Cho

- Hướng dẫn giải

* Ta có ${\sin }^2\alpha +{\cos }^2\alpha =1\Rightarrow {\sin }^2\alpha +\frac{16}{25}=1\Rightarrow {\sin }^2\alpha =\frac{9}{25}$

Mà $\sin \alpha <0\Rightarrow \sin \alpha =-\frac{3}{5}$

* Vì ${\sin }^2\beta +{\cos }^2\beta =1\Rightarrow \frac{9}{16}+{\cos }^2\beta =1\Rightarrow co{s}^2\beta =\frac{7}{16}$

 $\cos \beta >0\Rightarrow \cos \beta =\frac{\sqrt{7}}{4}$

* $\sin \left(\alpha +\beta \right)=\sin \alpha .\cos \beta +cos\alpha .\sin \beta =\frac{-3}{5}.\frac{\sqrt{7}}{4}+\frac{4}{5}.\frac{3}{4}=\frac{12-3\sqrt{7}}{20}$