2sin(x+y)*siny+cos(x+2y)可以化简为什么?请尽快回答

sin(x+y)cosy-cos(x+y)siny求解释 证明cosx-cos(x+2y)/2siny= sin(x+y)证明cosX-cos(x+2y)/2siny= sin(x+y) 证明sinx+cos(x+y)siny/cosx-sin(x+y)siny=tan(x+y) 请问,如何证明sinx+siny=2*sin(x+y/2)*cos(x-y/2) 为什么sin（x+y) - sinx = 2cos(x+y/2)siny/2 求推导sinx-siny=2cos（(x+y)/2)sin((x-y)/2) 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]| sin(x+y)-sinx=2siny/2·cos(x+y/2)怎么来的 证明 [sin(2x+y)/sinx]-2cos(x+y)=siny/sinx 求证sin（2x+y)/sinx-2cos(x+y)=siny/sinx 求证：sin(2x+y)/sinx-2cos(x+y)=siny/sinx 2sin(x+y)*siny+cos(x+2y)可以化简为什么?请尽快回答 cos(x+y)*cos(x-y)=cos*cosx+sin*siny 讨论大学函数连续性sin(x+y)-sinx=2siny/2 * cos(x+y/2)sin(x+y)-sinx=2siny/2 * cos(x+y/2)如何推导?y=delta(x) 已知x-y=派/3,且sin x-siny=1/2,求cos(x+y). sin x *cos y-cos x*siny 用Mathematica怎么化简