已知两点A(cos@,sin@),B(cos&,sin&),则绝对值的AB的最大值是?

已知两点A(cos@,sin@),B(cos&,sin&),则绝对值的AB的最大值是? 在三角形中,已知,cos C/cos B=(3a-c)/b 求：sin B 已知A(a,a^2)、B(b,b^2)(a≠b)两点的坐标,满足a^2sinθ+acosθ=1,b^2sinθ+bcosθ=1a^2sinθ+acosθ-1=0,b^2sinθ+bcosθ-1=0a=(-cosθ+√(cos^2θ+4sinθ)/(2sinθ)b=(-cosθ-√(cos^2θ+4sinθ)/(2sinθ)a+b=-cotθ.ab=-/sinθ/设直线Y=KX+C,坐标(a cos(A)*tan(B)*sin(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)= 已知a为第二象限角,则四个数中必大于零的是（多选题） A.sin(cos a) B.sin(sin a) C.cos(sin a)D.cos(cos a) 已知A(2.0),B(0.2),C(cos a ,sin a),且0 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 高一向量问题.已知向量a=(cosα,sinα),向量b=（cosβ,sinβ）,向量c=（cosγ,sinγ）已知向量a=(cosα,sinα),向量b=（cosβ,sinβ）,向量c=（cosγ,sinγ）且3cosα+4cosβ+5cosγ=0, 3sinα+4sinβ+5sinγ=0.（1）求证向量a 在三角形ABC中,已知sin(B+C/2)=4/5 求cos(A-B) cos²a-cos²b=c,则sin(a+b)sin(a-b)= 若sin²A+sin²B+cos²C sin²A+sin²B=cos²C 已知钝角三角形ABC,A,B为其中的两个钝角,则下列关系正确的是 A.sin A > sin BB.sin A ≤sin B C.sin A ＞cos BD.sinA < cos B 已知(sin A)^2+(sin B)^2+（sin C)^2=1,且A,B,C均为锐角,求cos Acos Bcos C的最大值 已知cosB = cos θ*sinA,cosC = sinθ*sinA.θ为已知角.则sin^2 A + sin^2 B +sin^2C等于多少?