若三角形ABC中a.b.c分别为内角A.B.C的对边,且1-2sinBsinC=cos2B连下面

1-2sinBsinC=cos2B+cos2C-cos2A
则1-2sinBsinC=(1-2sin^2B)+(1-2sin^2C)-( 1-2sin^2A)
化简得：-2sinBsinC=-2sin^2B-2sin^2C+2sin^2A,
sinBsinC=sin^2B+sin^2C-sin^2A
根据正弦定理得：bc=b^2+c^2-a^2,
所以cosA=( b^2+c^2-a^2)/(2bc)=1/2,
A=60度.
B＋C＝120,
sinB+sinC=sinB+sin(120-B)
=sinB+根号3/2*cosB+1/2*sinB
=根号3/2*cosB+3/2*sinB
=根号3*sin(B+30)
当B=60时,sinB+sinC最大取根号3.无最小值.