设三角形ABC的三个内角分别为A,B,C,三边长分别为a,b,c.求证：(a^2-b^2)/c^2=[sin(A-B)]/sinC
问题描述：
设三角形ABC的三个内角分别为A,B,C,三边长分别为a,b,c.求证：(a^2-b^2)/c^2=[sin(A-B)]/sinC
答
我是从右边算过来的a/sinA=b/sinB=c/sinCsinA=(a*sinC)/c,sinB=(b*sinC)/c[sin(A-B)]/sinC=(sinAcosB-cosAsinB)/sinC=[cosB*(a*sinC)/c-cosA*(b*sinC)/c]/sinC=(a/c)*cosB-b/c*cosA=(a/c)*(a^2+c^2-b^2)/(2ac)-(b/c)...