Question

求证|sinx-siny|=|2sin[(x-y)/2]cos[(x+y)/2]|

Answer 1

sinx-siny=sin[(x+y)/2+(x-y)/2]-sin[(x+y)/2-(x-y)/2]
=sin[(x+y)/2]cos[(x-y)/2]+cos[(x+y)/2]sin[(x-y)/2]-{sin[(x+y)/2]cos[(x-y)/2]-cos[(x+y)/2]sin[(x-y)/2]}
=2cos[(x+y)/2]sin[(x-y)/2]
这是和差化积，连绝对值都不用加的，运算量不用楼上这么大的！望采纳。

Answer 2

|2sin[(x-y)/2]cos[(x+y)/2]|
=I2[sin(x/2)cos(y/2)-sin(y/2)cos(x/2)][cos(x/2)cos(y/2)-sin(x/2)(sin(y/2)]
=I2[sin(x/2)cos(x/2)cos²(y/2)-sin²(x/2)sin(y/2)cos(y/2)-cos²(x/2)sin(y/2)cos(y/2)+sin(x/2)cos(x/2)sin²(y/2)]I
=Isinxcos²(y/2)-sin²(x/2)siny-cos²(x/2)siny+sinxsin²(y/2)I
=Isinx[sin²(y/2)+cos²(y/2)]-siny[sin²(x/2)+cos²(x/2)]I
=Isinx-sinyI
得证