cos (A/2)=2sin[(C-B)/2],求证sinC-sinB=1/2sinA

sin（A+B/2）=cos（C/2） 三角形ABC,求证cos(A+B)=-cosC,cos[(A+B)/2]=sin(C/2)和sin(3A+3B)=sin(3C),sin[(3A+3B)/2]=-cos[(3C)/2] cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC证明 cos^2(B)-cos^2(C)=sin^2(A),问三角形形状, 证明 sin^2A+sin^2B-sin^2A*sin^2B+cos^2A*cos^2证明 sin^2A+sin^2B-sin^2A*sin^2B+cos^2A*cos^2B=1 在三角形ABC 中,若sin A:sin B:sin C=3:2:4,则cos C的值 cos^2a-cos^2b=m,那么sin(a+b)sin(a-b)=? 证明sin(A+B)sin(A-B)=cos^2B-cos^2A 在三角形ABC中,sin²a+sin²C-sinAsinC=sin²B,求2cos²A+cos（A-C）的范围 △ABC的内切圆半径为R,外接圆半径为R,则r/(4R)得知等于A.sin（A/2)sin(B/2)sin(C/2)B.cos（A/2)cos(B/2)sin(C/2)C.sin（A/2）cos(B/2)cos(C/2)D.sin（A/2）sin(B/2)cos(C/2)求详解,最好标明公式的值 sinA+sinB=2sin[(A+B)/2]cos[(A 求证cos(a+b)cos(a－b)=cos^2b－sin^2a 求证cos(a+b)cos(a-b)=cos^2a-sin^2b 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)= .若f(x)=sinα－cosx,则f′(α)等于 A.sinα B.cosα C.sinα+cosα D.2sinα sin(a-b)*cos a-cos(a-b)*sin a=1/5,则cos 2b的值是 由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a sin b解题设a为锐角,证：1、2分之根3乘cos a + 2分之1乘sin a=cos(6分之π-a)2、cos a-sin a=根号2cos(4分之π+a) a=sin 2,b=cos a,c=tan b 比大小?