+B) + sin(A-B) }

    a^2 * 2 cos { [(A+B)+(A-B)]/2 } sin { [(A+B)-(A-B)]/2 } = b^2 * 2 sin{ [(A+B)+(A-B)]/2 } cos { [(A+B)-(A-B)]/2 }

    a^2 * 2 cosA sinB = b^2 * 2 sinA cosB

    a^2/b^2 = sinA cosB/(cosA sinB)

    (sinA/sinB)^2 = sinA cosB/(cosA sinB)

    ∵sinA≠0,sinB≠0,∴两边同乘以sinB/sinA

    sinA/sinB = cosB/cosA

    sinAcosA=sinBcosB

    1/2sin(2A)=1/2(sin2B)

    sin2A=sin2B

    0<2A,2B<180°

    ∴2A=2B,或者2A+2B=180°

    ∴A=B,或A+B=90°

    等腰三角形,或直角三角形