问题补充：
已知三角形ABC的三个内角A,B,C成等差数列,求(sinA)*2+(sinC)*2的取值范围
答案：
三角形ABC的三个内角A,B,C成等差数列 则A+C=2B 因为A+B+C=180° 3B=180° 所以B=60° A+C=120° (sinA)^2+(sinC)^2 =(sinA+sinC)^2-2sinAsinC ={2sin[(A+C)/2]cos[(A-C)/2]}^2+cos(A+C)-cos(A-C) =3{cos[(A-C)/2]}^2-1/2-cos(A-C) =3[1+cos(A-C)]/2-1/2-cos(A-C) =1+[cos(A-C)]/2 -120°
