cos(a-b)cos(b-c)+sin(a-b)sin(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 cos(a-b)cos(b-c)+sin(a-b)sin(b-c)= cos(A)*tan(B)*sin(C) cos^B-cos^C=sin^A,三角形的形状 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 cos²a-cos²b=c,则sin(a+b)sin(a-b)= 若sin a+sin b+ sin c=0,cos a+cos b+cos c=0,求cos(a-b) 若sin²A+sin²B+cos²C sin²A+sin²B=cos²C 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 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 化简COS(a+B)COS(a-B)+sin平方B 化简cos(a+b)cos(a-b)+sin^2b sin（A+B/2）=cos（C/2） matlab符号运算,系数的提取问题.比如一个符号多项式,A*cos(a)*cos(b)+B*sin(a)*sin(b)+C*cos(a)+D*sin(b),这个表达式中的变量包括4个,cos(a)*cos(b)、sin(a)*sin(b)、cos(a)、sin(b),要找到它们的系数A B C D,该怎么找 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') 求证cos(a+b)cos(a－b)=cos^2b－sin^2a