若sin/a=cos/b=cos/c,则△ABC是sinA/a=cosB/b=cosC/c,写错了,不好意思

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) 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 若sin/a=cos/b=cos/c,则三角形是? 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 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 cos(a-b)cos(b-c)+sin(a-b)sin(b-c)= sin²A+sin²B=cos²C .若f(x)=sinα－cosx,则f′(α)等于 A.sinα B.cosα C.sinα+cosα D.2sinα 若cos(A+B)cos(A-B)=1/3 则cos平方A－sin平方B 等于 若cos(a+b)cos(a-b)=1/3,则cos²a-sin²b的值 cos平方B-cos平方C=sin平方A,则此三角形的形状 sin a sin b +cos a cos b =0,则sin a cos a+sin b cos b的值 在三角形ABC 中,若sin A:sin B:sin C=3:2:4,则cos C的值 cos∧a-cos∧b=t则sin(a+b)sin(a-b)= 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 sin(a-b)*cos a-cos(a-b)*sin a=1/5,则cos 2b的值是