问题: 三角恒等式
已知cosx-cosy=1/2，sinx-siny=1/3，则sin(x+y)=_____。
解答:
解：1/3=sin[(x+y)-y]-sin[(x+y)-x]
```````=sin(x+y)cosy-cos(x+y)siny-sin(x+y)cosx+cos(x+y)sinx
```````=sin(x+y)(cosy-cosx)+cos(x+y)(sinx-siny)
```````=-sin(x+y)/2+cos(x+y)/3
1/3=-sin(x+y)/2+cos(x+y)/3……(1)
同理1/2=cos[(x+y)-y]-cos[(x+y)-x]
```````=cos(x+y)cosy+sin(x+y)siny-cos(x+y)cosx-sin(x+y)sinx
```````=sin(x+y)(siny-sinx)+cos(x+y)(cosy-cosx)
```````=-sin(x+y)/3-cos(x+y)/2
1/2=-sin(x+y)/3-cos(x+y)/2……(2)
联立(1)、(2)解得sin(x+y)=-12/13.
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