证：
sin(2x+y)/sinx -2cos(x+y)
=[ sin[(x+y)+x] -2cos(x+y)sinx ]/sinx
=[sin(x+y)cosx+cos(x+y)sinx-2cos(x+y)sinx]/sinx
=[sin(x+y)cosx-cos(x+y)sinx]/sinx
=sin[(x+y)-x]/sinx
=siny/sinx
等式成立.
求证：sin(2x+y)/sinx-2cos(x+y)=siny/sinx
1个回答
相关问题
-
求证sin（2x+y)/sinx-2cos(x+y)=siny/sinx
-
求证：sin（x-y）/（sinx-siny）=cos[（x-y）/2]/cos[(x+y)/2]
-
sinx-siny=2cos（x+y/2）sin（x-y/2）
-
sin(x+y)-sinx=2siny/2·cos(x+y/2)怎么来的
-
证明sinx+cos(x+y)siny/cosx-sin(x+y)siny=tan(x+y)
-
sinx=1/2 siny=1/3 ,sin(x+y)(sinx-y)=?
-
讨论大学函数连续性sin(x+y)-sinx=2siny/2 * cos(x+y/2)
-
证明cosx(cosx-cosy)+sinx(sinx-siny)=2sin(x-y)/2
-
两角和差公式sinx-siny=2sin[(x-y)/2]cos[(x+y)/2]推导.
-
(a)证明cos(x-y)-cos(x+y)=2*sinx*siny (b)由此,证明 2sinθ(sinθ+sin3θ