已知正三角形ABC边长为a,当一点P在三角形ABC的外接圆上的劣弧AB(AB上面有一弧)上移动时,求S三角形PAC +S

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  • 正三角形中心为O,半径r.

    a/sin60=2r

    r=a/2sin60=a/根号3

    设∠PAB=m

    ∠PAO=m+30

    PA=2rcos∠PAO=2acos(m+30)/根号3

    S三角形PAC+S三角形PAB

    =PA*ACsin(m+60)/2+PA*ABsinm/2

    =2acos(m+30)/根号3*asin(m+60)/2+2acos(m+30)/根号3*asinm/2

    =a^2/根号3*[cos(m+30)sin(m+60)+cos(m+30)sinm]

    cos(m+30)sin(m+60)+cos(m+30)sinm

    =(cosmcos30-sinm/2)(sinm/2+cosmsin60)+(cosmcos30-sinm/2)sinm

    =3(cosm)^2/4-(sinm)^2/4+sinmcosmcos30-(sinm)^2/2

    =3/4[(cosm)^2-(sinm)^2]+sin2mcos30/2

    =3cos2m/4+根号3sin2m/4

    =根号3/2[根号3/2cos2m+sin2m/2]

    =根号3/2*sin(2m+60)

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