求推导sinx-siny=2cos（(x+y)/2)sin((x-y)/2)

求推导sinx-siny=2cos（(x+y)/2)sin((x-y)/2) 两角和差公式sinx-siny=2sin[(x-y)/2]cos[(x+y)/2]推导. 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) 己知sinx+siny=1/3,求z=siny—cos^2 x的最大值. 己知sinx+siny=1/3,求z=siny—cos^2 x的最大值. 已知sinx+siny=1/3,求 siny-cos²x的最大值? sinx+siny=2/5,cosx+cosy=6/5,求cos(2x-2y) 已知cosx+cosy=1/2,sinx-siny=1/3,求cos(x+y) 已知sinx+siny=2/3,求2/3+siny-cos²a取值范围 已知sinx+siny=2/3,求2/3+siny-cos²a取值范围 若3sinx*2+2siny*2=1 且3(sinx+cosx)*2+2(siny+cosy)*2=1求cos(2x+2y) sinX+sinY=1 cosX+cosY=0 求cos(2X)+cos(2Y)的值 -sinx=cos((3π/2-x)从-sinx推导 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 已知sinx+siny=1/3,求t=siny-cos^x的最大值和最小值． 已知sinx+siny=1/3,求u=sinx-cos^2y的最值