若cosx cosy+sinx siny=1/2，sin2x+sin2y=2/3，求sin（x+y）值
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sin(x+y) = sinxcosy+cosxsiny
Sin2x=2Sinx•Cosx
Sin2y=2Siny•Cosy
2Sinx•Cosx+2Siny•Cosy=2/3
cosx cosy+sinx siny=1/2
sin2x+sin2y=sin[(x+y)+(x-y)]+sin[(x+y)-(x-y)]=sin(x+y)cos(x-y)+cos(x+y)sin(x-y)+sin(x+y)cos(x-y)-cos(x+y)sin(x-y)=2sin(x+y)cos(x-y)=2/3
∴sin（x+y）cos（x-y）=1/3 由cosxcosy+sinxsiny=1/2 cos(x-y)=1/2
∴sin(x+y)=2/3
Sin2x=2Sinx•Cosx
Sin2y=2Siny•Cosy
2Sinx•Cosx+2Siny•Cosy=2/3
cosx cosy+sinx siny=1/2
sin2x+sin2y=sin[(x+y)+(x-y)]+sin[(x+y)-(x-y)]=sin(x+y)cos(x-y)+cos(x+y)sin(x-y)+sin(x+y)cos(x-y)-cos(x+y)sin(x-y)=2sin(x+y)cos(x-y)=2/3
∴sin（x+y）cos（x-y）=1/3 由cosxcosy+sinxsiny=1/2 cos(x-y)=1/2
∴sin(x+y)=2/3
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