cos(2A)+cos(2C)=0怎么得出2cos(A+C)cos(A-C)=0

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 cos相乘变cos相加?!怎么转换过去的呀在?奇特的数学.cos(b+a) * cos(c+a) = 1/2cos(2a+b+c) + 1/2cos(b-a)这怎么弄的.简直奇妙了. 2cos(A-B)+2cos（B-C)+2cos（C-A）=-3 求cos（2A-2C）由判别式得：cos[(x-y)/2]>=1 没看懂~ 在三角形ABC中,为什么1-cos^2A-cos^2B-cos^2C-2cosAcosBcosC=0成立? 求证cos^2A+cos^2B+cos^2C+2*cosA*cosB*cosC=1 cos^2A+cos^2B+cos^2C=1三角形ABC是什么形状 求证cos(a+b)*cos(a-b)=(cosa)^2-(cosb)^2从(cosacosb)²-(1-cos²a)(1-cos²b) 怎么得到cos²a-cos²b 不是cos²a+cos²b-1吗？ 2cos^2(a)-cos(a)-1=0 4cos^a-2(1+根号3)cos+根号3=0怎么解 sin(45+a)cos(45+a)怎么得=1/2sin(90+2a)? 4（cos+1）方-2cos方=4 怎么化成cos方+4cos=0 在三角形ABC中求证sinA+sinB+sinC=4cos（A/2)COS(B/2)COS(C/2)证到这步然后怎么证:原式=4sin(B/2)COS(B/2)COS^2(C/2)+4sin(C/2)COS(C/2)COS^2(B/2)=? 从这步开始, 复制的别来步骤不要跳！！！ 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 证明(cosC+cosA)=2cos[(A+C)/2][cos(A-C)/2] 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 1－cosAcosB－cos²(C/2)＝0 ∵cos²(C/2)＝(cosC＋1)/2 怎么得来的呢 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