作业帮证明sinx+cos(x+y)siny/cosx-sin(x+y)siny=tan(x+y)
来源：学生作业帮助网 编辑：作业帮 时间：2024/10/01 21:56:14
xSJ@Y*c H#M
EfH0PHZ-
(bk xt51F(Y9;wӵAzy nxk>e0C_oksshRf8O(!? [ͮG ӻe7U@,0L!ʽ E*Ubv..f*0 (6 \wHkk _L{0TI1( ~<4^p#n 8+e%xY1U[x]A
证明sinx+cos(x+y)siny/cosx-sin(x+y)siny=tan(x+y)
证明sinx+cos(x+y)siny/cosx-sin(x+y)siny=tan(x+y)
证明sinx+cos(x+y)siny/cosx-sin(x+y)siny=tan(x+y)
sin(x+y)=sinxcosy+cosxsiny 具体推导：首先建立直角坐标系,在直角坐标系cos(x+y) = cosx cosy - sinx siny sin(x+y) = sinx cosy + cosx
∵sinx=sin[﹙x＋y﹚-y]
=sin﹙x＋y﹚cosy-cos﹙x﹢y﹚siny
cosx=[﹙x＋y﹚-y]
=cos﹙x＋y﹚cosy＋sin﹙x﹢y﹚siny
∴原式＝[sin﹙x＋y﹚cosy-cos﹙x﹢y﹚siny+cos﹙x﹢y﹚siny]／[cos﹙x＋y﹚cosy＋sin﹙x﹢y﹚siny-sin﹙x﹢y﹚siny]
=[sin﹙...
全部展开
∵sinx=sin[﹙x＋y﹚-y]
=sin﹙x＋y﹚cosy-cos﹙x﹢y﹚siny
cosx=[﹙x＋y﹚-y]
=cos﹙x＋y﹚cosy＋sin﹙x﹢y﹚siny
∴原式＝[sin﹙x＋y﹚cosy-cos﹙x﹢y﹚siny+cos﹙x﹢y﹚siny]／[cos﹙x＋y﹚cosy＋sin﹙x﹢y﹚siny-sin﹙x﹢y﹚siny]
=[sin﹙x＋y﹚cosy]／[cos﹙x＋y﹚cosy]
=sin﹙x＋y﹚／cos﹙x＋y﹚=tan﹙x+y﹚
收起
证明sinx+cos(x+y)siny/cosx-sin(x+y)siny=tan(x+y)
证明 [sin(2x+y)/sinx]-2cos(x+y)=siny/sinx
|sinx-siny|≤|x-y|如何证明
证明|x-y|≥|sinx-siny|
证明不等式|sinx-siny|《 |x-y|
请问,如何证明sinx+siny=2*sin(x+y/2)*cos(x-y/2)
证明sin(x+y)sin(x-y)=sinx-siny
证明：对x≠y,恒有|sinx-siny|
证明.（sinx+siny)/cosx-cosy=ctg(y-x)/2
导数的应用证明不等式|(sinx-siny)/(x-y)-cosy|
求证sin（2x+y)/sinx-2cos(x+y)=siny/sinx
求证：sin(2x+y)/sinx-2cos(x+y)=siny/sinx
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