cos(A+B)+sin(A-B)=(cosA+sinA)(cosB-sinB)

cos(a-b)cos(b-c)+sin(a-b)sin(b-c)= cos²a-cos²b=c,则sin(a+b)sin(a-b)= 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^B-cos^C=sin^A,三角形的形状 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 sin²A+sin²B=cos²C 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 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) cos(A)*tan(B)*sin(C) 在三角形中,已知,cos C/cos B=(3a-c)/b 求：sin B sin（A+B/2）=cos（C/2） 如何证明sin(a+b)=sin(a)cos(b)+cos(α)sin(b) 若sin²A+sin²B+cos²C 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^2a-cos^2b=m,那么sin(a+b)sin(a-b)=? cos∧a-cos∧b=t则sin(a+b)sin(a-b)= 证明sin(A+B)sin(A-B)=cos^2B-cos^2A