若sinx+sinz=2siny,求证:tan((x+y)/2)+tan((y+z)/2)=2tan((x+z)/2)

若sinx+sinz=2siny,求证:tan((x+y)/2)+tan((y+z)/2)=2tan((x+z)/2) 已知sinx+siny+sinz=0,cosx+cosy+cosz=0,求证cos(x-y)=cos(y-z)=cos(z-x) sinx+siny+sinz=0;cosx+cosy+cosz=0;求cos(x-y) 一道关于三角函数的不等式证明题已知0≤x＜y＜z≤π求证：（siny-sinx）/(y-x)>(sinz-siny)/(z-y) 已知x,y,z属于(0,派/2),sin^2x+sin^2y+sin^2z=1,求(sinx+siny+sinz)/(cosx+cosy+cosz)的最大值 证明sinx+siny+sinz-sin(x+y+z)=4sin((x+y)/2)sin((x+y)/2)sin((x+y)/2) 求证.(tanx+tany)/(tanx-tany)=(sinx+siny)/(si求证.(tanx+tany)/(tanx-tany)=(sinx+siny)/(sinx-siny) 均值不等式的疑问x+y+z = pi ,求 sinx+siny+sinz 的最大值这题用和差化积做是(3/2)*根号2,但是如果用均值不等式,sinx+siny+sinz>=3(sinxsinysinz)^(1/3).当x=y=z=pi/3时取等,此时最小值是(3/2)*根号2,这是怎么回事 1.已知8cos(2x+y)+5cosy=0,求tan(x+y)*tanx的值.2.已知关于x的方程x^2+px+q=0的两根是tanx,tany,求sin(x+y)/cos(x-y)的值.3.化简:sin(x-y)/(sinx*siny)+sin(y-z)/(siny*sinz)+sin(z-x)/(sinz*sinx). 已知sinx+siny+sinz=0,cosx+cosy+cosz=0 则cos（x-y）=______ 要详解 sinx+siny+sinz=0,cosx+cosy+cosz=0,求cos（y-z）的值. 已知sinx+siny+sinz=cosx+cosy+cosz=0,求tan(x+y+z)+tanxtanytanz的值 求证sin（2x+y)/sinx-2cos(x+y)=siny/sinx 求证：sin(2x+y)/sinx-2cos(x+y)=siny/sinx 已知(tanx)^2=2(tany)^2+1,求证(siny^)2=2(sinx^)2-1 求证sinx+siny=2sin(x+y)/2*cos(x-y)/2 求证:sinx+siny=2sin(x+y)/2cos(x-y)/2 求证|sinx-siny|=|2sin[(x-y)/2]cos[(x+y)/2]|