高一数学问题一道...在线等...O(∩_∩)O谢谢

2sinAsinBcosC
=sinA[sin(B+C)+sin(B-C)]
=(sinA)^2+sinAsin(B-C)
=(sinA)^2-(1/2)cos(A+B-C)+(1/2)cos(A-B+C)
=(sinA)^2-(1/2)cos(180°-2C)+(1/2)cos(180°-2B)
=(sinA)^2+(1/2)cos(2C)-(1/2)cos(2B)
=(sinA)^2+(1/2)[1-2(sinC)^2]-(1/2)[1-2(sinB)^2]
=(sinA)^2+(sinB)^2-(sinC)^2,

sin*A+sin*B-2sinAsinBcosC
=(sinC)^2=1/4.

用正弦定理和余弦定理好像更简便些：
(sinA)^2+(sinB)^2-2sinAsinBcosC
=a^2/(4R^2)+b^2/(4R^2)-2(a/2R)(b/2R)[(a^2+b^2-c^2)/(2ab)]
=a^2/(4R^2)+b^2/(4R^2)+(a^2+b^2-c^2)/(4R^2)
=c^2/(4R^2)
=(sinC)^2
=1/4.