a=sin 2,b=cos a,c=tan b 比大小?

cos(a-b)cos(b-c)+sin(a-b)sin(b-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 cos²a-cos²b=c,则sin(a+b)sin(a-b)= 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 sin²A+sin²B=cos²C sin（A+B/2）=cos（C/2） 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 cos平方a-cos平方b=A -cos｛a+b｝乘以cos｛a-b｝Bcos｛a+b｝乘以cos｛a-b｝C-sin｛a+b｝乘以sin｛a-b｝Dsin｛a+b｝乘以sin｛a-b｝要过程 若sin a+sin b+ sin c=0,cos a+cos b+cos c=0,求cos(a-b) 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 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 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] 求证cos(a+b)cos(a－b)=cos^2b－sin^2a 求证cos(a+b)cos(a-b)=cos^2a-sin^2b 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)