Question

求证sin²x-sin²y=sin(x+y)sin(x-y)

要过程 谢谢

Answer 1

由积化和差公式：
cos(x+y)=cosxcosy-sinxsiny; cos(x-y)=cosxcosy+sinxsiny;
两式相减得： 2sinxsiny=cos(x-y)-cos(x+y)
所以：
右边=sin(x+y)sin(x-y)=(cos2y-cos2x)/2
=(1-2sin²y-1+2sin²x)/2
=sin²x-sin²y=左边
所以命题得证；

Answer 2

答：
sin²x-sin²y
=(sinx+siny)(sinx-siny)
=2sin[(x+y)/2]cos[(x-y)/2] * 2cos[(x+y)/2]sin[(x-y)/2]
=sin(x+y)*sin(x-y)

所以：sin²x-sin²y=sin(x+y)*sin(x-y)

追问

=2sin[(x+y)/2]cos[(x-y)/2] * 2cos[(x+y)/2]sin[(x-y)/2]
=sin(x+y)*sin(x-y)
这两部为什么直接相等了？

追答

因为两倍角公式：sin2A=2sinAcosA

所以：
=2sin[(x+y)/2]cos[(x-y)/2] * 2cos[(x+y)/2]sin[(x-y)/2]

=2sin[(x+y)/2]cos[(x+y)/2] * 2cos[(x-y)/2]sin[(x-y)/2] 把两个cos的因子调换一下位置

=sin(x+y)*sin(x-y)

Answer 3

sin(x+y)sin(x-y)=(sinxcosy-sinycosx)(sinxcosy-sinycosx)=sinx^2cosy^2-siny^2cosx^2
=sinx^2cosy^2+sinx^2siny^2-sinx^2siny^2-siny^2cosx^2
=sinx^2(cosy^2+siny^2)-siny^2(sinx^2+cosx^2)=sinx^2-siny^2