sin^2[(B+C)/2]=[1-cos(B+C)]/2
sin^2[(B+C)/2]=[1-cos(B+C)]/2
sin(A+B/2)=cos(C/2)
已知b,c∈{0,π/2},b=sin(cosb),c=cos(sinc),比较b,c大小
在△ABC中,求证:sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1
求证在三角形ABC中,(1)sinA=sin(B+C) (2)cosa=-cos(B+C)
1.在三角形ABC中,已知b²sin²C+c²sin²B=2bccosB×cos
在三角形ABC 中,若sin A:sin B:sin C=3:2:4,则cos C的值
cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC证明
cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC
三角形ABC,求证cos(A+B)=-cosC,cos[(A+B)/2]=sin(C/2)和sin(3A+3B)=sin
在三角形ABC中,已知sin(B+C/2)=4/5 求cos(A-B)
cos(a-b)cos(b-c)+sin(a-b)sin(b-c)=