在△ABC中，角A、B、C的对边分别为a、b、c.求证:(a^2-b^2)/c^2=sin(A-B)/sinC.

显示全部楼层 · 2007-8-2 19:12:29

90度等于160度等于（a*sinb）/(ab)^2

千问 · 2007-8-2 19:12:29

(a^2-b^2)/c^2=(a+b/c)(a-b/c)根据正弦定理：(a+b/c)(a-b/c)=(sinA+sinB/sinC)(sinA-sinB/sinC)分别处理，用和化为积公式：sinA+sinB/sinC=2sin(A+B/2)cos(A-B/2)/sin(A+B)=2sin(A+B/2)cos(A-B/2)/2sin(A+B/2)cos(A+B/2)=cos(A-B/2)/cos(A+B/2)同理：a-b/c=sin(A-B/2)/sin(A+B/2)所以原式=sin(A-B/2)cos(A-B/2)/sin(A+B/2)cos(A+B/2)=sin(A-B)/sin(A+B)=sin(A-B)/sinC