sinA+sinB-sinC=4sin(A/2)sin(B/2)cos(C/2)
在三角形ABC中.已知sin^2A+sin^2B*sin^2C=sinB*sinC+sinC*sinA+sinA*sin
cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC证明
cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC
为什么sinA-sinB/sinA+sinB=cos[(A+B)/2]sin[(A-B)/2]/{sin[(A+B)/2
在三角形ABC中,求证:sin^A/2+sin^B/2+sin^C/2=1-2sinA/2sinB/2sinC/2
sinb/sina=cos(a+b),证明3sinb=sin(2a+b)
在△ABC中,求证:sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)
11.在△ABC中,求证sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)
证明sin(2a+b)/sina-2cos(a+b)=sinb/sina
求证sin(2A+B)/sinA-2cos(A+B)=sinB/sinA
2sinC-sinA/sinB=cosA-2cosC/cosB 如何整理求得sin(A+B)=2sin(B+C)
三角形ABC中,已知(sin^2 A-sin^2 B-sin^2 C)/(sinB sinC)=1 求A?