三角变换中的积化和差应用

因为$\frac{\sin\alpha +\sin\beta }{\cos\alpha +\cos\beta }=\frac{\sin\left[ \left( \frac{\alpha +\beta }{2} \right)\text{+}\left( \frac{\alpha -\beta }{2} \right) \right]+\sin\left[ \left( \frac{\alpha +\beta }{2} \right)-\left( \frac{\alpha -\beta }{2} \right) \right]}{\cos\left[ \left( \frac{\alpha +\beta }{2} \right)\text{+}\left( \frac{\alpha -\beta }{2} \right) \right]+\cos\left[ \left( \frac{\alpha +\beta }{2} \right)-\left( \frac{\alpha -\beta }{2} \right) \right]}=\frac{2\sin\frac{\alpha +\beta }{2}\cos\frac{\alpha -\beta }{2}}{2\cos\frac{\alpha +\beta }{2}\cos\frac{\alpha -\beta }{2}}=\frac{1}{3}$，

即$\tan\frac{\alpha +\beta }{2}=\frac{1}{3}$，则$\sin\left( \alpha +\beta  \right)=\frac{2\sin\frac{\alpha +\beta }{2}\cos\frac{\alpha +\beta }{2}}{\text{si}{{\text{n}}^{2}}\frac{\alpha +\beta }{2}\text{+co}{{\text{s}}^{2}}\frac{\alpha +\beta }{2}}=\frac{2\tan\frac{\alpha +\beta }{2}}{1+\text{ta}{{\text{n}}^{2}}\frac{\alpha +\beta }{2}}=\frac{\frac{2}{3}}{1+\frac{1}{9}}=\frac{3}{5}$.