如果sin(C/2)=cos[(A+C)/2] 则三角形ABC等腰三角形请问为什么..谢了

非线性方程解析解-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0-x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos(b)-y0*sin(a)*co 如果tan a=2,那么sin a*sin a+sin a*cos a+cos a*cos a的值是（ ）A:7/3B:7/5C:5/4D:5/3 如果sin(C/2)=cos[(A+C)/2] 则三角形ABC等腰三角形请问为什么..谢了 求非线性方程组的“解析解”-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0 -x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos sin（A+B/2）=cos（C/2） cos^B-cos^C=sin^A,三角形的形状 sin²A+sin²B=cos²C cos(a-b)cos(b-c)+sin(a-b)sin(b-c)= 1.cos^2 73°+cos^2 47°+cos73°cos47°=多少?2.如果A+B+C=π ,求证：cosA+cosB+cosC=1+4sin A/2 sin B/2 sin C/2 在三角形ABC 中,若sin A:sin B:sin C=3:2:4,则cos C的值 在三角形ABC中,sin²a+sin²C-sinAsinC=sin²B,求2cos²A+cos（A-C）的范围 三角形ABC,求证cos(A+B)=-cosC,cos[(A+B)/2]=sin(C/2)和sin(3A+3B)=sin(3C),sin[(3A+3B)/2]=-cos[(3C)/2] cos²a-cos²b=c,则sin(a+b)sin(a-b)= cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC证明 cos^2(B)-cos^2(C)=sin^2(A),问三角形形状, 求证sinA+sinC=2*sin[A+C)/2]*cos[(A-C)/2] 为什么sinA+sinC=2sin[(A+C)/2]cos[(A-C)/2], 证明sinA+sinC=2sin（A+C）/2 * cos（A-C)/2