已知A(a,0)B(0,b)C(cos,α sinα)三点共线,a,b大于0,α属于0,π/2,求1/a^2+1/b^2的最小值.c点是cosα,sinα

求证:a^2(cos^2b-cos^2c)+b^2(cos^c-cos^2a)+c^2(cos^2a-cos^2b)=0 已知A(2.0),B(0.2),C(cos a ,sin a),且0 已知三角形ABC的三个内角A,B,C的对边分别为a,b,c求证(a^2-b^2)/(cos A+ cos B)+(b^2-c^2)/(cos B+cos C)+(c^2-a^2)/(cos A+cos C)=0 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 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 下列不等式正确的是（ ）如果0＜m＜b＜a,那么,下列不等式成立的是（ ）A,cos(b+m/a+m)＜cos(b/a)＜cos(b-m/a-m)B,cos(b/a)＜cos(b-m/a-m)＜cos(b+m/a+m)C,cos(b-m/a-m)＜cos(b/a)＜cos(b+m/a+m)D,cos(b+m/a+m)＜cos(b-m/a-m)＜c 在三角形中,已知,cos C/cos B=(3a-c)/b 求：sin B 若sin a+sin b+ sin c=0,cos a+cos b+cos c=0,求cos(a-b) 已知a>b>0,c 已知a>b>0,c 已知a>b>0,c 已知a>b>0,c 已知a>b>0,c 已知a>b>0,c 已知a>b>0,c 已知a>b>0,c 已知a>b>0,c