化简cos(A-C)

化简cos(A-C) cos(2A)+cos(2C)=0怎么得出2cos(A+C)cos(A-C)=0 求证:a^2(cos^2b-cos^2c)+b^2(cos^c-cos^2a)+c^2(cos^2a-cos^2b)=0 三角形中a/cos A=b/cos B=c/cos C,说明是什么三角形, 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 cos^B-cos^C=sin^A,三角形的形状 cos(a+b)=为什么等于 -cos c 在锐角三角形内cos(B+C)=-cos(A), 求证a cos A + b cos B = c cos （A-B ） cos(a-pai)化简 a cos(B+C)=b cos(A+C)=c cos(A+B) 判断三角形ABC的形状 cos(A)*tan(B)*sin(C) 化简(sin a^2/cos a +cos a)*cot a 化简(sinθ-cosθ)/(tanθ-1)A.tanθ B.sinθ C.-sinθ D.cosθ 化简sin a+cos a 三角形中cos(A/2)^2 + cos(B/2)^2 + cos(C/2)^2 > 2 cos^2A+cos^2B+cos^2C=1三角形ABC是什么形状