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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