举报
最爱凌晨三点
我想了想,可以这样:
cosx=cos(x+y-y)=cos(x+y)cosy+sin(x+y)siny
=(cosxcosy-sinxsiny)cosy+sin(x+y)siny
=cosx(1-sinysiny)-sinxsinycosy+sin(x+y)siny
即
0=-cosxsinysiny-sinxsinycosy+sin(x+y)siny
由y的任意性,上式除以siny
即
0=-cosxsiny-sinxcosy+sin(x+y) 即sin(x+y)=sinxcosy+cosxsiny
siny=0即y=kPI,带入上式即 sin(x+kPi)=sinxcos(kPi) 容易验证k为奇数和偶数都成立
即sin(x+y)=sinxcosy+cosxsiny对任意x和y成立
取y=-t,带入即证得另一式。