sin^2B+sin^2C-sinBsinC=sin^2A,由正弦定理得
b²+c²-bc=a²
b²+c²-a²=bc=2bccosA
解得cosA=1/2
A=π/3
正弦定理a/2R=sinA
R=1
∴BC=√3
方法一：向量AB•向量AC=AB·AC·cosA
cosA为定值
当AB=AC时值最大,为[√(AB²+AC²)]/2
向量AB•向量AC=AB·AC·cosA
=√3×√3×1/2
=3/2
方法二：向量AB•向量AC=AB·AC·cosA
=2RsinC·2RsinB·cosA
=4R²sinC·sinB·cosA
4R²是定值,cosA是定值.
∴求向量AB•向量AC的最大值就是求sinC·sinB的最大值.
∴sinC·sinB=sinC·sin(2π/3-C)
=sinC(√3/2cosC+1/2sinC)
=√3/2sinCcosC+1/2sin²C
=√3/4sin2C+1/2[(1-cos2C)/2]
=√3/4sin2C+1/4-1/4cos2C
=√3/4sin2C-1/4cos2C+1/4
=1/2(√3/2sin2C-1/2cos2C)+1/4
=1/2(cosπ/6sin2C-sinπ/6cos2C)+1/4
=1/2sin(2C-π/6)+1/4
当C=π/3时,sin(2C-π/6)最大,1/2sin(2C-π/6)+1/4最大,为3/4
∴
向量AB•向量AC=AB·AC·cosA
=2RsinC·2RsinB·cosA
=4R²sinC·sinB·cosA
=4×3/4×1/2
=3/2