sin(B+C)=?cos(B+C)=?

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 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 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 若sin a+sin b+ sin c=0,cos a+cos b+cos c=0,求cos(a-b) sin（A+B/2）=cos（C/2） 在三角形ABC,如果tanB=cos(C-B)/sinA+sin(C-B)则cos(B+C)=? matlab初学者,麻烦给解个三角函数的方程,我的怎么没有解呀?syms a b c>> [a,b,c]=solve('cos(a)*sin(b)*cos(c)+sin(a)*sin(c)=0.2082','sin(a)*sin(b)*cos(c)-cos(a)*sin(c)=0.71937','cos(b)*cos(c)=0.6691') cos平方a-cos平方b=A -cos｛a+b｝乘以cos｛a-b｝Bcos｛a+b｝乘以cos｛a-b｝C-sin｛a+b｝乘以sin｛a-b｝Dsin｛a+b｝乘以sin｛a-b｝要过程 sinA=sin(B+C)怎么来的?cosA=-cos(B+C)怎么来的? 在三角形ABC中,tanA=-3/4求sin(B+C）cos(B+c) sin^2[(B+C)/2]=[1-cos(B+C)]/2 -2sin(B-C)cos(B+C)=sin2C-sin2B怎么化的?