-2sin(B-C)cos(B+C)=sin2C-sin2B怎么化的?

cos(a-b)cos(b-c)+sin(a-b)sin(b-c)= sin(B+C)=?cos(B+C)=? 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 sin（A+B/2）=cos（C/2） 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 cos^B-cos^C=sin^A,三角形的形状 sin²A+sin²B=cos²C cos²a-cos²b=c,则sin(a+b)sin(a-b)= 在三角形中,已知,cos C/cos B=(3a-c)/b 求：sin B sin^2[(B+C)/2]=[1-cos(B+C)]/2 请问：为什么sinB+sinC=2sin[(B+C)/2]cos[(B-C)/2]. -2sin(B-C)cos(B+C)=sin2C-sin2B怎么化的? 数学三角函数的提A,b,c,∈(0,90) ,sin a + sin c = sin b ,cos b + cos c= cos a ,则b-a 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] 若sin a+sin b+ sin c=0,cos a+cos b+cos c=0,求cos(a-b) 在三角形ABC 中,若sin A:sin B:sin C=3:2:4,则cos C的值 已知b,c∈{0,π/2},b=sin(cosb),c=cos(sinc),比较b,c大小 cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC证明