siny=1/3,sin(x＋y)=1,求sin(2x＋y)
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sin(x＋y)=1 所以可得：cos(x+y)=0
sin(2x+2y)=2sin(x+y)cos(x+y)=0
cos(2x+2y)=cos^2(x+y)-sin^2(x+y)=-1
siny=1/3 所以可得：cosy=±√[1-sin^2y]=±2√2/3
当：cosy=2√2/3 时有：
sin(2x+y)
=sin(2x+2y-y)
=sin(2x+2y)cosy-cos(2x+2y)siny
=-2√2/3
当cosy=-2√2/3 时有：
sin(2x+y)
=sin(2x+2y-y)
=sin(2x+2y)cosy-cos(2x+2y)siny
=2√2/3
sin(2x+2y)=2sin(x+y)cos(x+y)=0
cos(2x+2y)=cos^2(x+y)-sin^2(x+y)=-1
siny=1/3 所以可得：cosy=±√[1-sin^2y]=±2√2/3
当：cosy=2√2/3 时有：
sin(2x+y)
=sin(2x+2y-y)
=sin(2x+2y)cosy-cos(2x+2y)siny
=-2√2/3
当cosy=-2√2/3 时有：
sin(2x+y)
=sin(2x+2y-y)
=sin(2x+2y)cosy-cos(2x+2y)siny
=2√2/3
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