Question

已知cosx+cosy=4/5,sinx+siny=-3/5,求cos2x+cos2y和sin(x+y)

Answer 1

解答如下：
根据已知条件得：
(cosx+cosy)(sinx+siny)
=sinx cos x+siny cosy+sinx cosy+cosx siny
=1/2(sin2x+sin2y)+sin(x+y)
=12/25
(cosx+cosy)^2-(sinx+siny)^2=cos2x+cos2y+2cos(x+y)=7/25
(cosx+cosy)-(sinx+siny)=1/5
两边完全平方得：
(cosx^2-sinx^2+cosy^2-siny^2)+2cos(x+y)-(sin2x+sin2y+2sin(x+y))
=cos2x+cos2y+2cos(x+y)-(sin2x+sin2y)-2sin(x+y)
(cosx+cosy)-(sinx+siny)=(cosx-sinx)+(cosy-siny)=1/5
1-sin2x+1-sin2y+2(cosxcosy-cosxsiny-sinxcosy+sinxsiny)